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'

sed is a stream editor. It searches and replaces greedily. The only way to do what you asked for is using an intermediate substitution pattern and changing it back in the end.

echo 'abcd' | sed -e 's/ab/xy/;s/cd/ab/;s/xy/cd/'

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.