访问存储在此查询返回的结果中的数据 [英] Access data stored in the result returned from this query

问题描述

我有以下查询:

$sql = "SET @rownum := 0;
              SELECT * FROM (
                SELECT @rownum := @rownum + 1 AS rank, totalpoints, useridFB, username
                FROM user_test ORDER BY totalpoints DESC
                ) as result WHERE useridFB=".$uid."";   

通过它，我获得了通过在我的网络上玩一些游戏而获得积分的用户的排名.

With it I'm getting the rank of a user that has earned points by playing some games in my web.

当我将查询粘贴到 phpMyAdmin 时，查询效果很好，我得到了用户的正确排名.

The query works great when I paste it in phpMyAdmin, I get the correct rank for the user.

那有什么问题呢?好吧，我无法显示存储在结果中的任何数据.

What's the problem then? Well, I cannot display any data stored in the result.

我已经试过了:

$result = mysql_query($sql); 
while ($row = mysql_fetch_assoc($result)) {
    echo $row['totalpoints'];
    echo $row['rank'];
    echo $row['useridFB'];
    echo $row['username'];
} 

但它返回一个错误:

警告:mysql_fetch_assoc():提供的参数不是有效的 MySQL 结果资源...

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in ...

没有打印结果.

我做错了什么?非常感谢！

What am I doing wrong? Thanks a lot!

推荐答案

您没有检查 mysql_query() 的返回值.最有可能的问题是您通常无法通过 mysql_query() - "SET @rownum := 0;" 运行多个查询作为第一个查询，但在 mysql_query() 失败后快速检查 mysql_error() 将是一个好主意.

You're not checking the return value of mysql_query(). Most likely, the problem is that you cannot normally run multiple queries via mysql_query() - "SET @rownum := 0;" being the first query, but a quick check with mysql_error() after the failed mysql_query() would be a good idea.

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