[PHP] Advice with some code

[Steve Finkelstein]

Hi all,

I'm having a brain freeze with some simple code that I wrote and now trying
to refactor.

I have a block of code that looks like this:

    public function backup()
      {
          $fname = "$this->dbName.sql.$this->zip";

          $this->cmd = "mysqldump -Q -u $this->dbUser -p$this->dbPass
$this->dbName 2>&1 > {$this->dest}/{$this->dbName}.{$this->lastId}.sql";



        $res = shell_exec("$this->cmd");
          error_log("First res: ".$res);

            if(!$res) {
                  $this->cmd = ($this->zip==="bz2") ?
                        "bzip2
{$this->dest}/{$this->dbName}.{$this->lastId}.sql 2>&1"  :
                        "gzip
{$this->dest}/{$this->dbName}.{$this->lastId}.sql 2>&1";

                    $res = shell_exec("$this->cmd");
                  error_log("second error: ".$res);
                    return !$res;
            }

            return FALSE;
    }

Now instead of that FALSE, is there a way I can pass FALSE with a particular
error message?  This is because on the other end I have code that looks like
this:

      if($mysqlDump->backup()) {
            $success = array('success' => '1');
            $sqlres = mysql_query($sql, $link) or
die(json_message('error',mysql_error()));
            shell_exec('/usr/bin/touch /tmp/build_transfer');
            mysql_close($link);
            return  '(' . json_encode($success) . ')';
      } else {
            $fail = array('fail' => $res);
            return '(' . json_encode($fail) . ')';
      }

I'd ultimately like to be able to deliver a failure message from the return
value...and properly catch that message so I can send it back in JSON format
to the client browser to report what the error is.

Think I should approach this with some try{..} catch code? Am I overlooking
something really simple? :-)

Thanks for your advice.