Removing everything outside brackets in a cell

justin_n84

Board Regular
Joined
Jan 29, 2017
Messages
59
Hi!!

I have a bunch of cells which contain data similar to "[FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif]0.120 (3.048)". I want to delete everything out side the brackets. Is there an easy way to do this?

Thanks in advance
[/FONT]
 
OK sounds like youre going my route with 2 replaces. For the first replace you want to replace anything followed by an open bracket with just an open bracket. This is done by entering *( in the search box and ( in the replace box. For the second search, type )* in the search box and ) in the replace box. I have highlighted the exact text to use in red.
 
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Excel Facts

What is the last column in Excel?
Excel columns run from A to Z, AA to AZ, AAA to XFD. The last column is XFD.
For those interested, here is a simple formula solution...

=-REPLACE(C1,1,FIND("(",C1)-1,"")
 
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Very good Rick !

It looks like that works because Excel is able to interpret the entry
(3.048)
as a number, not a text string, especially when encouraged to do so by use of the - symbol at the front.

So this approach does not work in a slightly different scenario, such as
0.120[3.048]
or
0.120X3.048X

<tbody>
</tbody>

Edit to add - don't know why my post has this box around the last couple of lines - wasn't intended by me and now don't know how to get rid of it :)
 
Last edited:
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It looks like that works because Excel is able to interpret the entry
(3.048)
as a number, not a text string, especially when encouraged to do so by use of the - symbol at the front.
One of the ways Excel can display negative numbers is when the number is enclosed in parentheses, so I used the REPLACE function to create that text string... if a numerical text string is involved in a mathematical expression, Excel will convert the text number to a real number in order to complete the mathematical operation. Placing a minus sign in front of a number is the same as multiplying that number by minus one which is a mathematical operation. So, in combination, my formula is multiplying a negative text number by minus one... the negative text number becomes a real negative number in order to complete the operation of multiplying it by minus one... and a minus number times a minus number produces a positive result (the product of the two values being multiplied).
 
Last edited:
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