mysqldump, exec() and escapeshellcmd()

[Toot4fun]

I'm trying to create a PHP script that will use mysqldump to backup my databases.  I have the following code:

 

$strCommand = "mysqldump --host=$strHostName --user=$strUserName --password=$strPassword $strDBName | gzip > $strBackupFile";
$strOutput = exec(escapeshellcmd($strCommand) . " 2>&1",$arrOutput,$intReturnValue);
if($intReturnValue != 0){
	etc...

 

However, I get an error that says "mysqldump: Couldn't find table: "|"".  I have to assume it's because of the escaping (\), but if I run it without the escapeshellcmd(), if I have an error, the error is written directly to the screen, not stored in $arrOutput or $intReturnValue.  If I use escapeshellcmd(), $intReturnValue is populated as is my output array $arrOutput.

 

What gives??  Hopefully I have supplied enough information.

 

Thanks!!

[trq]

I don't belive mysqldump outputs anything usefull to stdout. Nothing that could be piped to gzip anyway. And in fact, if you look at the man page...

 

mysqldump [options] db_name [tables]

 

The optional argument after the db_name tables. Hence, its getting confused and looking for a table called |

 

You'll need to do this in a few steps.

[Toot4fun]

If it wasn't for the fact that

 

$strCommand = "mysqldump --host=$strHostName --user=$strUserName --password=$strPassword $strDBName | gzip > $strBackupFile";
$strOutput = system($strCommand, $intReturnValue);

 

dumps and zips the output, I would agree with you.  But if I run the same command using system() as shown above, I end up with a gzip file.  The problem is that if I force an error (bad password), the error is written directly to the screen - $strOutput is empty and $intReturnValue is equal to 0.