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I'm trying to figure out how to identify IF a list of items in one cell contains a value or string.

EXAMPLE

Cell A1 contains sites, sheets, docs, slides.

I want cell B1 to display a 1 'if' cell A1 contains the string sites.

FORMULA

=if(A1 ?????? "sites", 1,0)

I'm not sure what to replace the ?????? within the above formula OR if this formula is possible. Any ideas on how to accomplish the desired outcome are greatly appreciated.

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149

You can use REGEXMATCH:

=IF(REGEXMATCH(A1, "sites"), 1, 0)

To explain, REGEXMATCH returns true if and only if the argument is a substring of your string.

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  • 12
    Using regexmatch would be simpler: if(regexmatch(A1, "sites"), 1, 0).
    – user79865
    Jul 19 '16 at 1:09
  • 3
    The "if" isn't needed since regexmatch already returns a boolean, so just "REGEXMATCH(A1, "sites")"
    – David
    Jan 18 '19 at 17:17
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    The if is there so they can convert the true to a 1 or 0 if false
    – Aurielle Perlmann
    Jan 18 '19 at 17:18
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    If parameters should be semicolon, not comma: =IF(REGEXMATCH(A1, "sites"); 1; 0)
    – Emil Stenström
    Dec 17 '19 at 12:35
  • 2
    "REGEXMATCH returns true if and only if the argument is a substring of your string." this is a highly misleading statement. Strings are interpreted as regular expressions so there are many cases where this does not work.
    – AnnanFay
    Jun 17 '20 at 19:11
20

If you, as Ryan Shillington suggests, want to know the number of cells within a range that has "sites" in it, you might try: 

=COUNTIF(A1:A100;"*sites*")

The * acts as a wildcard notion, so it will look for a match anywhere within the cell, not just an exact match for the whole cell.

I've also done this with a cell reference instead of "sites", but then you have to add wildcard-asterixes to the text in the cell.

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    I haven't tested it but I expect you could add the asterisks in the formula when referencing a cell by doing =COUNTIF(A1:A100; CONCATENATE("*", B1, "*"))
    – Matthew
    Dec 17 '20 at 18:43
  • I find this to be the simplest solution and can confirm that it works.
    – Patrick Walton
    Mar 1 at 17:49
20

This returns a #VALUE! error if the search term isn't found:

=IF(SEARCH("sites", B2) > 0, 1, 0)

This uses the same logic but catches the error and returns a 1/0 for success/failure:

=IF(IFERROR(SEARCH("sites", B2) > 0, 0), 1, 0)
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    The first formula will never return 0 because SEARCH(search_for, text_to_search, [starting_at]) will return an error when search_for do not be found.
    – Rubén ♦
    Jul 15 '16 at 23:05
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    How do I use search with COUNTIF?
    – Aaron Franke
    Feb 8 '20 at 0:10
10

A solution without REGEX:

=if(iserror(FIND("sites",B2)),0,1)

capitalization matters and if logic must be reversed, given that answer is true when the string is not found.

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  • This is basically the same as this answer, but introducing the case sensitive thing. By the way the sample data provided by the OP includes sheets not Sheets.
    – Rubén ♦
    Jan 16 '19 at 19:04
4

This will return TRUE or FALSE:

=IFERROR(SEARCH("keyword",A1)>0, FALSE)

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3

You're probably doing this because you want to know the count or average # of rows that have the word "sites" in it. Assuming you have 100 rows of entries in column A, you can use this function to find out how many rows have the word "sites" in it:

=QUERY(A1:A100, "select count(A) where A like '%sites%'")

This is much faster than creating a column of 1s and zeros and then operating on that new column.

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  • Can this work over a box instead of just a column?
    – Adam Starrh
    Sep 11 '19 at 18:53
  • Yup. instead of A1:A100 you could make it A1:C100 and then write a query with columns A, B and C.
    – Ryan Shillington
    Sep 11 '19 at 19:30
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If you go with the asterisks option and want a cell reference, you can use the CONCATENATE function to add the asterisks

=COUNTIF(A1:A100,CONCATENATE("*",A1,"*"))

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  • CONCATENATE can be harder to read than using & to join the strings: =COUNTIF(A1:A100, "*" &A1 &"*"))
    – RedYeti
    Feb 13 at 8:46
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=IF(QUERY(A1, "where A contains 'sites'")<>"", 1, )

0

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