ExcelBanter

ExcelBanter (https://www.excelbanter.com/)
-   Excel Discussion (Misc queries) (https://www.excelbanter.com/excel-discussion-misc-queries/)
-   -   Text fomula (https://www.excelbanter.com/excel-discussion-misc-queries/12949-text-fomula.html)

gregork

Text fomula
 
In cell B1 I could have any one of the following text examples: big apple,
small apple, orange, apple, pear etc. In A1 I want a formula that will
return "apple" if B1 contains any of the text apple. ie if it read "big
apple" in B1 I want it to return "apple" in A1.

Thanks
GK



Dave O

Check out the FIND() and SEARCH() functions. Then you could write an
IF() in A1 that says 'if the result of the FIND is greater than zero,
write "apple" in A1'.


JE McGimpsey

One way:

=IF(COUNTIF(B1,"*apple*"),"apple","")

In article ,
"gregork" wrote:

In cell B1 I could have any one of the following text examples: big apple,
small apple, orange, apple, pear etc. In A1 I want a formula that will
return "apple" if B1 contains any of the text apple. ie if it read "big
apple" in B1 I want it to return "apple" in A1.

Thanks
GK


Tom Ogilvy

=if(counif(B1,"*apple*)0,"apple","")

--
Regards,
Tom Ogilvy

"gregork" wrote in message
...
In cell B1 I could have any one of the following text examples: big

apple,
small apple, orange, apple, pear etc. In A1 I want a formula that will
return "apple" if B1 contains any of the text apple. ie if it read "big
apple" in B1 I want it to return "apple" in A1.

Thanks
GK





Tom Ogilvy

Left of a double quote:
=if(counif(B1,"*apple*")0,"apple","")

--
Regards,
Tom Ogilvy



gregork

Thanks Guys.....does exactly what I wanted.

Cheers
GK
"Tom Ogilvy" wrote in message
...
Left of a double quote:
=if(counif(B1,"*apple*")0,"apple","")

--
Regards,
Tom Ogilvy





Jack Sons

Tom,

Suppose in A1:F6 there can be all kinds of things (text, numbers, blanks,
formulae), but I want to check if "apple", "pear", "cherry"; "coconut" and
"apricot" each occur at least once. If so a formula in H1 should return
TRUE, if not FALSE.
I tried
=COUNT(MATCH({"apple", "pear", "cherry"; "coconut","apricot"},A1:F6,0))=5
but it won't work, also when array entered.
Your help will be appreciated.

Jack Sons
The Netherlands



"Tom Ogilvy" schreef in bericht
...
Left of a double quote:
=if(counif(B1,"*apple*")0,"apple","")

--
Regards,
Tom Ogilvy





R.VENKATARAMAN

try type in H1

=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y"))=1,TRUE,FALSE)

your data is in d5 to d16.

do you get what you want.

may be tortuous to write .may be bugs in it. try the experiment

====================================
Jack Sons wrote in message
...
Tom,

Suppose in A1:F6 there can be all kinds of things (text, numbers, blanks,
formulae), but I want to check if "apple", "pear", "cherry"; "coconut" and
"apricot" each occur at least once. If so a formula in H1 should return
TRUE, if not FALSE.
I tried
=COUNT(MATCH({"apple", "pear", "cherry"; "coconut","apricot"},A1:F6,0))=5
but it won't work, also when array entered.
Your help will be appreciated.

Jack Sons
The Netherlands



"Tom Ogilvy" schreef in bericht
...
Left of a double quote:
=if(counif(B1,"*apple*")0,"apple","")

--
Regards,
Tom Ogilvy







R.VENKATARAMAN

sorry. this is not working on subsequent experiments
apologise.

R.VENKATARAMAN $$$ wrote in message news:...
try type in H1


=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y"))=1,TRUE,FALSE)

your data is in d5 to d16.

do you get what you want.

may be tortuous to write .may be bugs in it. try the experiment

====================================
Jack Sons wrote in message
...
Tom,

Suppose in A1:F6 there can be all kinds of things (text, numbers,

blanks,
formulae), but I want to check if "apple", "pear", "cherry"; "coconut"

and
"apricot" each occur at least once. If so a formula in H1 should return
TRUE, if not FALSE.
I tried
=COUNT(MATCH({"apple", "pear", "cherry";

"coconut","apricot"},A1:F6,0))=5
but it won't work, also when array entered.
Your help will be appreciated.

Jack Sons
The Netherlands



"Tom Ogilvy" schreef in bericht
...
Left of a double quote:
=if(counif(B1,"*apple*")0,"apple","")

--
Regards,
Tom Ogilvy









R.VENKATARAMAN

strangely it works with

=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y")),TRUE,FALSE)

i.e without =1

But I could not explain the logic
=============================


R.VENKATARAMAN $$$ wrote in message
...
sorry. this is not working on subsequent experiments
apologise.

R.VENKATARAMAN $$$ wrote in message news:...
try type in H1



=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y"))=1,TRUE,FALSE)

your data is in d5 to d16.

do you get what you want.

may be tortuous to write .may be bugs in it. try the experiment

====================================
Jack Sons wrote in message
...
Tom,

Suppose in A1:F6 there can be all kinds of things (text, numbers,

blanks,
formulae), but I want to check if "apple", "pear", "cherry"; "coconut"

and
"apricot" each occur at least once. If so a formula in H1 should

return
TRUE, if not FALSE.
I tried
=COUNT(MATCH({"apple", "pear", "cherry";

"coconut","apricot"},A1:F6,0))=5
but it won't work, also when array entered.
Your help will be appreciated.

Jack Sons
The Netherlands



"Tom Ogilvy" schreef in bericht
...
Left of a double quote:
=if(counif(B1,"*apple*")0,"apple","")

--
Regards,
Tom Ogilvy















Jack Sons

R,

Thank you very much, it indeed works, also for D5:E16 (what I was looking
for, more that one dimension).

What also works is

=COUNT(MATCH({"apple", "pear", "cherry"; "coconut","apricot"},D5:D16,0))=5
but
=COUNT(MATCH({"apple", "pear", "cherry"; "coconut","apricot"},D5:E16,0))=5
does not work.

I got stuck with this problem, I really want it solved, also because the
list or array way (how should I call it?), with {......} is so elegant and
easier to use.

Can you or anybody out there in this august newsgroup help (and explain)?

Jack.

"R.VENKATARAMAN" $$$ schreef in bericht
...
strangely it works with

=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y")),TRUE,FALSE)

i.e without =1

But I could not explain the logic
=============================


R.VENKATARAMAN $$$ wrote in message
...
sorry. this is not working on subsequent experiments
apologise.

R.VENKATARAMAN $$$ wrote in message news:...
try type in H1



=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y"))=1,TRUE,FALSE)

your data is in d5 to d16.

do you get what you want.

may be tortuous to write .may be bugs in it. try the experiment

====================================
Jack Sons wrote in message
...
Tom,

Suppose in A1:F6 there can be all kinds of things (text, numbers,

blanks,
formulae), but I want to check if "apple", "pear", "cherry";
"coconut"

and
"apricot" each occur at least once. If so a formula in H1 should

return
TRUE, if not FALSE.
I tried
=COUNT(MATCH({"apple", "pear", "cherry";

"coconut","apricot"},A1:F6,0))=5
but it won't work, also when array entered.
Your help will be appreciated.

Jack Sons
The Netherlands



"Tom Ogilvy" schreef in bericht
...
Left of a double quote:
=if(counif(B1,"*apple*")0,"apple","")

--
Regards,
Tom Ogilvy

















RagDyeR

Why even use the IF()?

=OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pears" ),COUNTIF(D5:D16,"cherry")
)

OR, even simpler:

=OR(COUNTIF(D5:D16,{"apple","pears","cherry"}))
--

Regards,

RD
--------------------------------------------------------------------
Please keep all correspondence within the Group, so all may benefit !
--------------------------------------------------------------------



"R.VENKATARAMAN" $$$ wrote in message
...
strangely it works with

=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y")),TRUE,FALSE)

i.e without =1

But I could not explain the logic
=============================


R.VENKATARAMAN $$$ wrote in message
...
sorry. this is not working on subsequent experiments
apologise.

R.VENKATARAMAN $$$ wrote in message news:...
try type in H1



=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y"))=1,TRUE,FALSE)

your data is in d5 to d16.

do you get what you want.

may be tortuous to write .may be bugs in it. try the experiment

====================================
Jack Sons wrote in message
...
Tom,

Suppose in A1:F6 there can be all kinds of things (text, numbers,

blanks,
formulae), but I want to check if "apple", "pear", "cherry"; "coconut"

and
"apricot" each occur at least once. If so a formula in H1 should

return
TRUE, if not FALSE.
I tried
=COUNT(MATCH({"apple", "pear", "cherry";

"coconut","apricot"},A1:F6,0))=5
but it won't work, also when array entered.
Your help will be appreciated.

