Question

我有3张桌子：

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

Table: albumconnect
Columns: id, imageid, albumid

Table: albumimages
Columns: id, path

我正试图用一个JOIN查询替换所有那些不必要的查询：

<?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>';
        };

?>

现在根据我在线阅读的内容提出的JOIN查询是：

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

然而，当我尝试var_dump内容时，它会打印出Null，所以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。我没有测试它。正如@Difster所说，SQL引擎并不知道它应该引用哪个id。因此，定义表别名并使用它们为引用的列添加前缀。然后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：

以后不再使用的sql语句中的列是可选的。因此，如果您以后不再需要，则不需要选择所有列。

也许您也成为alc或ali表的NULL值行。这意味着并非所有专辑都有图像。然后，您必须向我们提供表中的值，以便我们为您提供适当的WHERE条件，例如WHERE ali IS NOT NULL。我的答案只是你的出发点。

编辑2：

此版本也可以。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>';
}

为什么我的3表JOIN MySQL查询不起作用？

1 个答案: