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-   -   Having trouble with returning a certian value using =IF function (https://www.excelbanter.com/excel-discussion-misc-queries/165960-having-trouble-returning-certian-value-using-%3Dif-function.html)

Stozy

Having trouble with returning a certian value using =IF function
 
Hello, i am a unprofessional Excel user. I use Excel from home to catalog my
professionally graded sports cards. I collect PSA graded cards that are
given a value between 1 - 10 depending on the quality of the card. 1 being
basically junk, and 10 being the best (Pristien, Gem.....).

PSA also offers qualifiers grades if you have a beautiful card that for
example was factory miscut.... a grade of 9Q will be given to a card that is
in Gem condition, but has one of a couple uncontrolable flaws. But the
problem is that a unqualified 9 is worth 9pts toward a set to where 9Q is
worth less.

I am using a drop menu assinging a value to each card in the individual set
between 1 and 10, and 2Q and 9Q.

My problem is that Excel trys to recalculate my formula using Q9 (cell
range) instead of 9Q in my drop down menu.

The formula i am trying to use is
=IF(C12=9Q,"7.77","1")
C12 obviously is where the drop menu has been selected to 9Q.
7.77 is what i want displayed in a seperate area for calculation of the set.
1 is what would be displayed if anything other than 9Q is selected from the
menu.
Excel returns this as a proposed correction
=IF(C12=Q9,"7.77","1")
Which does not work with the drop menu, it is looking at cell Q9 instead of
returning a value if 9Q.

It works just fine if i assign a different number value to the 8 possible
qualifier grades, but looks very messy.... I really want Q9, Q8 ect. to
display properly and still return a different value on my calculation portion
of the sheet (Looks much more professional)

Digging through the Excel website and help section in Excel has not helped
much on this issue :/

Am i missing a better simple function that i should use? Or is there
another formula i should be using? Help would be great! Thanks in advance

Thanks, Stozy

carlo

Having trouble with returning a certian value using =IF function
 
On Nov 14, 3:14 pm, Stozy wrote:
Hello, i am a unprofessional Excel user. I use Excel from home to catalog my
professionally graded sports cards. I collect PSA graded cards that are
given a value between 1 - 10 depending on the quality of the card. 1 being
basically junk, and 10 being the best (Pristien, Gem.....).

PSA also offers qualifiers grades if you have a beautiful card that for
example was factory miscut.... a grade of 9Q will be given to a card that is
in Gem condition, but has one of a couple uncontrolable flaws. But the
problem is that a unqualified 9 is worth 9pts toward a set to where 9Q is
worth less.

I am using a drop menu assinging a value to each card in the individual set
between 1 and 10, and 2Q and 9Q.

My problem is that Excel trys to recalculate my formula using Q9 (cell
range) instead of 9Q in my drop down menu.

The formula i am trying to use is
=IF(C12=9Q,"7.77","1")
C12 obviously is where the drop menu has been selected to 9Q.
7.77 is what i want displayed in a seperate area for calculation of the set.
1 is what would be displayed if anything other than 9Q is selected from the
menu.
Excel returns this as a proposed correction
=IF(C12=Q9,"7.77","1")
Which does not work with the drop menu, it is looking at cell Q9 instead of
returning a value if 9Q.

It works just fine if i assign a different number value to the 8 possible
qualifier grades, but looks very messy.... I really want Q9, Q8 ect. to
display properly and still return a different value on my calculation portion
of the sheet (Looks much more professional)

Digging through the Excel website and help section in Excel has not helped
much on this issue :/

Am i missing a better simple function that i should use? Or is there
another formula i should be using? Help would be great! Thanks in advance

Thanks, Stozy


Didn't read the whole thing.

But I think you need to tell Excel that you are working with text and
not with a cell, hence:

=IF(C12="9Q","7.77","1")

Where as the other "" are not really used, because they are numbers,
right? So you could
even do that:

=IF(C12="9Q",7.77,1)

hth

Carlo


Stozy

Having trouble with returning a certian value using =IF functi
 
Yes, that was one of the issues, but ofcourse it has open up another
problem....
Gonna mess with it for a while and see if cant muster up enough brain cells
to get this one :D

Thanxs, Stozy

"carlo" wrote:

On Nov 14, 3:14 pm, Stozy wrote:
Hello, i am a unprofessional Excel user. I use Excel from home to catalog my
professionally graded sports cards. I collect PSA graded cards that are
given a value between 1 - 10 depending on the quality of the card. 1 being
basically junk, and 10 being the best (Pristien, Gem.....).

PSA also offers qualifiers grades if you have a beautiful card that for
example was factory miscut.... a grade of 9Q will be given to a card that is
in Gem condition, but has one of a couple uncontrolable flaws. But the
problem is that a unqualified 9 is worth 9pts toward a set to where 9Q is
worth less.

I am using a drop menu assinging a value to each card in the individual set
between 1 and 10, and 2Q and 9Q.

My problem is that Excel trys to recalculate my formula using Q9 (cell
range) instead of 9Q in my drop down menu.

The formula i am trying to use is
=IF(C12=9Q,"7.77","1")
C12 obviously is where the drop menu has been selected to 9Q.
7.77 is what i want displayed in a seperate area for calculation of the set.
1 is what would be displayed if anything other than 9Q is selected from the
menu.
Excel returns this as a proposed correction
=IF(C12=Q9,"7.77","1")
Which does not work with the drop menu, it is looking at cell Q9 instead of
returning a value if 9Q.

It works just fine if i assign a different number value to the 8 possible
qualifier grades, but looks very messy.... I really want Q9, Q8 ect. to
display properly and still return a different value on my calculation portion
of the sheet (Looks much more professional)

Digging through the Excel website and help section in Excel has not helped
much on this issue :/

Am i missing a better simple function that i should use? Or is there
another formula i should be using? Help would be great! Thanks in advance

Thanks, Stozy


Didn't read the whole thing.

But I think you need to tell Excel that you are working with text and
not with a cell, hence:

=IF(C12="9Q","7.77","1")

Where as the other "" are not really used, because they are numbers,
right? So you could
even do that:

=IF(C12="9Q",7.77,1)

hth

Carlo




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