Question

我正在尝试制作一个“锦标赛”应用程序，并且无法查看来自6个不同表格的数据！

我有6个表格，我可以同时放入数据，但我无法一起查看！并试图在互联网上搜索几个小时，但我无法得到它......

无论如何。这是一场4胜的比赛，包括胜利，平局，输球和积分。我有6个表：

比赛名称 - ID，名称
团队 - ID，team1，team2，team3，team4
胜利 - ID，team1w，team2w，team3w，team4w
绘制 ID，team1d，team2d，team3d，team4d
失落 - ID，team1l，team2l，team3l，team4l
积分 - ID，team1p，team2p，team3p，team4p

这就是表格中的内容......

这是我的意见，它运作正常..

<?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。

我从一个教程中得到了这个代码，我曾经在我的网页上进行简单的现场编辑。虽然mabey我可以使用相同的输出代码，但在该网站上它只是来自单个表的新闻。

我已经阅读了一些关于UNION，LEFT JOIN，FULL JOIN等的内容......但是我没有得到它。

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 />";
                    }
                ?>

任何人都知道我会做什么或不做什么？

感谢您的回答！

编辑：我只是得到一个空表的输出！只输出点数，即0。

Answer 1

您需要加入表格。从你的问题来看，似乎你已经对表格的关系部分进行了排序（虽然我并不完全确定我一直都很好地遵循它），所以它实际上是一个关于连接使用什么样的问题的问题：

内部联接 - 这将仅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

依此类推（我假设tournament.ID是其他人所关注的专栏）

现在，您可以将其更改为外部联接，这将返回第一个表中的行，无论其他行中是否存在匹配的行：

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

我不能完全确定我能在这个答案中更好地解释我在之前写过的Q&A中解释过的任何方式。这个答案有意义吗？

查看MySQL中6个表的数据

1 个答案: