Replace matching regex in a file with custom word

Question

I am trying to replace version number (like 00.00.00-00) in a file with actual version number in shell.

For this I use the following command:

sed -i 's/[0-9]*\.+[0-9]*\.+[0-9]*\-+[0-9]*/MyNewVersion/g' file.txt

Example of the file.txt:

Some text Bla Bla Bla-00.00.00-00.x86_SomeOtherInterestingThings

Unfortunately it does not give me desired result. Any advice?

Remove all +.

Cyrus
– Cyrus

2017-09-20 08:36:07 +00:00
Commented Sep 20, 2017 at 8:36 — Cyrus
– Cyrus, Commented Sep 20, 2017 at 8:36

lpg · Accepted Answer · 2017-09-20 08:36:29Z

1

First of all, you would need -r flag in order to mark the pattern as a regular expression. Next, you would need to move the quantifier + next to the number ranges, so I believe you should be using this command:

sed -i -r 's/[0-9]+\.[0-9]+\.[0-9]+-[0-9]+/MyNewVersion/g' file.txt

edited Sep 20, 2017 at 8:36

answered Sep 20, 2017 at 8:35

lpg

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2 Comments

Wiktor Stribiżew Over a year ago

Well, even without -r that would be parsed as a regex, -r will make it be parsed as a POSIX ERE pattern (where + quantifier does not need escaping, in BRE POSIX, in GNU sed, + can be used if escaped). No need to escape -, BTW.

lpg Over a year ago

Removed backslash before -

Kent · Accepted Answer · 2017-09-20 08:48:57Z

0

based on the example input/pattern, this line should help:

sed 's/[-0-9.]\+\([.]x86_\)/-NEW\1/'

answered Sep 20, 2017 at 8:48

Kent

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