On the php manual for shell_exec, it shows that the function returns the output as a string. If you expect output from the program you launch, you need to capture this like so:

$execQuery = "echo -n test_command";
$output = shell_exeeec($execQuery);
echo $output;

Your question doesn't show trying to capture any data. If you also make sure to connect stdout and stderr when you run your command, you should get a better idea is what is going on. To use your example:

$out = shell_exec("/usr/bin/oneuser create test10 test10 2>&1");
var_dump($out);

That should help you see what is going on. As Shadur suggests, it seems likely that these programs expect an interactive terminal that can enter passwords in order to run. Even if don't need input, they might expect interactive shells. And he's right that su doesn't play nice in this context. There is, however, a correct tool for the job.

You can setup sudo to such that your http user can execute your program as username without a password but NOT be able to do anything else by running visudo or whatever you use to edit your sudoers file and adding this line:

http    ALL=(username) /usr/bin/oneadmin

Then in php your command would look something like this:

$execQuery = "sudo -u username /usr/bin/oneadmin postgres -c '/usr/bin/oneuser create test10 test10'";
$out = shell_exec ("$execQuery 2>&1");
echo $out

On the php manual for shell_exec, it shows that the function returns the output as a string. If you expect output from the program you launch, you need to capture this like so:

$execQuery = "echo -n test_command";
$output = shell_exec($execQuery);
echo $output;

Your question doesn't show trying to capture any data. If you also make sure to connect stdout and stderr when you run your command, you should get a better idea is what is going on. To use your example:

$output = shell_exec("/usr/bin/oneuser create test10 test10 2>&1");
var_dump($output);

That should help you see what is going on. As Shadur suggests, it seems likely that these programs expect an interactive terminal that can enter passwords in order to run. Even if don't need input, they might expect interactive shells. And he's right that su doesn't play nice in this context. There is, however, a correct tool for the job.

You can setup sudo to such that your http user can execute your program as username without a password but NOT be able to do anything else by running visudo or whatever you use to edit your sudoers file and adding this line:

http    ALL=(username) /usr/bin/oneadmin

Then in php your command would look something like this:

$execQuery = "sudo -u username /usr/bin/oneadmin postgres -c '/usr/bin/oneuser create test10 test10'";
$out = shell_exec ("$execQuery 2>&1");
echo $out

On the php manual for shell_exec, it shows that the function returns the output as a string. If you expect output from the program you launch, you need to capture this like so:

$execQuery = "echo -n test_command";
$output = shell_exeeec($execQuery);
echo $output;

Your question doesn't show trying to capture any data. If you also make sure to connect stdout and stderr when you run your command, you should get a better idea is what is going on. To use your example:

$out = shell_exec("/usr/bin/oneuser create test10 test10 2>&1");
var_dump($out);

That should help you see what is going on. As Shadur suggests, it seems likely that these programs expect an interactive terminal that can enter passwords in order to run. Even if don't need input, they might expect interactive shells. And he's right that su doesn't play nice in this context. There is, however, a correct tool for the job.

You can setup sudo to such that your http user can execute your program as username without a password but NOT be able to do anything else by running visudo or whatever you use to edit your sudoers file and adding this line:

http    ALL=(username) /usr/bin/oneadmin

Then in php your command would look something like this:

$execQuery = "sudo -u username /usr/bin/oneadmin postgres -c '/usr/bin/oneuser create test10 test10'";
$out = shell_exec ("$execQuery 2>&1");
echo $out

On the php manual for shell_exec, it shows that the function returns the output as a string. If you expect output from the program you launch, you need to capture this like so:

$execQuery = "echo -n test_command";
$output = shell_exeeec($execQuery);
echo $output;

Your question doesn't show trying to capture any data. If you also make sure to connect stdout and stderr when you run your command, you should get a better idea is what is going on. To use your example:

$out = shell_exec("/usr/bin/oneuser create test10 test10 2>&1");
var_dump($out);

That should help you see what is going on. As Shadur suggests, it seems likely that these programs expect an interactive terminal that can enter passwords in order to run. Even if don't need input, they might expect interactive shells. And he's right that su doesn't play nice in this context. There is, however, a correct tool for the job.

You can setup sudo to such that your http user can execute your program as username without a password but NOT be able to do anything else by running visudo or whatever you use to edit your sudoers file and adding this line:

http    ALL=(username) /usr/bin/oneadmin

Then in php your command would look something like this:

$execQuery = "sudo -u username /usr/bin/oneadmin postgres -c '/usr/bin/oneuser create test10 test10'";
$out = shell_exec ("$execQuery 2>&1");
echo $out