Question

<?php

    function get_user_id($username) {
        return mysql_result(mysql_query("Select id From users Where username = '" . mysql_real_escape_string($username) . "'"), 0);
    }

    $sql = "select * from rating 
    WHERE user_id=" . get_user_id($myusername) . "
            ORDER BY punkte ASC";
    $query = mysql_query($sql);
    while ($row = mysql_fetch_array($query)) {
        $catid = $row['rating_id'];
        $catname = $row['song_id'];
        echo "<li id='item_$catid' class='ui-state-default'><span class='ui-icon ui-icon-arrowthick-2-n-s'></span>$catname</li>";
    }
    ?>

更新

抱歉，我发现了这个错误，非常愚蠢：

    $catid = $row['rating_id'];
    $catname = $row['song_id'];

应该是：

$catid = $row['song_id'];
$catname = $row['song_name'];

所以，谢谢大家！一如既往：在向Stackoverflow发布问题之前，你无法弄明白：）

Answer 1

对于您的新问题，您可以使用内部联接来解决它：

select s.song_name from rating r
inner join songs s on s.song_name_id = r.song_id

Answer 2

内部联接

http://www.w3schools.com/sql/sql_join_inner.asp

Answer 3

是，

$ catid = $ row ['rating_id']; //评级ID与歌曲无关

要

$ catid = $ row ['song_id']; $ catname = $ row ['song_name'];

SQL语句仅更新一个用户的数据

3 个答案: