Question

学习如何为我的体育联盟网站加入MySQL表格。这似乎是一个相对简单的问题。如果有人能帮助澄清我的思考过程，那将非常感激：

我有一个联赛计划页面，可以为“test_league”MySQL表格中的每个游戏生成带有游戏信息的HTML表格行。其中一个HTML列旨在通过提取主队的位置数据来显示该游戏的位置，该位置数据存储在同一数据库的“user_schools”表中。

我确切地知道自己想做什么，我似乎无法正确理解语法。这是代码：

<?php

// Connect to the database:
require ('../mysqli_connect.php');

// Make the query for games from the schedule database:
$q = "SELECT game_date, game_time, away_team, home_team, home_score, away_score, arbiter_id FROM test_league";
$r = mysqli_query($db, $q);

// Make the query to determine where the games will be played:
$location = mysqli_query($db, "SELECT football_location FROM user_schools WHERE home_team=school_name INNER JOIN test_league");
$map = mysqli_query($db, "SELECT football_map FROM user_schools WHERE home_team=school_name INNER JOIN test_league");

// Declare two variables to help determine the background color of each row:
$i = 0;
$bgcolor = array('row1', 'row2');

// Begin function to print each game as a row:
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {

    echo '<tr class="' . $bgcolor[$i++ % 2] .'"><td>' . $row['game_date'] . '</td><td>' . $row['game_time'] . '</td><td class="alignleft"><a href="">' . $row['away_team'] . '</a> vs<br><a href="">' . $row['home_team'] . '</a></td>';

    // Determine if the score has been reported:
    if ($row['home_score'] == '0' && $row['away_score'] == '0') {
        echo '<td><a href=""><img src="images/report_icon.png" alt="Report Score" /></a></td>';
    } else {
        echo '<td>' . $row['home_score'] . '<br>' . $row['away_score'] . '</td>';
    }

    echo '<td><a href="' . $map . '">' . $location . '</a></td><td><a href="">' . $row['arbiter_id'] . '</a></td></tr>';

}

mysqli_free_result ($r);

mysqli_close($db);

?>

如果需要澄清，“home_team”位于“test_league”，“team_name”位于“user_schools”。

我已经看过使用表连接的维恩图插图，但似乎无法执行这个看似简单的任务。非常欢迎您的建议！

Answer 1

实际上，我没有执行此源代码，而是查看SQL。

某事似乎是错的。试试这个查询。

SELECT football_location 
FROM user_schools A
INNER JOIN test_league B
WHERE A.home_team=A.school_name 
;

和

SELECT football_map 
FROM user_schools A
INNER JOIN test_league B
WHERE A.home_team=A.school_name 
;

JOIN之后必须放置的位置。并使用别名A，B来澄清。

Answer 2

select us.football_location from user_schools us left join test_league tl on us.school_name = tl.home_team

它将为您提供所有位置。

如果你想在where子句中使用任何特定的put条件，例如。

select us.football_location from user_schools us left join test_league tl on us.school_name = tl.home_team where us.id = 'id here'

Answer 3

您在代码中执行了三个单独的SELECT，这取消了对JOIN的使用。您拥有代码的方式可以在不进行JOIN的情况下获得必要的信息，只需使用WHERE即可。 JOIN允许您在一次调用中从多个相关表中选择列。使用JOIN在一个查询中获取所有信息的查询将如下所示：

SELECT tl.game_date, tl.game_time, tl.away_team, tl.home_team, tl.home_score,tl.away_score, tl.arbiter_id, us.football_location, us.football_map
FROM test_league tl
    INNER JOIN user_schools us ON (tl.home_team = us.school_name)
ORDER BY tl.game_date, tl.game_time, tl.home_team, tl.away_team;

我添加了表别名，以便您可以看到列的来源。请注意，最后两列来自 user_schools 表，而不是主 test_league 表。

您的PHP代码将在此行中进行更改，因为您现在拥有 $ row 变量中的所有信息，您将其回显而不是 $ location 和 $ map ：

echo '<td><a href="' . $row['map'] . '">' . $row['football_location'] . '</a></td><td><a href="">' . $row['arbiter_id'] . '</a></td></tr>';

Answer 4

试试这个加入：

tl是test_league表的别名，us是user_schools表的别名。

SELECT
    tl.*,
    us.football_location,
    us.football_map
FROM
    test_league tl
LEFT OUTER JOIN user_schools us ON (tl.home_team = us.school_name)

我在SQL Fiddle上添加了一个演示。 See the results on SQL Fiddle

$q = "SELECT
        tl.*,
        us.football_location,
        us.football_map
    FROM
        test_league tl
    LEFT OUTER JOIN user_schools us ON (tl.home_team = us.school_name)";
$r = mysqli_query($db, $q);

$i = 0;
$bgcolor = array('row1', 'row2');

while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
    echo '<tr class="' . $bgcolor[$i++ % 2] .'"><td>' . $row['game_date'] . '</td><td>' . $row['game_time'] . '</td><td class="alignleft"><a href="">' . $row['away_team'] . '</a> vs<br><a href="">' . $row['home_team'] . '</a></td>';
    if ($row['home_score'] == '0' && $row['away_score'] == '0') {
        echo '<td><a href=""><img src="images/report_icon.png" alt="Report Score" /></a></td>';
    } else {
        echo '<td>' . $row['home_score'] . '<br>' . $row['away_score'] . '</td>';
    }
    echo '<td><a href="' . $row['football_map'] . '">' . $row['football_location'] . '</a></td><td><a href="">' . $row['arbiter_id'] . '</a></td></tr>';
}

PHP / MySQL连接语法：比较单独表中的两列

4 个答案: