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#1
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Want to create an if fuction that will give a result of "Yes" if any part of
the cell cotains 16Z. |
#2
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Try
=IF(ISERROR(SEARCH("16Z",A1)),"","Yes") If you want (for example) to say No if it isn't there try =IF(ISERROR(SEARCH("16Z",A1)),"No","Yes") If you want it to be case sensitive change SEARCH to FIND Mike "Supe" wrote: Want to create an if fuction that will give a result of "Yes" if any part of the cell cotains 16Z. |
#3
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Look at this:
=IF(ISERROR(FIND("16Z",A1)),"No","Yes") Regards, Per "Supe" skrev i meddelelsen ... Want to create an if fuction that will give a result of "Yes" if any part of the cell cotains 16Z. |
#4
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=IF(COUNTIF(A1,"*16z*"),"yes","no")
"Supe" wrote: Want to create an if fuction that will give a result of "Yes" if any part of the cell cotains 16Z. |
#5
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Sure, you can use the IF function in combination with the SEARCH function to achieve this. Here are the steps:
The formula will check if the cell contains the text "16Z" and return "Yes" if it does, and "No" if it doesn't. The SEARCH function is used to find the position of the text "16Z" within the cell, and the ISNUMBER function checks if the result is a number (which means the text was found). If the text is found, the IF function returns "Yes", otherwise it returns "No". Note that this formula is case-sensitive, so it will only find "16Z" and not "16z" or "16zZ". If you want to make it case-insensitive, you can use the FIND function instead of SEARCH and wrap the text to search for in the UPPER function to convert it to uppercase.
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