Error Warning: mysql_fetch_assoc() in php [duplicate]

Question

Possible Duplicate:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

In db.php I have:

<?php
class connect {

    private $host = "localhost";
    private $user = "root";
    private $pass = "";
    private $database = "databasename";
    private $connect = null;

    function connect() {
        $this->connect = mysql_connect($this->host, $this->user, $this->pass) or die("Can't connect database");
        mysql_select_db($this->database, $this->connect);
    }

    function getData() {
        $data = array();
        $sql = 'Select * From test';
        $query = mysql_query($sql);
        while($row = mysql_fetch_assoc($query)) {
            $data[] = array($row['id'], $row['name']);
        }
        return $data;
    }

}
?>

In index.php I have:

<?php
include 'db.php';
$connect = new connect();
$connect->connect();
$data = $connect->getData();
$str = '';
foreach ($data as $dt) {
    $str .= $dt[1];
}
echo $str;
?>

I am getting the following error: => error: <b>Warning</b>: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource from db.php.

What am I doing wrong?

I tried to copy paste your code, No Error Found ! Please check your database name, database password, database user or table name. — Zul
– Zul, Commented Dec 19, 2011 at 3:57
As a rule of thumb, you shouldn't name a local variable the same as your class: 'connect'. Change 'private $connect' to (for instance) 'private $connection'. Then you should use that variable in your query: $query = mysql_query($sql, $this->connection); — dar7yl
– dar7yl, Commented Dec 19, 2011 at 11:40

check123 · Accepted Answer · 2011-12-19 03:51:18Z

6

Try to find what is the error:

  function getData() {
    $data = array();
    $sql = 'Select * From test';
    $query = mysql_query($sql);
    if(!$query) 
    {
     echo 'Error: ' . mysql_error(); /* Check what is the error and print it */
     exit;
    }

    while($row = mysql_fetch_array($query)) {  /* Better use fetch array instead */
        $data[] = array($row['id'], $row['name']);
    }
    return $data;
}

edited Dec 19, 2011 at 3:51

answered Dec 19, 2011 at 3:43

check123

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2 Comments

spatel Over a year ago

Don't know why this was downvoted. Something is wrong with the query or else it would return a result resource object. Only way to find out is to display the error message.

Jessedc Over a year ago

I bet the error message is one of the two things I mentioned :P

Jessedc · Accepted Answer · 2011-12-19 03:55:40Z

2

That error is telling you that your query executed by $query = mysql_query($sql); is returning an error. It's not returning zero results, it's returning an error which suggests that your database named 'databasename' or table within that named 'test' doesn't exist.

answered Dec 19, 2011 at 3:55

Jessedc

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Comments

Dan · Accepted Answer · 2011-12-19 03:56:34Z

0

It sounds like no results are returned from the query or a general query error, do the columns and table in the query exist and is your connection to the database all okay?

answered Dec 19, 2011 at 3:56

Dan

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