mysql_fetch_array（）返回'提供的参数不是有效的MySQL结果资源'

Question

I am trying the following code:

<?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);

?> 

And get this error:

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

What am I missing?

Answer 1

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

That would be something like:

<?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);
?>