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php - 一个表而不是另一个表中的 MySQL PHP ID

转载 作者:行者123 更新时间:2023-11-30 22:00:27 24 4
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| Fixture_ID | League_ID | Home_Team | Away_Team 
| 1 | 1 | 1 | 2
| 2 | 1 | 2 | 3
| 3 | 1 | 3 | 1

| Result_ID | Fixture_ID | Home_Goals | Away_Goals 
| 1 | 1 | 2 | 0

| Team_ID | Team_Name   |
| 1 | Team A
| 2 | Team B
| 3 | Team C

如何加入表格以仅显示尚未输入结果但在显示赛程(在下拉列表中)时输出实际团队名称(A 队对 B 队)的赛程?

以下代码适用于输出所有灯具:

echo '<td> <select name ="fixture_id">';    

// TRY TO SHOW FIXTURES WITH NO RESULTS
$stmt = $pdo->prepare('SELECT f.*, t1.Team_Name AS Home, t2.Team_Name AS Away
FROM Fixture f
INNER JOIN Team t1 ON f.Home_Team = t1.Team_ID
INNER JOIN Team t2 ON f.Away_Team = t2.Team_ID');


$stmt->execute();
foreach ($stmt as $row) {
echo '<option>' . $row['Home'] . ' v ' . $row['Away'] . '</option>';
}

?>

最佳答案

您的 SQL 应该如下所示:

SELECT  f.*, t1.Team_Name AS Home, t2.Team_Name AS Away
FROM Fixture f
INNER JOIN Team t1 ON f.Home_Team = t1.Team_ID
INNER JOIN Team t2 ON f.Away_Team = t2.Team_ID
LEFT JOIN Result r ON f.Fixture_ID = r.Fixture_ID
WHERE r.id IS NULL;

关于php - 一个表而不是另一个表中的 MySQL PHP ID,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43588684/

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