How to check if a shell command exists from PHP

I need something like this in php:

If (!command_exists('makemiracle')) {
  print 'no miracles';
  return FALSE;
}
else {
  // safely call the command knowing that it exists in the host system
  shell_exec('makemiracle');
}

Are there any solutions?

---

On Linux/Mac OS Try this:

function command_exist($cmd) {
    $return = shell_exec(sprintf("which %s", escapeshellarg($cmd)));
    return !empty($return);
}

Then use it in code:

if (!command_exist('makemiracle')) {
    print 'no miracles';
} else {
    shell_exec('makemiracle');
}

Update: As suggested by @camilo-martin you could simply use:

if (`which makemiracle`) {
    shell_exec('makemiracle');
}

---

Windows uses where , UNIX systems which to allow to localize a command. Both will return an empty string in STDOUT if the command isn't found.

PHP_OS is currently WINNT for every supported Windows version by PHP.

So here a portable solution:

/**
 * Determines if a command exists on the current environment
 *
 * @param string $command The command to check
 * @return bool True if the command has been found ; otherwise, false.
 */
function command_exists ($command) {
  $whereIsCommand = (PHP_OS == 'WINNT') ? 'where' : 'which';

  $process = proc_open(
    "$whereIsCommand $command",
    array(
      0 => array("pipe", "r"), //STDIN
      1 => array("pipe", "w"), //STDOUT
      2 => array("pipe", "w"), //STDERR
    ),
    $pipes
  );
  if ($process !== false) {
    $stdout = stream_get_contents($pipes[1]);
    $stderr = stream_get_contents($pipes[2]);
    fclose($pipes[1]);
    fclose($pipes[2]);
    proc_close($process);

    return $stdout != '';
  }

  return false;
}

---

你可以使用is_executable来检查它是否可执行，但你需要知道命令的路径，你可以使用which命令来获取它。

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