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.