I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3 digits from the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or spaces that are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3 digits from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or spaces that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

Biff,
can you tell me what the " does in the solution?

your solutions worked perfectly! thank you!
very much appreciated :)

jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3 digits from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or spaces that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

and sorry I didn't combine my questions but for the 2nd solution also.... if
'from the right is need, why go with LEFT? and how does the -2 work into it?
I just want to be able to understand it so I can explain it to some one else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3 digits from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or spaces that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

This formula is simply replacing the 3 characters from the right with the 2
characters from the right. If:

A1 = 66014

Replace 3 characters from the right = 014
With 2 characters from the right = 14

Result = 6614

The result of the Substitute function is always a TEXT value. The +0 coerces
the result to be NUMERIC.

The second formula:

=LEFT(A1,FIND("(",A1)-2)

I'm assuming that the portion of the string that you want to extract is
variable in length and that this portion is always present: (S)

TRAD EC JEANS (S) | CL11941033

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of that.
In this example the "(" is the 15th character. So we subtract 2, one for the
"(" and one for the space before the "(". That results in 13. So the formula
is returning the first 13 characters starting from the left.

Biff

"Jane" wrote in message
...
and sorry I didn't combine my questions but for the 2nd solution also....
if
'from the right is need, why go with LEFT? and how does the -2 work into
it?
I just want to be able to understand it so I can explain it to some one
else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3 digits from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or spaces that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

Ooops! Correction needed:

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of that.

Should be:

In that case all we need to do is find the location
of the opening parenthesis "(" in the string and extract
everything to the LEFT of that.

Biff

"T. Valko" wrote in message
...
=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

This formula is simply replacing the 3 characters from the right with the
2 characters from the right. If:

A1 = 66014

Replace 3 characters from the right = 014
With 2 characters from the right = 14

Result = 6614

The result of the Substitute function is always a TEXT value. The +0
coerces the result to be NUMERIC.

The second formula:

=LEFT(A1,FIND("(",A1)-2)

I'm assuming that the portion of the string that you want to extract is
variable in length and that this portion is always present: (S)

TRAD EC JEANS (S) | CL11941033

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of that.
In this example the "(" is the 15th character. So we subtract 2, one for
the "(" and one for the space before the "(". That results in 13. So the
formula is returning the first 13 characters starting from the left.

Biff

"Jane" wrote in message
...
and sorry I didn't combine my questions but for the 2nd solution also....
if
'from the right is need, why go with LEFT? and how does the -2 work into
it?
I just want to be able to understand it so I can explain it to some one
else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3 digits
from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or spaces
that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

that was so clear - thank you so much for the explanation!
take care, Jane

"T. Valko" wrote:

Ooops! Correction needed:

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of that.

Should be:

In that case all we need to do is find the location
of the opening parenthesis "(" in the string and extract
everything to the LEFT of that.

Biff

"T. Valko" wrote in message
...
=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

This formula is simply replacing the 3 characters from the right with the
2 characters from the right. If:

A1 = 66014

Replace 3 characters from the right = 014
With 2 characters from the right = 14

Result = 6614

The result of the Substitute function is always a TEXT value. The +0
coerces the result to be NUMERIC.

The second formula:

=LEFT(A1,FIND("(",A1)-2)

I'm assuming that the portion of the string that you want to extract is
variable in length and that this portion is always present: (S)

TRAD EC JEANS (S) | CL11941033

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of that.
In this example the "(" is the 15th character. So we subtract 2, one for
the "(" and one for the space before the "(". That results in 13. So the
formula is returning the first 13 characters starting from the left.

Biff

"Jane" wrote in message
...
and sorry I didn't combine my questions but for the 2nd solution also....
if
'from the right is need, why go with LEFT? and how does the -2 work into
it?
I just want to be able to understand it so I can explain it to some one
else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3 digits
from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or spaces
that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

You're welcome. Thanks for the feedback!

Biff

"Jane" wrote in message
...
that was so clear - thank you so much for the explanation!
take care, Jane

"T. Valko" wrote:

Ooops! Correction needed:

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of
that.

Should be:

In that case all we need to do is find the location
of the opening parenthesis "(" in the string and extract
everything to the LEFT of that.

Biff

"T. Valko" wrote in message
...
=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

This formula is simply replacing the 3 characters from the right with
the
2 characters from the right. If:

A1 = 66014

Replace 3 characters from the right = 014
With 2 characters from the right = 14

Result = 6614

The result of the Substitute function is always a TEXT value. The +0
coerces the result to be NUMERIC.

The second formula:

=LEFT(A1,FIND("(",A1)-2)

I'm assuming that the portion of the string that you want to extract is
variable in length and that this portion is always present: (S)

TRAD EC JEANS (S) | CL11941033

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of
that.
In this example the "(" is the 15th character. So we subtract 2, one
for
the "(" and one for the space before the "(". That results in 13. So
the
formula is returning the first 13 characters starting from the left.

Biff

"Jane" wrote in message
...
and sorry I didn't combine my questions but for the 2nd solution
also....
if
'from the right is need, why go with LEFT? and how does the -2 work
into
it?
I just want to be able to understand it so I can explain it to some
one
else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3 digits
from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or spaces
that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

Hi Biff,
I shared your solutions with a co-worker who also had a similar situation.
She was going to approach the problem with =RIGHT but include LEN.. she
wanted me to ask you if that might also work?
thank you, Jane

"T. Valko" wrote:

You're welcome. Thanks for the feedback!

Biff

"Jane" wrote in message
...
that was so clear - thank you so much for the explanation!
take care, Jane

"T. Valko" wrote:

Ooops! Correction needed:

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of
that.

Should be:

In that case all we need to do is find the location
of the opening parenthesis "(" in the string and extract
everything to the LEFT of that.

Biff

"T. Valko" wrote in message
...
=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

This formula is simply replacing the 3 characters from the right with
the
2 characters from the right. If:

A1 = 66014

Replace 3 characters from the right = 014
With 2 characters from the right = 14

Result = 6614

The result of the Substitute function is always a TEXT value. The +0
coerces the result to be NUMERIC.

The second formula:

=LEFT(A1,FIND("(",A1)-2)

I'm assuming that the portion of the string that you want to extract is
variable in length and that this portion is always present: (S)

TRAD EC JEANS (S) | CL11941033

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of
that.
In this example the "(" is the 15th character. So we subtract 2, one
for
the "(" and one for the space before the "(". That results in 13. So
the
formula is returning the first 13 characters starting from the left.

Biff

"Jane" wrote in message
...
and sorry I didn't combine my questions but for the 2nd solution
also....
if
'from the right is need, why go with LEFT? and how does the -2 work
into
it?
I just want to be able to understand it so I can explain it to some
one
else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3 digits
from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or spaces
that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

She was going to approach the problem with =RIGHT
but include LEN.. she wanted me to ask you if that
might also work?

To do what exactly?

Biff

"Jane" wrote in message
...
Hi Biff,
I shared your solutions with a co-worker who also had a similar situation.
She was going to approach the problem with =RIGHT but include LEN.. she
wanted me to ask you if that might also work?
thank you, Jane

"T. Valko" wrote:

You're welcome. Thanks for the feedback!

Biff

"Jane" wrote in message
...
that was so clear - thank you so much for the explanation!
take care, Jane

"T. Valko" wrote:

Ooops! Correction needed:

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of
that.

Should be:

In that case all we need to do is find the location
of the opening parenthesis "(" in the string and extract
everything to the LEFT of that.

Biff

"T. Valko" wrote in message
...
=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

This formula is simply replacing the 3 characters from the right
with
the
2 characters from the right. If:

A1 = 66014

Replace 3 characters from the right = 014
With 2 characters from the right = 14

Result = 6614

The result of the Substitute function is always a TEXT value. The +0
coerces the result to be NUMERIC.

The second formula:

=LEFT(A1,FIND("(",A1)-2)

I'm assuming that the portion of the string that you want to extract
is
variable in length and that this portion is always present: (S)

TRAD EC JEANS (S) | CL11941033

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of
that.
In this example the "(" is the 15th character. So we subtract 2, one
for
the "(" and one for the space before the "(". That results in 13. So
the
formula is returning the first 13 characters starting from the left.

