在MySQL中对MySQL进行排序...? [英] Sorting MySQL results in PHP...?

问题描述

我的网站上有一个页面，上面列出了一堆产品，每个产品都有用户评分.我使用一个查询为每个产品提取一些数据点(详细信息查询")，并使用第二个查询返回每个产品的平均用户评分(评分查询").

I have a page on my site that lists a bunch of products, each with a user rating. I use one query to pull some data points for each product ("details query"), and a second query that returns the average user rating for each product ("ratings query").

我想将每个产品的用户评分附加到详细信息查询"结果集中，然后按评分降序排序.我已经在Stack Overflow，php.net等上阅读了很多条目，我认为我需要使用带有自定义函数的usort()，但是每次我将MySQL结果传递给usort()时，我都会得到一个php错误，说我传递给usort()的对象不是数组.例如，我尝试过:

I want to append the user rating for each product onto the "details query" result set, then sort by rating in descending order. I've read a bunch of entries on Stack Overflow, php.net etc. and I think I need to use usort() with a custom function, but every time I pass my MySQL result to usort() I get a php error saying the object I'm passing to usort() isn't an array. For example, I've tried:

$data = mysql_fetch_array($details_query);
usort($data,"compare");

执行上述操作将引发错误，指出$ data不是数组.我究竟做错了什么?

Doing the above will throw an error saying $data isn't an array. What am I doing wrong?

此外，如果有人对如何完成此操作有其他建议，我将非常感谢.由于某种原因，我在这段时间过得很艰难...

Also, if anyone has any other suggestions on how to get this done I'd really appreciate. I'm having a really tough time with this for some reason...

谢谢！

推荐答案

您误用了MySQL函数-mysql_fetch_array不接受SQL查询，而是接受MySQL结果资源:

You're mis-using the MySQL functions - mysql_fetch_array doesn't take a SQL query, it takes a MySQL result resource:

http://php.net/manual/en/function. mysql-fetch-array.php

所以您想要类似的东西:

So you want something like:

$all_data = array();

$sql = "SELECT blah FROM blah";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
    $all_data[] = $row;
}

// $all_data is now a 2-D array with all your data.
usort($all_data, "compare");

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