Question

我有这两个表，我想从第二个表中获取第一个表中两个外键的值。

团队表：

ID   | Name          | No_Players | Country  | F_Year | Team_Logo
-----+---------------+------------+----------+--------+----------------------
1    | Real Madrid   |  22        | Spain    | 1940   | Http://Xyz.jpg  
2    | Arsenal       |  22        | England  | 1950   | Http://XXYYZZ.jpg  
3    | FCB.Barcelona |  22        | Spain    | 1960   | Http://YYZZz.jpg  
4    | Bayern Meunekh|  22        | Germany  | 1940   | Http://GGG.jpg

匹配表格：

ID | Date        |First_Team     | Second_Team | M_Country |M_City   | M_Period | Result
------------------------------------------------------------------------------------------
1  |15-02-2016   | Real Madrid   | Arsenal     | Spain     |  Madrid | 120      | 1-1
2  |19-03-2016   | FCB.Barecolna | Madrid      | Spain     |  Madrid | 90       | 4-5

我想获取Team_Logo对于PHP脚本中json表单中匹配表中的每个团队。我试图获得团队徽标，但不幸的是查询无效。

有人可以帮我吗？

<?php
require ('config.php');
$conn = mysqli_connect($servername, $username, $password, $db);
$query = "select * from matches,team where matches.first_team=team.Name OR matches.second_team=team.Name";
$result = mysqli_query($conn, $query);
$rows = array();
echo mysqli_error($conn);

while($row = mysqli_fetch_assoc($result)) {
    $rows[]=$row;
}

echo json_encode($rows);
?>

Answer 1

SELECT
    matches.*,
    t1.Team_Logo AS logo1,
    t2.Team_Logo AS logo2
FROM matches
JOIN team AS t1
    ON t1.Name = matches.First_Team
JOIN team As t2
    ON t2.Name = matches.Second_Team

另请注意，您不应选择*，因为团队表中的Id列将覆盖匹配表的Id列。这就是为什么您应该只包含您想要选择的列。

Answer 2

您应该尝试此查询：

$query="select m.*,t.* from matches as m INNER JOIN on team as t1 ON m.First_Team=t1.Name JOIN team As t2
ON m.Second_Team=t2.Name";

Answer 3

使用关系数据库结构。

为两个表添加主键

将第二个表的主键的引用键作为外键添加到第二个表中。

并参考以下查询。

sql="SELECT * FROM Matches as m JOIN Team as t ON t.ID = m.t_id where t.id=1 OR t.id=2

从外键表中检索字段

3 个答案: