Escaping in sed command

Question

I'm trying to replace all instances of $timestamp = mysqli_real_escape_string($db_conx, $_POST['Timestamp']); with $timestamp = date("Y-m-d H:i:s"); in all files under /home/user/public_html/

I have tried many variations of the below command but can't seem to get the escaping right!

find /home/user/public_html/ -type f -print0 | xargs -0 sed -i 's/\$timestamp = mysqli_real_escape_string(\$db_conx, \$_POST['\''Timestamp'\'']);/\$timestamp = date(\"Y-m-d H:i:s\");/g'

Thanks, all the other threads I could find on this were hard to follow but I was able to get it thanks to that thread — Shane Buckley, Jan 08 '17 at 22:23

score 0 · Answer 1 · answered Jan 08 '17 at 22:22

0

I got it thanks to @phk

find /home1/user/public_html/ -type f -print0 | xargs -0 sed -i 's/\$timestamp = mysqli_real_escape_string(\$db_conx, \$_POST\['\''Timestamp'\''\]);/\$timestamp = date("Y-m-d H:i:s");/g'

answered Jan 08 '17 at 22:22

Shane Buckley

3

As often (if not always), -print0 and xargs -0 are useless and the command can be replaced by find /home1/user/public_html/ -type f -exec sed -i 's/\$timestamp = mysqli_real_escape_string(\$db_conx, \$_POST\['\''Timestamp'\''\]);/\$timestamp = date("Y-m-d H:i:s");/g' {} + – jlliagre Jan 09 '17 at 02:21

Escaping in sed command

1 Answers1