How to replace multiple patterns at once with sed?
Maybe something like this:
sed 's/ab/~~/g; s/bc/ab/g; s/~~/bc/g'
Replace ~
with a character that you know won't be in the string.
I always use multiple statements with "-e"
$ sed -e 's:AND:\n&:g' -e 's:GROUP BY:\n&:g' -e 's:UNION:\n&:g' -e 's:FROM:\n&:g' file > readable.sql
This will append a '\n' before all AND's, GROUP BY's, UNION's and FROM's, whereas '&' means the matched string and '\n&' means you want to replace the matched string with an '\n' before the 'matched'
Here is a variation on ooga's answer that works for multiple search and replace pairs without having to check how values might be reused:
sed -i '
s/\bAB\b/________BC________/g
s/\bBC\b/________CD________/g
s/________//g
' path_to_your_files/*.txt
Here is an example:
before:
some text AB some more text "BC" and more text.
after:
some text BC some more text "CD" and more text.
Note that \b
denotes word boundaries, which is what prevents the ________
from interfering with the search (I'm using GNU sed 4.2.2 on Ubuntu). If you are not using a word boundary search, then this technique may not work.
Also note that this gives the same results as removing the s/________//g
and appending && sed -i 's/________//g' path_to_your_files/*.txt
to the end of the command, but doesn't require specifying the path twice.
A general variation on this would be to use \x0
or _\x0_
in place of ________
if you know that no nulls appear in your files, as jthill suggested.