Call and run php script from shell script

Question

I'm trying something simple yet can't figured it out. Trying to run one shell script which call php script from the site directory. My shell script is simple test.sh:

#!/bin/sh

LIST="/path/to/my/site/dir"
CONFIG="/usr/bin/php "

for i in $LIST
do
    . "${CONFIG}${i}/test.php"        
done

My test.php doesn't have errors and when I run it directly in browser it's working. It again simple script like

<?php
  // source code
?>

When I run ./test.sh result is

./test.sh: 8: .: Can't open /usr/bin/php /path/to/my/site/dir/test.php

Line 8 is . "${CONFIG}${i}/test.php"

I've tried also to add at the beginning of PHP file this line but the result is same

#!/usr/bin/php

UPDATE: Path to php

$ which -a php
/usr/bin/php

I've also made chmod +x test.php

Answer 1

The problem in your code is the line:

. "${CONFIG}${i}/test.php"  

Remove the .

---

Here is another example:

$ ls -l

-rwxrwxr-x 1 bg bg 67 Oct 20 09:42 index.php
-rwxrwxr-x 1 bg bg 68 Oct 20 09:43 test.sh

index.php

<?php
    shell_exec('echo Hello > /tmp/hello.txt');
?>

test.sh

#!/bin/bash
/usr/bin/php index.php

Answer 2

The problem here is that you are quoting the entiore command you are trying to run as a single variable. As a result, you're not running php with foo.php as an argument but instead are attempting to execute a file called php foo.php. Here's a simpler example to show you what I mean:

$ var1="echo "
$ var2="foo"
$ set -x ## debugging info
$ "$var1$var2"
+ 'echo foo'      ### The shell tries to execute both vars as a single command
bash: echo foo: command not found

$ "$var1" "$var2"
+ 'echo ' foo     ### The shell tries to execute 'echo ' (echo with a space)
bash: echo foo: command not found  

So, the right way is to remove the space and quote each variable separately:

$ var1="echo"
$ var2="foo"
$ "$var1" "$var2"

If you do that though, you'll hit the next error. The . is the source command. That tries to read a shell script and execute it in the current session. You are giving it a php script instead. That won't work, you need to execute it, not source it.

Finally, always avoid using CAPITAL variable names. The shell's reserved variables are capitalized so it's a good idea to always use lower case variable names for your scripts.

Putting all this together (with a few other minor improvements), what you want is something like:

#!/bin/sh

list="/path/to/my/site/dir"
config="/usr/bin/php"

for i in "$list"
do
    "$config" "$i"/test.php
done

Answer 3

Usually the command php is used for interpretation of PHP scripts in the shell.

$ php /path/script-name.php

I made simple test.sh and it works:

$ cat ./text.sh
#!/bin/bash
sudo php /var/www/wiki/maintenance/update.php

$ chmod +x ./test.sh
$ ./test.sh

It works.

After that I made complicated script as your example:

$ cat ./text.sh

#!/bin/sh

LIST=/var/www/wiki/maintenance
CONFIG=/usr/bin/php

for i in $LIST
do
    ${CONFIG} ${i}/update.php
done

$ sudo ./test.sh

It works!

$ cat ./text.sh

#!/bin/sh

LIST="/var/www/wiki/maintenance"
CONFIG="/usr/bin/php "

for i in $LIST
do
    ${CONFIG}${i}/update.php
done

$ sudo ./test.sh

Works also!