PHP: connected to database, but mysql_query fails

Question

I have very strange problem with PHP which I am starting to learn .. I have created tables in MySQL database with some data, and now I want to show them in webpage.

This is my source where I have this problem:

<?php
    // Here I open connection
    $con = mysql_connect("localhost","root","aaaaaa");
    // set the mysql database
    $db = mysql_select_db("infs", $con);

    // I check the connection
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    else {
        // It always goes here
        echo "Connected to database!";
    }

    // I am testing very simple SQL query.. there should be no problem
    $result = mysql_query("SELECT * FROM cathegories", $con, $db);

    if (!$result) {
        // but it always dies
        $message  = 'Invalid query: ' . mysql_error() . "\n";
        $message .= 'Whole query: ' . $query;
        die($message);
    }

    mysql_close($con);
?>

What is wrong?

Thanks in advance!

Simply use $result = mysql_query("SELECT * FROM cathegories"); — Shankar Narayana Damodaran
– Shankar Narayana Damodaran, Commented Apr 20, 2014 at 14:50
You are mixing mysql and mysqli library calls! Don't do this! The former is deprecated. Use mysqli only. — Jonathon Reinhart
– Jonathon Reinhart, Commented Apr 20, 2014 at 14:56
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the pink box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — h2ooooooo
– h2ooooooo, Commented Apr 20, 2014 at 15:04

el3ien · Accepted Answer · 2014-04-20 15:03:42Z

1

You are mixing mysql and mysqli. Try something like:

<?php   
    $con= new mysqli("localhost","user","passwd","database");
    if ($con->connect_errno){
      echo "could not connect";
    }

    $select = "SELECT * FROM tablename";

    if($result = $con->query($select)){
        while($row = $result->fetch_object()){
            echo $row->rowname."<br>";
        }
    }
    else { echo 'no result'; }
    $con->close();  
?>

answered Apr 20, 2014 at 15:03

el3ien

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Comments

user3241507 · Accepted Answer · 2014-04-20 14:52:43Z

1

// Here I open connection
$con = mysql_connect("localhost","root","aaaaaa");
// set the mysql database
$db = mysql_select_db("infs", $connection);

change to

// Here I open connection
$con = mysql_connect("localhost","root","aaaaaa");
// set the mysql database
$db = mysql_select_db("infs", $con);

answered Apr 20, 2014 at 14:52

user3241507

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1 Comment

Mike M. Over a year ago

the main issue from you was you used $connection to connect while you never defined it but did define $con as connector whats the error you get now?

Ashley · Accepted Answer · 2014-04-20 14:51:29Z

0

mysql_query only takes two parameters - the actual SQL and then the link identifier (I assume in your case that's stored in $con; therefore remove $db from the third parameter).

You don't even need the second $con parameter really.

Where's the actual logic to connect to the database initially? Just because mysqli_connect_errno() doesn't return an error it doesn't mean the connection actually exists and that $con is available in the current scope.

I'd var_dump($con) before the mysql query to make sure it's a valid connection.

answered Apr 20, 2014 at 14:51

Ashley

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1 Comment

Firzen Over a year ago

I have tried to check connection with var_dump($con), and it prints: "resource(2) of type (mysql link) ". This should be right as I think..

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