简体   繁体   English

在同一张桌子上的两个联接

[英]Two JOINS on same table

I'm working on a website which shows the soccer score with two teams and two scores in my DB. 我在一个网站上工作,该网站在我的数据库中显示了两支球队的足球得分和两分。

This is the code I have: 这是我的代码:

<?php 
        $result = mysql_query( "SELECT * FROM resultat 
        LEFT JOIN brukere ON resultat.dommer = brukere.id
        INNER JOIN lag ON resultat.lag1 = lag.id
            INNER JOIN lag ON resultat.lag2 = lag.id
        ORDER BY slutt DESC
        " )
        or die("SELECT Error: ".mysql_error());
        $num_rows = mysql_num_rows($result);if ($result) {while ($row = mysql_fetch_array($result)) {
        $_SESSION['id']=$row['id'];
        ?>
        <tr>
                <td><?php echo $row['klubb']; ?></td>
                <td><?php echo $row['klubb']; ?></td>
                <td><?php echo $row['stilling1'] . " - " . $row['stilling2']; ?></td>
                <td><?php echo $row['roed'] . " rødt og " . $row['gult'] . " gult"; ?></td>
                <td><?php echo $row['bane']; ?></td>
                <td><?php echo $row['navn']; ?></td>
            </tr>
        <?php
        }
        }
        ?>

And I get: 我得到:

SELECT Error: Not unique table/alias: 'lag' 选择错误:不是唯一的表/别名:'滞后'

When I try: 当我尝试:

INNER JOIN lag s ON resultat.lag1 = lag.id
INNER JOIN lag c ON resultat.lag2 = lag.id

I get: 我得到:

SELECT Error: Unknown column 'lag.id' in 'on clause' SELECT错误:“ on子句”中的未知列“ lag.id”

How do I do it? 我该怎么做? Two joins from the same table 来自同一表的两个联接

You need to use the alias you just gave to your table lag . 您需要使用刚赋予表lag的别名。 You assigned s at first join and c at second so you should use them 您在第一次加入时分配了s ,在第二次加入时分配了c ,因此您应该使用它们

INNER JOIN lag s ON resultat.lag1 = s.id
INNER JOIN lag c ON resultat.lag2 = c.id

You can try to combine JOIN like this: 您可以尝试将JOIN这样合并:

INNER JOIN lag ON resultat.lag1 = lag.id OR resultat.lag2 = lag.id

Replace OR with AND if you need 如果需要,请用AND替换OR

This is a self-join and you need the alias in your expession: 这是一个自联接,并且在expession中需要别名:

     INNER JOIN lag s ON resultat.lag1 = s.id
     INNER JOIN lag c ON resultat.lag2 = c.id 

In general you use self-joins with adjacency-list and to walk the tree down a dimension: 通常,您将自连接与邻接列表一起使用,并将树沿维度移动:

    SELECT t1.name AS lev1, t2.name as lev2, t3.name as lev3, t4.name as lev4
    FROM category AS t1
    LEFT JOIN category AS t2 ON t2.parent = t1.category_id
    LEFT JOIN category AS t3 ON t3.parent = t2.category_id
    LEFT JOIN category AS t4 ON t4.parent = t3.category_id
    WHERE t1.parent is NULL 

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM