1

I'm just trying to replace a setting in a php.ini file using a bash script.

This is my case:

I have a very specific setting and the default is:

;date.timezone =

Now, I'm trying to replace this by sed to this:

date.timezone = "America/Mexico_City"

And the command that I'm trying to use is this:

sed -i '/^;date\.timezone[[:space:]]=.*$/date.timezone = "America/Mexico_City"' php.ini

But I get this:

sed: -e expression #1, char 35: extra characters after command

I don't really know why but if I leave just a character it works (not the expected way but it doesn't return an error). Example:

sed -i '/^;date\.timezone[[:space:]]=.*$/d' php.ini

This just removes the line ;date.timezone =

I'm really new to sed and I'd really appreciate any help.

Thank you

6

You are trying to substitute one pattern for another, so you need to use the sed s command, of the form

's/pattern/replacement/'

Also, you need to escape any / characters within the pattern or replacement strings (in your case "America\/Mexico_City"). So

sed 's/^;date\.timezone[[:space:]]=.*$/date.timezone = "America\/Mexico_City"/'

Alternatively, you can avoid escaping slashes by using an different pattern-replacement delimiter such as s#pattern#replacement# e.g.

sed 's#^;date\.timezone[[:space:]]=.*$#date.timezone = "America/Mexico_City"#'

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