php - 为什么我的 3 表 JOIN MySQL 查询不起作用？

Question

我有 3 张 table :

Table: album
Columns: id, name, description, author, path, image

Table: albumconnect
Columns: id, imageid, albumid

Table: albumimages
Columns: id, path

我正在尝试用单个 JOIN 查询替换所有这些不必要的查询:

<?php

        $albumID = $_SERVER['QUERY_STRING'];
        $realAlbumID = substr($albumID, 1);
        $realestAlbumID = str_replace('%20', ' ', $realAlbumID);

        $sql = "SELECT * FROM album WHERE id='$realestAlbumID'";
        $result = mysqli_query($conn, $sql);
        $getResult = mysqli_fetch_assoc($result);
        $albumPath = $getResult['path'];

        $sql2 = "SELECT * FROM albumconnect WHERE albumid='$realestAlbumID'";
        $result2 = mysqli_query($conn, $sql2);

        while ($row = $result2->fetch_assoc()){
            $imageId = $row['imageid']; 
            $sql3 = "SELECT * FROM albumimages WHERE id='$imageId'";
            $result3 = mysqli_query($conn, $sql3);
            $getResult3 = mysqli_fetch_assoc($result3);
            $imagePath = $getResult3['path'];
            echo '<div class="imageContainerAlbums"><li class="listAlbums"><img class="specificAlbumThumnails" src="'.$albumPath.$imagePath.'" alt="Random image" /></li></div>';
        };

?>

现在，我根据在线阅读的内容提出的 JOIN 查询是这样的:

$sql5 = "SELECT * FROM album
                JOIN albumconnect ON albumconnect.albumid=album.id
                JOIN albumimages ON albumimages.id=albumconnect.imageid
                WHERE id='$realestAlbumID'";
        $result5 = mysqli_query($conn, $sql5);

但是，当我尝试 var_dump 内容时，它会打印 Null，因此我认为我的查询不正确，但我无法找出正确的方法。

Answer 1

应该是这样的。我没有测试过。正如@Difster正确所说，SQL引擎不知道它应该引用哪个id。因此，定义表别名并用它们为引用的列添加前缀。然后也为列名定义唯一的别名。否则你的sql语句几乎是完美的。

<?php

$albumID = $_SERVER['QUERY_STRING'];
$realAlbumID = substr($albumID, 1);
$realestAlbumID = str_replace('%20', ' ', $realAlbumID);

$sql = "SELECT 
            alb.name AS album_name,
            alb.description AS album_description,
            alb.author AS album_author,
            alb.path AS album_path,
            alb.image AS album_image,
            ali.path AS image_path
        FROM album AS alb
        LEFT JOIN albumconnect AS alc ON alc.albumid = alb.id
        LEFT JOIN albumimages AS ali ON ali.id = alc.imageid
        WHERE alb.id = '$realestAlbumID'";
$result = mysqli_query($conn, $sql);

while ($row = $result->fetch_assoc()) {
    $albumPath = $row['album_path'];
    $imagePath = $row['image_path'];
    echo '<div class="imageContainerAlbums">';
    echo '<li class="listAlbums">';
    echo '<img class="specificAlbumThumnails" src="' . $albumPath . $imagePath . '" alt="Random image" />';
    echo '</li>';
    echo '</div>';
}

编辑1:

稍后不使用的 sql 语句中的列是可选的。因此，如果以后不需要它们，则无需选择所有列。

也许您也会成为 alc 或 ali 表中具有 NULL 值的行。这意味着并非所有相册都有图像。然后，您必须向我们提供表中的值，以便我们可以为您提供适当的进一步 WHERE 条件，例如 WHERE ali IS NOT NULL。我的这个回答只是你的起点。

编辑2:

这个版本也不错。我刚刚更改了sql语句。

<?php

$albumID = $_SERVER['QUERY_STRING'];
$realAlbumID = substr($albumID, 1);
$realestAlbumID = str_replace('%20', ' ', $realAlbumID);

$sql = "SELECT 
            alb.name AS album_name,
            alb.description AS album_description,
            alb.author AS album_author,
            alb.path AS album_path,
            alb.image AS album_image,
            ali.path AS image_path
        FROM albumimages AS ali 
        LEFT JOIN albumconnect AS alc ON alc.imageid = ali.id 
        LEFT JOIN album AS alb ON alb.id = alc.albumid 
        WHERE alb.id = '$realestAlbumID'";
$result = mysqli_query($conn, $sql);

while ($row = $result->fetch_assoc()) {
    $albumPath = $row['album_path'];
    $imagePath = $row['image_path'];
    echo '<div class="imageContainerAlbums">';
    echo '<li class="listAlbums">';
    echo '<img class="specificAlbumThumnails" src="' . $albumPath . $imagePath . '" alt="Random image" />';
    echo '</li>';
    echo '</div>';
}