Biff

"Jane" wrote in message
...
and sorry I didn't combine my questions but for the 2nd solution
also....
if
'from the right is need, why go with LEFT? and how does the -2 work
into
it?
I just want to be able to understand it so I can explain it to some
one
else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3
digits
from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or
spaces
that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

Hi there,
So sorry I was clearer... for problem #w, she had asked if there was a way
to use =RIGHT and the LEN function.

Since that morning note, she actually tried:
=LEFT(C5,LEN(C5)-16) and it produced the same results. She wanted to know if
this is a flexible enough solution to use in teh same way we used your
solution below.
take care, Jane

"T. Valko" wrote:

She was going to approach the problem with =RIGHT
but include LEN.. she wanted me to ask you if that
might also work?

To do what exactly?

Biff

"Jane" wrote in message
...
Hi Biff,
I shared your solutions with a co-worker who also had a similar situation.
She was going to approach the problem with =RIGHT but include LEN.. she
wanted me to ask you if that might also work?
thank you, Jane

"T. Valko" wrote:

You're welcome. Thanks for the feedback!

Biff

"Jane" wrote in message
...
that was so clear - thank you so much for the explanation!
take care, Jane

"T. Valko" wrote:

Ooops! Correction needed:

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of
that.

Should be:

In that case all we need to do is find the location
of the opening parenthesis "(" in the string and extract
everything to the LEFT of that.

Biff

"T. Valko" wrote in message
...
=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

This formula is simply replacing the 3 characters from the right
with
the
2 characters from the right. If:

A1 = 66014

Replace 3 characters from the right = 014
With 2 characters from the right = 14

Result = 6614

The result of the Substitute function is always a TEXT value. The +0
coerces the result to be NUMERIC.

The second formula:

=LEFT(A1,FIND("(",A1)-2)

I'm assuming that the portion of the string that you want to extract
is
variable in length and that this portion is always present: (S)

TRAD EC JEANS (S) | CL11941033

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right of
that.
In this example the "(" is the 15th character. So we subtract 2, one
for
the "(" and one for the space before the "(". That results in 13. So
the
formula is returning the first 13 characters starting from the left.

Biff

"Jane" wrote in message
...
and sorry I didn't combine my questions but for the 2nd solution
also....
if
'from the right is need, why go with LEFT? and how does the -2 work
into
it?
I just want to be able to understand it so I can explain it to some
one
else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3
digits
from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or
spaces
that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

=LEFT(C5,LEN(C5)-16) and it produced the same results

As long as *every* entry is the same format that will work just fine. For
example, that method would not work on these:

TRAD EC JEANS (S) | CL119410
TRAD EC JEANS (S) CL1941033
TRAD EC JEANS (S) | CL11

That's when using: =LEFT(A1,FIND("(",A1)-2), is the best solution. Also,
since you only posted a single example I offered what I believe to be the
most versatile solution based on the limited information available.

Biff

"Jane" wrote in message
...
Hi there,
So sorry I was clearer... for problem #w, she had asked if there was a way
to use =RIGHT and the LEN function.

Since that morning note, she actually tried:
=LEFT(C5,LEN(C5)-16) and it produced the same results. She wanted to know
if
this is a flexible enough solution to use in teh same way we used your
solution below.
take care, Jane

"T. Valko" wrote:

She was going to approach the problem with =RIGHT
but include LEN.. she wanted me to ask you if that
might also work?

To do what exactly?

Biff

"Jane" wrote in message
...
Hi Biff,
I shared your solutions with a co-worker who also had a similar
situation.
She was going to approach the problem with =RIGHT but include LEN.. she
wanted me to ask you if that might also work?
thank you, Jane

"T. Valko" wrote:

You're welcome. Thanks for the feedback!

Biff

"Jane" wrote in message
...
that was so clear - thank you so much for the explanation!
take care, Jane

"T. Valko" wrote:

Ooops! Correction needed:

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right
of
that.

Should be:

In that case all we need to do is find the location
of the opening parenthesis "(" in the string and extract
everything to the LEFT of that.

