php - 查看MySQL中6个表的数据

Question

我正在尝试制作一个“锦标赛”应用程序，但在查看来自 6 个不同表格的数据时卡住了!

我有6个tabel可以同时放入数据，但是不能一起查看!并尝试在互联网上搜索几个小时，但我就是找不到...

任何方式。这是一个 4 队比赛，有胜、平、负和积分。我有 6 个表:

• 锦标赛名称 - 身份证、姓名
• 团队 - ID, team1, team2, team3, team4
• 获胜 - ID, team1w, team2w, team3w, team4w
• 画画 ID, team1d, team2d, team3d, team4d
• 松动 - ID, team1l, team2l, team3l, team4l
• 积分 - ID, team1p, team2p, team3p, team4p

这就是表格中的内容...

这是我的输入，它工作正常..

<?php
                if(isset($_POST['submit'])) {
                    $cupname = $_POST['cupname'];
                    $team1 = $_POST['team1'];
                    $team2 = $_POST['team2'];
                    $team3 = $_POST['team3'];
                    $team4 = $_POST['team4'];

                    $zero = "0";

                    $result = mysql_query("INSERT INTO cupname (name) VALUES ('$cupname')");
                    $result = mysql_query("INSERT INTO teams (team1, team2, team3, team4) VALUES ('$team1', '$team2', '$team3', '$team4')");
                    $result = mysql_query("INSERT INTO wins (team1w, team2w, team3w, team4w) VALUES ($zero, $zero, $zero, $zero)");
                    $result = mysql_query("INSERT INTO draws (team1d, team2d, team3d, team4d) VALUES ($zero, $zero, $zero, $zero)");
                    $result = mysql_query("INSERT INTO looses (team1l, team2l, team3l, team4l) VALUES ($zero, $zero, $zero, $zero)");
                    $result = mysql_query("INSERT INTO points (team1p, team2p, team3p, team4p) VALUES ($zero, $zero, $zero, $zero)");
                echo "<h1>Turnering og lag lagt til!</h1>";
                }

                ?>

这是制作“联赛/锦标赛”的开始。所有数据都在表格中。

我从一个教程中获得了这段代码，我曾经对我的网页进行简单的现场编辑。虽然也许我可以使用相同的输出代码，但在该站点上它只是来自单个表的新闻。

我读过一些关于 UNION、LEFT JOIN、FULL JOIN 等的内容……但我没有理解。

这是我的“输出”代码:

<?php
                $result = mysql_query("SELECT * FROM cupname ORDER BY id DESC");
                $result = mysql_query("SELECT * FROM teams ORDER BY id DESC");
                $result = mysql_query("SELECT * FROM wins ORDER BY id DESC");
                $result = mysql_query("SELECT * FROM draws ORDER BY id DESC");
                $result = mysql_query("SELECT * FROM looses ORDER BY id DESC");
                $result = mysql_query("SELECT * FROM points ORDER BY id DESC");
                    while($row = mysql_fetch_array($result)) {
                        $i=$i + 1;
                        echo "<table>";
                        echo "<tr>";
                        echo $row['name'];
                        echo "</tr>";
                        echo "<tr><td><b>Lag</b></td>
                        <td><b>Seiere</b></td>
                        <td><b>Uavgjort</b></td>
                        <td><b>Tap</b></td>
                        <td><b>Poeng</b></td>
                        </tr><tr><td>";
                        echo $row['team1'];
                        echo "</td><td>";
                        echo $row['team1w']; 
                        echo "</td><td>"; 
                        echo  $row['team1d']; 
                        echo "</td><td>"; 
                        echo  $row['team1l']; 
                        echo "</td><td>"; 
                        echo  $row['team1p']; 
                        echo "</td></tr><tr><td>"; 
                        echo  $row['team2']; 
                        echo  "</td><td>"; 
                        echo  $row['team2w']; 
                        echo "</td><td>"; 
                        echo  $row['team2d']; 
                        echo "</td><td>"; 
                        echo  $row['team2l']; 
                        echo "</td><td>"; 
                        echo  $row['team2p']; 
                        echo "</td></tr><tr><td>"; 
                        echo  $row['team3']; 
                        echo  "</td><td>"; 
                        echo  $row['team3w']; 
                        echo "</td><td>"; 
                        echo  $row['team3d']; 
                        echo "</td><td>"; 
                        echo  $row['team3l']; 
                        echo "</td><td>";
                        echo  $row['team3p']; 
                        echo "</td></tr><tr><td>"; 
                        echo  $row['team4']; 
                        echo  "</td><td>"; 
                        echo  $row['team4w']; 
                        echo "</td><td>"; 
                        echo  $row['team4d']; 
                        echo "</td><td>"; 
                        echo  $row['team4l']; 
                        echo "</td><td>"; 
                        echo  $row['team1p']; 
                        echo "</td></tr></table><br /><hr /><br />";
                    }
                ?>

有人知道我应该做什么或不做什么吗？

感谢您的回答!

编辑:我只是得到空表的输出!只输出点，即 0。

Answer 1

您需要加入表格。从你的问题来看，你似乎已经对表的关系部分进行了排序(尽管我不完全确定我是否完全遵循它)，所以这实际上是一个使用什么连接的问题:

内部联接 - 这将只返回所有表中数据匹配的记录:

select
    tourn.id,
    tourn.name
from
    tournament
        join teams team
            on tourn.id=team.id
        join wins win
            on tourn.id=win.id

等等(我假设 tournament.ID 是其他相关的列)

现在，您可以将其更改为外连接，无论其他表中是否存在匹配的行，它都会从您的第一个表中返回行:

select
    tourn.id,
    tourn.name
from
    tournament
        right outer join teams team
            on tourn.id=team.id
        right outer join wins win
            on tourn.id=win.id

我不完全确定我是否可以用我在 Q&A 中解释的任何方式更好地解释此答案中的很多内容我之前写过的。这个答案有意义吗？