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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T00:23:52+00:00 2026-05-25T00:23:52+00:00

I’m running a grep to find any *.sql file that has the word select

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I’m running a grep to find any *.sql file that has the word select followed by the word customerName followed by the word from. This select statement can span many lines and can contain tabs and newlines.

I’ve tried a few variations on the following:

$ grep -liIr --include="*.sql" --exclude-dir="\.svn*" --regexp="select[a-zA-Z0-
9+\n\r]*customerName[a-zA-Z0-9+\n\r]*from"

This, however, just runs forever. Can anyone help me with the correct syntax please?

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  1. Editorial Team
    Editorial Team
    2026-05-25T00:23:52+00:00Added an answer on May 25, 2026 at 12:23 am

    Without the need to install the grep variant pcregrep, you can do a multiline search with grep.

    $ grep -Pzo "(?s)^(\s*)\N*main.*?{.*?^\1}" *.c
    

    Explanation:

    -P activate perl-regexp for grep (a powerful extension of regular expressions)

    -z Treat the input as a set of lines, each terminated by a zero byte (the ASCII NUL character) instead of a newline. That is, grep knows where the ends of the lines are, but sees the input as one big line. Beware this also adds a trailing NUL char if used with -o, see comments.

    -o print only matching. Because we’re using -z, the whole file is like a single big line, so if there is a match, the entire file would be printed; this way it won’t do that.

    In regexp:

    (?s) activate PCRE_DOTALL, which means that . finds any character or newline

    \N find anything except newline, even with PCRE_DOTALL activated

    .*? find . in non-greedy mode, that is, stops as soon as possible.

    ^ find start of line

    \1 backreference to the first group (\s*). This is a try to find the same indentation of method.

    As you can imagine, this search prints the main method in a C (*.c) source file.

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