Biff

"T. Valko" wrote in message
...
=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

This formula is simply replacing the 3 characters from the right
with
the
2 characters from the right. If:

A1 = 66014

Replace 3 characters from the right = 014
With 2 characters from the right = 14

Result = 6614

The result of the Substitute function is always a TEXT value. The
+0
coerces the result to be NUMERIC.

The second formula:

=LEFT(A1,FIND("(",A1)-2)

I'm assuming that the portion of the string that you want to
extract
is
variable in length and that this portion is always present: (S)

TRAD EC JEANS (S) | CL11941033

In that case all we need to do is find the location of the
opening
parenthesis "(" in the string and extract everything to the right
of
that.
In this example the "(" is the 15th character. So we subtract 2,
one
for
the "(" and one for the space before the "(". That results in 13.
So
the
formula is returning the first 13 characters starting from the
left.

Biff

"Jane" wrote in message
...
and sorry I didn't combine my questions but for the 2nd solution
also....
if
'from the right is need, why go with LEFT? and how does the -2
work
into
it?
I just want to be able to understand it so I can explain it to
some
one
else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3
digits
from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or
spaces
that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

ah gotcha Biff. makes total sense.... I agree that yours is more versatile.
Since there are often so many moving parts that, I agree, it is the best
solution - will pass on the info.

Again, very much appreciate your help and time!!
take good care, Jane

"T. Valko" wrote:

=LEFT(C5,LEN(C5)-16) and it produced the same results

As long as *every* entry is the same format that will work just fine. For
example, that method would not work on these:

TRAD EC JEANS (S) | CL119410
TRAD EC JEANS (S) CL1941033
TRAD EC JEANS (S) | CL11

That's when using: =LEFT(A1,FIND("(",A1)-2), is the best solution. Also,
since you only posted a single example I offered what I believe to be the
most versatile solution based on the limited information available.

Biff

"Jane" wrote in message
...
Hi there,
So sorry I was clearer... for problem #w, she had asked if there was a way
to use =RIGHT and the LEN function.

Since that morning note, she actually tried:
=LEFT(C5,LEN(C5)-16) and it produced the same results. She wanted to know
if
this is a flexible enough solution to use in teh same way we used your
solution below.
take care, Jane

"T. Valko" wrote:

She was going to approach the problem with =RIGHT
but include LEN.. she wanted me to ask you if that
might also work?

To do what exactly?

Biff

"Jane" wrote in message
...
Hi Biff,
I shared your solutions with a co-worker who also had a similar
situation.
She was going to approach the problem with =RIGHT but include LEN.. she
wanted me to ask you if that might also work?
thank you, Jane

"T. Valko" wrote:

You're welcome. Thanks for the feedback!

Biff

"Jane" wrote in message
...
that was so clear - thank you so much for the explanation!
take care, Jane

"T. Valko" wrote:

Ooops! Correction needed:

In that case all we need to do is find the location of the opening
parenthesis "(" in the string and extract everything to the right
of
that.

Should be:

In that case all we need to do is find the location
of the opening parenthesis "(" in the string and extract
everything to the LEFT of that.

Biff

"T. Valko" wrote in message
...
=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

This formula is simply replacing the 3 characters from the right
with
the
2 characters from the right. If:

A1 = 66014

Replace 3 characters from the right = 014
With 2 characters from the right = 14

Result = 6614

The result of the Substitute function is always a TEXT value. The
+0
coerces the result to be NUMERIC.

The second formula:

=LEFT(A1,FIND("(",A1)-2)

I'm assuming that the portion of the string that you want to
extract
is
variable in length and that this portion is always present: (S)

TRAD EC JEANS (S) | CL11941033

In that case all we need to do is find the location of the
opening
parenthesis "(" in the string and extract everything to the right
of
that.
In this example the "(" is the 15th character. So we subtract 2,
one
for
the "(" and one for the space before the "(". That results in 13.
So
the
formula is returning the first 13 characters starting from the
left.

