Escaping Regex in SED

Question

I need to use Sed to do a search and replace. I'm replacing /**# for define('WP_POST_REVISIONS', 3);\\n\\n/**# .

But I can't figure out the proper escaping. Even after escaping the (obviously needed) single quotes, I still get a bash: syntax error near unexpected token ')'

What is the proper escaping in this case?

Answer 1

It is not sed escaping, but bash escaping.

Escaping does not work within single-quotes ( ' )

You can use double-quotes ( " ), if you have no special characters like "$\\ in the parameter (or escape them there if necessary):

find /start/path -name *.html -exec sed -ie "s/abc/define('WP_POST_REVISIONS', 3);/g" '{}' \;

Or quote using $' , which supports escaping:

find /start/path -name *.html -exec sed -ie $'s/abc/define(\'WP_POST_REVISIONS\', 3);/g' '{}' \;

Answer 2

try to replace your:

find /start/path -name *.html -exec sed -ie 's|/**#|define(\'WP_POST_REVISIONS\', 3);|g' '{}' \;

with:

find /start/path -name '*.html' -print0  \
  | xargs -0 -n 1 sed -ie 's|/\*\*#|define('\''WP_POST_REVISIONS'\'', 3);\n/\*\*#|g'

and tell us what it gives you

(I tried to guess you were looking for the actual string "/**#" in your file(s) ... please give us examples of what you are really looking for, if it isn't that actual string)