【问题标题】:Print Record In Descending ORDER using GROUP BY and ORDER BY Clause使用 GROUP BY 和 ORDER BY 子句以降序打印记录
【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

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

我的代码输出是

这里它只适用于我的变量$id$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


    【解决方案1】:

    是的。我是通过以下代码做到的

    <?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 
        inner join
    (
      select max(notification_date) notification_date,
        client_id
      from og_ratings
      group by client_id
      ORDER BY notification_date DESC
    ) r2
      on r.notification_date = r2.notification_date
      and r.client_id = r2.client_id
    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)";
    $result = mysql_query($q_opinion) or die;
    $rating = array();
    while($row = mysql_fetch_assoc($result))
    {
      $rating[] = $row['client_id'];
      $action[] = $row['atitle'];
      $opinion[] = $row['opinion'];
      $date[] = $row['notification_date'];
      $lrating[] = $row['ltitle'];
      $srating[] = $row['stitle'];
    }
    for ($i=0; $i<count($rating); $i++) {
        if ($rating[$i] == "")continue;
         ?>
        <table border="1">
        <tr>
              <td><?= $rating[$i] ?> </td>
               <td><?= $date[$i] ?> </td>
              <td><?= $opinion[$i] ?> </td>
             <td><?= $action[$i] ?> </td>
              <td><?= $lrating[$i] ?> </td>
               <td><?= $srating[$i] ?> </td>
        </tr>
        </table>
    <?php   
    }
    ?>
    

    【讨论】:

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