将变量从PHP传递给Ubuntu shell脚本

Question

I have a shell script like this (in /usr/local/bin/esm-script/import-master.php):

#! /bin/bash 
echo "this is the file name $1."
script -c 'PGPASSWORD="pwd123" /usr/bin/psql --host localhost --port 5432 --username "postgres" --no-password --quiet "dbESM" < "$1"' /dev/null

Now I'm calling it through a PHP script like this:

$NewFile = "/var/www/...master-data.sql"; //full path
$strImport = '/usr/local/bin/esm-script/import-master.sh ' . $NewFile;  
$strMsg = shell_exec($strImport);
echo "$strMsg<br />";
echo 'done!';

However, when I run the PHP code, this is the message I get on the browser:

this is the file name /var/www/ESM-Backend/uploads/master-data.sql. Script started, file is /dev/null Script done, file is /dev/null sh: 1: cannot open : No such file Script started, file is /dev/null
done! 

I'm not a shell scripting person so I don't know if I'm missing something.

I've checked that the folder with the sql file has the correct permissions (775) and has data (insert statements).

So why does this not work? Any ideas and guidelines are really appreciated.

EDIT I hard-coded the file in the shell script file like this:

script -c 'PGPASSWORD="#UR!23" /usr/bin/psql --host localhost --port 5432 --username "postgres" --no-password --quiet "dbESM" < "/var/www/ESM-Backend/uploads/master-data.sql"' /dev/null

And it works. But I need it to run with the file passed through PHP.

Answer 1

I think the problem might be with the way you are passing the argument to your shell script i.e. $NewFile. From a relevant SO Post, you might want to try enclosing the argument(s) in double quotes like this:

$strImport = '/usr/local/bin/esm-script/import-master.sh "'.$NewFile.'"';

I'm assuming the permissions are set correctly and you are able to use shell_exec() normally to execute shell scripts via PHP. Hope this helps.