Error Warning: mysql_fetch_assoc() in php [duplicate] - php

Question

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Closed 10 years ago.
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?

Answer 1

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;
}

Answer 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.

Answer 3

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?