PHP and MySQL: Simple query fails

Question

I have a lot of sql query on my php code, and all of them work well except this one.

$sql_pay = "SELECT * FROM transaccion_cobro WHERE codigo='A2131231' AND idEstadoTransaccion=2";
$query_pay=mysql_query($sql_pay,$db);
$num_pay = mysql_num_rows($query_pay);

This is the error I get:

mysql_num_rows() expects parameter 1 to be resource, boolean given

I guess that mysql_query fails and returns false. But I don't know why. If I execute that query on "MySQL Query Browser" it returns the expected values. Also, the rest of the queries are correctly executed.

Does anyone have any idea of what could be happening here?

UPDATE:

Problem solved thanks to @ICanHasCheezburger comment

`mysql_query($sql_pay,$db) or die(mysql_error());` put this code for display errors for fixing.... — Mahmoud.Eskandari, Feb 18 '14 at 08:05
@M.Eskandari 'die()' is not for fixing, it is used for debugging the query :). — Benz, Feb 18 '14 at 08:06
@Benz . I known This. I say `show errors`>>`check errors for fix` . :| — Mahmoud.Eskandari, Feb 18 '14 at 08:07
@ICanHasCheezburger The problem was that the user had no permission for that specific table. Thanks to your comment I Could find the solution. Thanks. — Rumpelstinsk, Feb 18 '14 at 08:24

score 0 · Answer 1 · edited May 23 '17 at 12:24

Try this

$sql_pay = "SELECT * FROM transaccion_cobro WHERE codigo='A2131231' AND idEstadoTransaccion=2";
$query_pay=mysql_query($sql_pay,$db);
if ( $query_pay ) {
    $num_pay = mysql_num_rows($query_pay);
} else {
    echo "Query Failed: " . mysql_error($db);
}

Warning

This extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used. See also MySQL: choosing an API guide and related FAQ for more information. Alternatives to this function include:

Have a look at this Why shouldn't I use mysql_* functions in PHP?

score -2 · Answer 2 · answered Feb 18 '14 at 08:07

$query_pay=mysql_query($sql_pay,$db); Make it $query_pay=mysqli_query($db, $sql_pay); and $num_pay = mysql_num_rows($query_pay); Make it $num_pay = mysqli_num_rows($query_pay);

Then it should work because mysqli_query necessarily takes $db or connection parameter, for mysql_query it is optional.

PHP and MySQL: Simple query fails

2 Answers2