Re: coding MySQL error traps [WAS: mysql if empty]

Paul Novitski wrote:

$sql = "SELECT Client FROM booked WHERE Name = 'larry'"; $result = mysql_query($sql) OR die('[MYSQL ERROR] - ['.mysql_errno().']<br />'.mysql_error()); while ( list($client) = mysql_fetch_row($result) ) {

        echo "{$client}<br />\n";

}

I agree with this logic overall. The above is of course just oversimplified demo code, but I'd like to mention a few details I would change before putting this into practice. The syntax "$result = mysql_query($sql) OR die();" is nicely compact but obfuscates the program logic. PHP is not actually ORing two values, it's halting execution halfway through the statement if mysql_query() returns true. This constitutes a hack because it depends entirely on the way the parser processes code rather than on explicit elements of the language. To help keep legacy code from crashing with PHP version X I'd break this into two statements: run the query, then act on the result. The parser won't care, and your code will be more easily readable by us humans. Displaying mysql_error() is great for the developer but in a public application will expose the names of tables and fields to the public who shouldn't really see your wires and pipes. mysql_error() doesn't usually contain the entire query and doesn't always contain the segment of the query that actually caused the error, so I always add the full query to the error message to save time in debugging. Finally, die()ing on a mysql error is pretty harsh; a friendlier application would return the error state to parent layers, translate the error into advice a non-technical user can deal with, and display it in a way that doesn't crash the page. Rather than trying to remember to go through the code later fixing these bits, I suggest adopting a convention early on that suits both development and publication, such as: _______________________________ define('bDebug', true); ... $bResult = mysql_query($sql);

        if (!$bResult) return ReportSQLError('checking user name', mysql_errno(), mysql_error(), $sql);

... function ReportSQLError($context, $errno, $errorMsg, $sql) {

        if (bDebug)
        {
                die("MYSQL ERROR $errno $context:<hr />\n$errorMsg<hr />\n$sql");
        }
        else
        {
                return $generate_friendly_error_message;
        }

} _______________________________ Regards, Paul __________________________ Paul Novitski Juniper Webcraft Ltd. http://juniperwebcraft.com

Yes, this was intended to only show the OP a very simple way to get done what they were wanting to do. -- Enjoy, Jim Lucas Different eyes see different things. Different hearts beat on different strings. But there are times for you and me when all such things agree. - Rush