Jack Sons
The Netherlands



"Tom Ogilvy" schreef in bericht
...
Left of a double quote:
=if(counif(B1,"*apple*")0,"apple","")

--
Regards,
Tom Ogilvy
















RagDyeR

I replied to R without reading the entire thread.

For your question, replace the "OR" and try this:

=AND(COUNTIF(D5:D16,{"apple","pears","cherry"}))

--

HTH,

RD
==============================================
Please keep all correspondence within the Group, so all may benefit!
==============================================


"Jack Sons" wrote in message
...
R,

Thank you very much, it indeed works, also for D5:E16 (what I was looking
for, more that one dimension).

What also works is

=COUNT(MATCH({"apple", "pear", "cherry"; "coconut","apricot"},D5:D16,0))=5
but
=COUNT(MATCH({"apple", "pear", "cherry"; "coconut","apricot"},D5:E16,0))=5
does not work.

I got stuck with this problem, I really want it solved, also because the
list or array way (how should I call it?), with {......} is so elegant and
easier to use.

Can you or anybody out there in this august newsgroup help (and explain)?

Jack.

"R.VENKATARAMAN" $$$ schreef in bericht
...
strangely it works with


=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y")),TRUE,FALSE)

i.e without =1

But I could not explain the logic
=============================


R.VENKATARAMAN $$$ wrote in message
...
sorry. this is not working on subsequent experiments
apologise.

R.VENKATARAMAN $$$ wrote in message news:...
try type in H1




=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y"))=1,TRUE,FALSE)

your data is in d5 to d16.

do you get what you want.

may be tortuous to write .may be bugs in it. try the experiment

====================================
Jack Sons wrote in message
...
Tom,

Suppose in A1:F6 there can be all kinds of things (text, numbers,

blanks,
formulae), but I want to check if "apple", "pear", "cherry";
"coconut"

and
"apricot" each occur at least once. If so a formula in H1 should

return
TRUE, if not FALSE.
I tried
=COUNT(MATCH({"apple", "pear", "cherry";

"coconut","apricot"},A1:F6,0))=5
but it won't work, also when array entered.
Your help will be appreciated.

Jack Sons
The Netherlands



"Tom Ogilvy" schreef in bericht
...
Left of a double quote:
=if(counif(B1,"*apple*")0,"apple","")

--
Regards,
Tom Ogilvy


















R.VENKATARAMAN

yes elegant. thanks

RagDyeR wrote in message
...
Why even use the IF()?


=OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pears" ),COUNTIF(D5:D16,"cherry")
)

OR, even simpler:

=OR(COUNTIF(D5:D16,{"apple","pears","cherry"}))
--

Regards,

RD
--------------------------------------------------------------------
Please keep all correspondence within the Group, so all may benefit !
--------------------------------------------------------------------



"R.VENKATARAMAN" $$$ wrote in message
...
strangely it works with


=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y")),TRUE,FALSE)

i.e without =1

But I could not explain the logic
=============================


R.VENKATARAMAN $$$ wrote in message
...
sorry. this is not working on subsequent experiments
apologise.

R.VENKATARAMAN $$$ wrote in message news:...
try type in H1




=IF(OR(COUNTIF(D5:D16,"apple"),COUNTIF(D5:D16,"pea rs"),COUNTIF(D5:D16,"cherr
y"))=1,TRUE,FALSE)

your data is in d5 to d16.

do you get what you want.

may be tortuous to write .may be bugs in it. try the experiment

====================================
Jack Sons wrote in message
...
Tom,

Suppose in A1:F6 there can be all kinds of things (text, numbers,

blanks,
formulae), but I want to check if "apple", "pear", "cherry";

"coconut"
and
"apricot" each occur at least once. If so a formula in H1 should

return
TRUE, if not FALSE.
I tried
=COUNT(MATCH({"apple", "pear", "cherry";

"coconut","apricot"},A1:F6,0))=5
but it won't work, also when array entered.
Your help will be appreciated.

Jack Sons
The Netherlands



"Tom Ogilvy" schreef in bericht
...
Left of a double quote:
=if(counif(B1,"*apple*")0,"apple","")

--
Regards,
Tom Ogilvy



















All times are GMT +1. The time now is 03:10 PM.

Powered by vBulletin® Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
ExcelBanter.com