Escaping SED command

Question

I am trying to run a script to replace a section of code in a lot of .php files. I need to replace:

`popupOpen(\'/home/open.php?c='.urlencode($row['id']).'&s=\', 650, 600)` 

with this:

`popupOpen(\'/home/open.php?c='.urlencode($row['id']).'&s=\', 1024, 750)`

The command I am using is:

find . -name "index.php" -print | xargs sed -i "s/popupOpen(\\\'\/home\/open.php?c='.urlencode\($row\['id']\).'\&s=\\\', 650, 600)/popupOpen(\\\'\/home\/open.php?c='.urlencode\($row\['id']\).'\&s=\\\', 1024, 750)/g"

I just cant seem to escape it properly. What am I missing?

Answer 1

You can do without xargs :

find .name "index.php" -exec sed -i '' sed "whatever" {} \;

and in this case I think this will work:

find . -name "index.php" -exec sed "s|popupOpen(\\\\'/home/open\.php\?c=\'\.urlencode(\\\$row\[\'id\'\])\.\'&s=\\\\', 650, 600)|popupOpen(\\\\'/home/open\.php\?c=\'\.urlencode(\\\$row\[\'id\'\])\.\'\&s=\\\\', 1024, 750)|g" {} \;