Using grep to find a string

Question

I am trying to locate a file that (should) contains the string "1896.345", somewhere under /home/dir/.

Any ideas of how I can do it? Thanks

January · Accepted Answer · 2012-09-11 14:25:17Z

5

cd home/dir

grep -r -l -i "1896\.345" .

January

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answered Sep 11 '12 at 13:32

Shagun Sodhani

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2

A dot instead of the asterisk would avoid any accidental shell expansion. – Thor Sep 11 '12 at 13:35
2

I have added an escape backslash to your answer, since otherwise strings like "1896-345" would also be matched. – January Sep 11 '12 at 13:54
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This will print the actual match, what the OP asks for is the name of the file, so grep -rl would do. – Thor Sep 11 '12 at 14:01

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Thor · Accepted Answer · 2012-09-11 13:32:40Z

0

Use find and xargs:

find /home/dir -type f -print0 | xargs -0 grep -l 1896.345

answered Sep 11 '12 at 13:25

Thor

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I assumed the question meant searching for text within the file, in which case you can use "find /home/dir -type f -exec grep 1896.345 {} \;" - but be warned that this may take a lot of time if there are many files, and will only work if the text is in a plain text file (not eg, LibreOffice document). – chronitis Sep 11 '12 at 13:30
You're right, I misread. Should work now. – Thor Sep 11 '12 at 13:31
Yeah, but why spawn a separate grep instance for every file that you want to search, when grep -r does the job? That solution makes sense only if you want to limit your search by some other limits (like file creation date, file type, permissions etc.) – January Sep 11 '12 at 13:52
xargs passes as many files as fit into MAXARG to grep. But you're about it being redundant when grep -r pattern . will do. – Thor Sep 11 '12 at 13:59

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2 Answers 2