PHP exec() not returning error message in output

Question

I am trying to get certain output for svn command in XML format. Output is ok when I type valid parameters. However, when I type in wrong password, output does not show error message. This is the PHP code:

exec('/usr/bin/svn --username something --password something --non-interactive log -r HEAD --xml --verbose http://a51.unfuddle.com/svn/a51_activecollab/', $output); 

Here is output I get in the terminal:

<?xml version="1.0"?> <log> svn: OPTIONS of 'http://a51.unfuddle.com/svn/a51_activecollab': authorization failed: Could not authenticate to server: rejected Basic challenge (http://a51.unfuddle.com) 

And here is the output I get from the $output variable with var_dump:

array(2) { [0]=> string(21) "<?xml version="1.0"?>" [1]=> string(5) "<log>" } 

As you can see the $output variable does not return third line of output, where terminal does. Please help me to get the same output as I get in terminal (I even tried with shell_exec() or system() methods but they return the same output as exec()) How do I get full output? Thank you in advance!

Answer 1

You need to capture the stderr too.

Redirecting stderr to stdout should do the trick. Append 2>&1 to the end of your command.

e.g.

exec("/usr/bin/svn --username something --password something --non-interactive log -r HEAD --xml --verbose http://a51.unfuddle.com/svn/a51_activecollab/ 2>&1", $output); 

If it is a complex command (e.g. one with a pipe, like doing a mysqldump and passing it to gzip and then redirecting to file mysqldump ... | gzip > db.sql.gz) create a subshell to capture the overall standard-error and redirect it to standard-output:

exec('( error_command | cat >/dev/null ) 2>&1', $output) #     ^                                ^   ^ #     `-- sub-shell with the command --´   `-- stderr to $output