I need parse uri in shell scripts. So, I tried to use php in bash as below.
#!/bin/sh
uri="http://www.google.com?key=value
key="host"
value=$(php -r "$parse = parse_url('$uri'); echo $parse['$key']")
It has showing the following error.
PHP Parse error: syntax error, unexpected '=' in Command line code on line 1
Some body can help how to use embedded php in bash ?
A cheap way of debugging this is to use echo to see what you're passing in to php:
echo "$parse = parse_url('$uri'); echo $parse['$key']"
shows
= parse_url('http://www.google.com?key=value'); echo ['host']
You're already using $uri to mean "the value of the shell variable uri", so it's not surprising that $parse is also considered a shell variable and expanded to its value (unset, nothing).
Use \$ when you want a literal dollar sign in your double quoted string:
value=$(php -r "\$parse = parse_url('$uri'); echo \$parse['$key']")
You can use it easily, but you must be careful because of escaping in bash.
I recommend to use single quotes (you do not need to escape anything) and exit from the quotes when you want to do something special. Your example:
php -r '$parse=parse_url("'$url'"); echo $parse["'$part'"];'
Note that
$parse'$url'" instead.Update:
Just for the clarification - parse error happened because $parse was interpreted as bash variable (empty string) so the php command incorrectly started with =.
php test.phpcan also be called on command line same way you do with a bash script. Seems you're also more comfortable with PHP anyway.