sed regexp for: [111]

Question

INPUT

RANDOMSTRING BLA-BLA [111] RANDOMSTRING BLA-BLA

OUTPUT

RANDOMSTRING BLA-BLA [200] RANDOMSTRING BLA-BLA

[111] could contain various numbers with 3 digits: 120 or 300 or 400 or 101, etc.

So all I want ~this:

sed "s/Howtos<\/a> \[111] \//`grep '^<a href="' list.txt | wc -l`/" original.txt

"Regexp'ing the 111" that's the only thing I can't figure out

The output of:

grep '^<a href="' list.txt | wc -l

would be e.g.: 200 in the example

I'm not sure if this is a typo or not, but I think you need to escape both the opening [ and the closing ], if you want those literal chars. — Telemachus
– Telemachus, Commented May 8, 2011 at 19:32
I cannot make out what you are trying to accomplish here. Do you want to rip out the 3 digit numbers for use in something? Do you want to replace them with something else? Is it possible that digits could appear in the random strings before and after the one you want to work with? Are the fields consistent (e.g. Is it always the third space separated item from the left of the string?) — Caleb
– Caleb, Commented May 9, 2011 at 9:19

ennuikiller · Accepted Answer · 2011-05-08 19:33:52Z

2

regex for 3 digit number: [0-9]{3} So to replace 111 with 200:

sed -E 's/[0-9]{3}/200/g'

you need the -E flag for extended regexs

answered May 8, 2011 at 19:27

ennuikiller

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