Linux – sed replace text – escape characters

escapinglinuxreplacesearch-and-replacesed

I have a mysqldump file in which I want to replace the following text <a href="/tag/ with <a href="http://www.mydomain.com/tag/ but can't find a way to correctly escape all special characters.

I'm using the following command (+ a few other variants)
cat wordpress_posts.sql | sed 's%/<a href="\/\tag\/\/<a href="\http:/\/\www.mydomain.com/\tag/\/%g' > wordpress_posts_new.sql
but it's not working.

Can anybody help?

Update 1:
Turns out that the source string in the mysqldump isn't <a href="/tag/ but <a href=\"/tag/ (note the extra backslash after the equal symbol)
Here's a pastebin of one line of the SQL file which contains the string I want to replace: http://pastebin.com/8G5mcxpJ

I tried all 3 suggested versions of the sed command, but none would replace the above string with <a href=\"http://www.mydomain.com/tag/
(yes I added the backslash after the equal symbol)

Best Answer

No need to pipe cat to sed:

$ sed 's/<a href="\/tag\//<a href="http:\/\/www.mydomain.com\/tag\//g' wordpress_posts.sql

Remove the percentage sign
You only need to escape the slash
Specify an in-place editing (-i) if you want

Related Solutions

Bash – Permission Denied when using Sed

Do not run the command you are trying to run

If you try to redirect output from sed back into the same file it will blank it out, deleting all of the files contents. Try something like this:

sed s/world/cat/ <test.properties >newtest.properties && sudo mv newtest.properties test.properties

You get permission denied because the redirection part of the command is not run via sudo, but run as your normal user.

The first command works because you are simply reading the first file and writing to a file you own, so you regular user could do that.

Bash – Using sed to remove both an opening and closing square bracket around a string

This is easy, if you follow the manual carefully: all members inside a character class lose special meaning (with a few exceptions). And ] loses its special meaning if it is placed first in the list. Try:

$ echo '[123]' | sed 's/[][]//g'
123
$

This says:

inside the outer [ brackets ], replace any of the included characters, namely:
- ] and
- [
replace any of them by the empty string — hence the empty replacement string //,
replace them everywhere (globally) — hence the final g.

Again, ] must be first in the class whenever it is included.

Best Answer

Related Solutions

Bash – Permission Denied when using Sed

Bash – Using sed to remove both an opening and closing square bracket around a string

Related Topic