mysql_fetch_array()返回“提供的参数不是有效的MySQL结果资源" [英] mysql_fetch_array() returns 'supplied argument is not a valid MySQL result resource'

问题描述

我正在尝试以下代码:

<?php

    $link = mysql_connect('localhost', 'root', 'geheim');
    if (!$link) {
        die('Could not connect: ' . mysql_error());
    }
    echo 'Connected successfully';



    $query = "SELECT * FROM Auctions";
    $result = mysql_query($query);

    while($row = mysql_fetch_array($result, MYSQL_ASSOC))
    {
        foreach($row as $field=>$value)
        {
            echo "$field: {$value} <br />";
        }
    }
    mysql_close($link);

?> 

并得到此错误:

Warning: mysql_fetch_array(): supplied argument is not a
    valid MySQL result resource in
    C:\Programme\EasyPHP 2.0b1\www\test.php on line 14

我想念什么?

推荐答案

您尚未选择数据库-请使用 mysql_select_db()

You haven't selected a database - use mysql_select_db()

那将是这样的:

<?php
    $link = mysql_connect('localhost', 'root', 'geheim');
    if (!$link) {
        die('Could not connect: ' . mysql_error());
    }
    echo 'Connected successfully';

    $db_selected = mysql_select_db('foo', $link);
    if (!$db_selected) {
        die ('Error selecting database: '. mysql_error());
    }
    echo 'Using database successfully';

    $query = "SELECT * FROM Auctions";
    $result = mysql_query($query);
    while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
        foreach($row as $field=>$value) {
            echo "$field: {$value} <br />";
        }
    }
    mysql_close($link);
?> 

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