php -l: suppress output on valid files

Question

When using the php -l myFile.php command (PHP 5.5.30), if the file has a syntax error then I get the proper warnings and stack trace, etc.

However, if the file has no syntax warnings I get the message

No syntax errors detected in myFile.php

Is there a way to have the command have no output when the syntax is valid? I only care if a file has invalid syntax - I don't need a message saying it's valid.

Answer 1

The "no syntax errors..." message is sent out on the stdout while the syntax errors are sent out on stderr. You can redirect those to somewhere like /dev/null if you don't want them.

php -l file.php 1> /dev/null

that will output the errors if there were any or nothing if no errors. You do lose the "Errors parsing..." message, but will get the errors if there was a problem.

Answer 2

php -ln script.php >/dev/null || php -ln script.php

EDIT:

chronic php -ln script.php

Answer 3

You can use chronic to suppress all output if the command succeeds (returns 0):

chronic php -l myFile.php

---

DESCRIPTION

chronic runs a command, and arranges for its standard out and standard error to only be displayed if the command fails (exits nonzero or crashes). If the command succeeds, any extraneous output will be hidden.

On Debian, it's in the moreutils package.

Answer 4

Don't check the output, check the return code.

$ php -l good.php &> /dev/null; echo $?
0

$ php -l bad.php &> /dev/null; echo $?
255

So:

if ! php -l somescript.php &> /dev/null; then
  echo 'OH NOES!'
fi

Or if you're feelin fancy:

if ! foo=$(php -l somescript.php 2>&1); then
  echo $foo
fi