dtx6087 2011-12-19 03:40
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错误警告:php中的mysql_fetch_assoc()[重复]

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?

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3条回答 默认 最新

  • doubengman2072 2011-12-19 03:43
    关注

    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;
    }
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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