How can I debug why simplest MySQL query returns false?

Question

I work with xampp. I performed MySQL connection:

$connection = mysql_connect($host , $user , $passw);
mysql_select_db($db, $connection);

I received output with echo command (by check the boolean returned values) that connection is established and the database $db is found.

But the simplest query like:

$query = mysql_query("SELECT * FROM 'users'");

returns false. How can I debug why?Thanks.

Did u try to run query in phpmyadmin or database you are using ? — Milap
– Milap, Commented Mar 8, 2012 at 14:58
@ashes999 Thanks for an idea. I tried it right now and still I receives return $query = false. — tatiana_c
– tatiana_c, Commented Mar 8, 2012 at 14:59
@Marcus Adams The "SELECT 1" also returns false. But I tried now also with `` , still $query = false. — tatiana_c
– tatiana_c, Commented Mar 8, 2012 at 15:04

Your Common Sense · Accepted Answer · 2019-04-24 13:13:00Z

18

An obligatory update: as mysql ext is no more, here are answers for two remaining MySQL APIs which I written on my site based on the experience from answering 1000s questions on Stack Overflow:

How to report errors in mysqli
How to connect to MySQL using PDO (with the aim of the proper error reporting).

In short, for mysqi the following line have to be added before mysqli_connect() call:

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

while for PDO the proper error mode have to be set, for example

$pdo->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );

As of the old mysql ext,

To get an error from mysql_query() you have to use mysql_error() function.
So always run all your queries this way, at least until you develop a more advanced query handler:

$query = "SELECT * FROM 'users'";
$result = mysql_query($query) or trigger_error(mysql_error()." ".$query);

the problem with your current query is 'users' part. Single quotes have to be used to delimit strings while for the identifiers you have to use backticks:

SELECT * FROM `users`

In order to see these errors during development, add these lines at the top of your code to be sure you can see every error occurred

ini_set('display_errors',1);
error_reporting(E_ALL);

on the production server, however, the value on the first line should be changed from 1 to 0

edited Apr 24, 2019 at 13:13

answered Mar 8, 2012 at 14:58

Your Common Sense

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3 Comments

tatiana_c Over a year ago

It's great. It gives me the errors. I see for now that database is not selected. This is exactly command that I looked for. Thanks!

knt5784 Over a year ago

mysql_* functions are removed in PHP 7. Do you have an alternative?

Your Common Sense Over a year ago

@knt5784 thank you, I added alternatives for PDO and mysqli

Ynhockey · Accepted Answer · 2012-03-08 15:06:17Z

4

Use the mysql_error() function:

$query = mysql_query("SELECT * FROM 'users'") or die(mysql_error());

EDIT: Per Col. Shrapnel's comment: you should never use die() outside of a test environment. In general it's bad practice when writing code that's even intended for production.

Here is some more information: http://www.phpfreaks.com/blog/or-die-must-die

edited Mar 8, 2012 at 15:06

answered Mar 8, 2012 at 14:58

Ynhockey

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9 Comments

Your Common Sense Over a year ago

Never use die(). That's extremely bad practice.

Ynhockey Over a year ago

In a production environment I would completely agree, but this seems to clear be a test environment. Still, thanks for the comment and I'll clarify it in the answer.

Your Common Sense Over a year ago

you are using different codes on the development and production? Really you are?

Marcus Adams Over a year ago

In case you're wondering why, die(mysql_error()) shows the MySQL error to the end user, which, besides being user unfriendly, is also a potential security risk.

Your Common Sense Over a year ago

Anyway, how it is supposed to track errors on the production?

|

Community · Accepted Answer · 2017-05-23 11:54:47Z

1

Based on Your Common Sense answer this is an object oriented style:

$query = "SELECT * FROM 'users'";
$result = $mysqli -> query($query) or trigger_error($mysqli -> error." ".$query);

edited May 23, 2017 at 11:54

CommunityBot

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answered Dec 12, 2015 at 13:05

mrroot5

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