【发布时间】:2015-12-03 16:06:41
【问题描述】:
我是 php 新手。我有一个问题,我无法按降序获取我的记录。
这是我的代码
<?php
include ("connection.php");
$q_opinion="SELECT r.client_id,c.id,t.id,a.id,o.id,c.name as opinion, r.notification_date, t.title as ttitle,a.title as atitle,o.title as otitle, l.title as ltitle, s.title as stitle, pr.opinion_id, pc.id, pr.client_id as pr_client, pc.address, pc.liaison_one, city.id, pc.head_office_id, city.city, pc.title as cname
FROM og_ratings r
LEFT JOIN og_companies c
ON r.client_id = c.id
LEFT JOIN og_rating_types t
ON r.rating_type_id = t.id
LEFT JOIN og_actions a
ON r.pacra_action = a.id
LEFT JOIN og_outlooks o
ON r.pacra_outlook = o.id
LEFT JOIN og_lterms l
ON r.pacra_lterm = l.id
LEFT JOIN og_sterms s
ON r.pacra_sterm = s.id
LEFT JOIN pacra_client_opinion_relations pr
ON pr.opinion_id = c.id
LEFT JOIN pacra_clients pc
ON pc.id = pr.client_id
LEFT JOIN city
ON city.id = pc.head_office_id
WHERE r.client_id IN (SELECT opinion_id FROM pacra_client_opinion_relations WHERE client_id = 50)
Group By r.client_id
ORDER BY r.client_id DESC
";
$result = mysql_query($q_opinion) or die;
$rating = array();
while($row = mysql_fetch_assoc($result))
{
$id[] = $row['client_id'];
$action[] = $row['atitle'];
$opinion[] = $row['opinion'];
}
for ($i=0; $i<count($rating); $i++)
{
if ($rating[$i] == "")continue;
?>
<table border="1">
<tr>
<td><?= $id[$i] ?> </td>
<td><?= $opinion[$i] ?> </td>
<td><?= $action[$i] ?> </td>
</tr>
</table>
<?php
}
?>
现在我解释一下我的代码和我的问题
我有多个表,我使用左连接加入它们。 首先,我将解释我的子查询。此查询包含多个 id 的结果:
之后我有一张表og_ratings,我在其中记录了id's
在og_ratings 表中,client_id 列用作opinion_id 的外键
我的代码运行良好,但我的降序有问题
当我运行此代码时,我的降序子句仅适用于 id 而不是数据
这里它只适用于我的变量$id 和$opinion 它不适用于$action。我想在$action 上应用降序。
希望你能理解我的问题。请帮帮我。
【问题讨论】:
标签: php mysql join subquery sql-order-by