Biff

"Jane" wrote in message
...
and sorry I didn't combine my questions but for the 2nd solution
also....
if
'from the right is need, why go with LEFT? and how does the -2
work
into
it?
I just want to be able to understand it so I can explain it to
some
one
else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed 3
digits
from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/ or
spaces
that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)

You're very welcome!

Biff

"Jane" wrote in message
...
ah gotcha Biff. makes total sense.... I agree that yours is more
versatile.
Since there are often so many moving parts that, I agree, it is the best
solution - will pass on the info.

Again, very much appreciate your help and time!!
take good care, Jane

"T. Valko" wrote:

=LEFT(C5,LEN(C5)-16) and it produced the same results

As long as *every* entry is the same format that will work just fine. For
example, that method would not work on these:

TRAD EC JEANS (S) | CL119410
TRAD EC JEANS (S) CL1941033
TRAD EC JEANS (S) | CL11

That's when using: =LEFT(A1,FIND("(",A1)-2), is the best solution. Also,
since you only posted a single example I offered what I believe to be the
most versatile solution based on the limited information available.

Biff

"Jane" wrote in message
...
Hi there,
So sorry I was clearer... for problem #w, she had asked if there was a
way
to use =RIGHT and the LEN function.

Since that morning note, she actually tried:
=LEFT(C5,LEN(C5)-16) and it produced the same results. She wanted to
know
if
this is a flexible enough solution to use in teh same way we used your
solution below.
take care, Jane

"T. Valko" wrote:

She was going to approach the problem with =RIGHT
but include LEN.. she wanted me to ask you if that
might also work?

To do what exactly?

Biff

"Jane" wrote in message
...
Hi Biff,
I shared your solutions with a co-worker who also had a similar
situation.
She was going to approach the problem with =RIGHT but include LEN..
she
wanted me to ask you if that might also work?
thank you, Jane

"T. Valko" wrote:

You're welcome. Thanks for the feedback!

Biff

"Jane" wrote in message
...
that was so clear - thank you so much for the explanation!
take care, Jane

"T. Valko" wrote:

Ooops! Correction needed:

In that case all we need to do is find the location of the
opening
parenthesis "(" in the string and extract everything to the
right
of
that.

Should be:

In that case all we need to do is find the location
of the opening parenthesis "(" in the string and extract
everything to the LEFT of that.

Biff

"T. Valko" wrote in message
...
=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

This formula is simply replacing the 3 characters from the
right
with
the
2 characters from the right. If:

A1 = 66014

Replace 3 characters from the right = 014
With 2 characters from the right = 14

Result = 6614

The result of the Substitute function is always a TEXT value.
The
+0
coerces the result to be NUMERIC.

The second formula:

=LEFT(A1,FIND("(",A1)-2)

I'm assuming that the portion of the string that you want to
extract
is
variable in length and that this portion is always present:
(S)

TRAD EC JEANS (S) | CL11941033

In that case all we need to do is find the location of the
opening
parenthesis "(" in the string and extract everything to the
right
of
that.
In this example the "(" is the 15th character. So we subtract
2,
one
for
the "(" and one for the space before the "(". That results in
13.
So
the
formula is returning the first 13 characters starting from the
left.

Biff

"Jane" wrote in message
...
and sorry I didn't combine my questions but for the 2nd
solution
also....
if
'from the right is need, why go with LEFT? and how does
the -2
work
into
it?
I just want to be able to understand it so I can explain it
to
some
one
else
if asked and alter if needed for future use.

Thank you again! Jane

"T. Valko" wrote:

Try these:

1st problem:

=SUBSTITUTE(A1,RIGHT(A1,3),RIGHT(A1,2))+0

2nd problem:

=LEFT(A1,FIND("(",A1)-2)

Biff

"Jane" wrote in message
...
I believe these problems have the same solution.

1st problem:
I would like to be able to remove the zero that is placed
3
digits
from
the
RIGHT. The values a
66014
8004
8015

The values should look like:
6614
804
815

2nd problem in the same mproject:
I would like to be able to remove the values, text, and/
or
spaces
that
are
16 places from the RIGHT.
The entry is:
TRAD EC JEANS (S) | CL11941033

The entry should look like:
TRAD EC JEANS

thank you for your help! :)