php - 如何使用 PHP 和 SQL 生成嵌套的 JSON 响应
问题描述
我有四张表,ALBUM,ARTIST,SONG,GENRES
ALBUM (ID(PK),ALBUM_NAME,ARTIST_ID(FK),GENRES_ID(FK),ALBUM_POSTER)
SONG (ID(PK),SONG_NAME,ALBUM_ID(FK),ARTIST_ID(FK),SONG_LOCATION)
ARTIST (ID(PK),ARTIST_NAME )
GENRES (ID(PK), GENRES_NAME)
我已经创建了波纹管 php 文件,以便将 json 文件创建为“所需输出”所示,但我得到了“实际输出”,如下所示。如实际输出所示,所有专辑中的所有歌曲都包含在每个专辑的“song_detail”部分中。
php
$sql = "SELECT DISTINCT ALBUM_ID, ALBUM_NAME,ARTIST_NAME,GENRES_NAME,ALBUM_POSTER "
. "FROM ALBUM,ARTIST,SONG,GENRES "
. "WHERE ALBUM.ID=SONG.ALBUM_ID AND SONG.ARTIST_ID=ARTIST.ID AND GENRES.ID=ALBUM.GENRES_ID";
$result = $db->query($sql);
$response["albums"] = array();
if ($result->num_rows > 0)
{
while ($row = $result->fetch_assoc())
{ $album = array();
$album["album_id"] = $row["ALBUM_ID"];
$album["album_name"] = $row["ALBUM_NAME"];
$album["artist_name"] = $row["ARTIST_NAME"];
$album["genres_name"] = $row["GENRES_NAME"];
$album["album_poster"] = "http://www.".$row["ALBUM_POSTER"];
$album['song_details'] = array();
$sql2 = "SELECT*"
. " FROM SONG,ALBUM where SONG.ALBUM_ID = ALBUM.ID";
$result2 = $db->query($sql2);
while ($row2 = $result2->fetch_assoc())
{
$album['song_details'][] = array(
'songName' => $row2["SONG_NAME"],
'song_location' => "http://www.".$row2["SONG_LOCATION"]);
}
array_push($response["albums"], $album);
}
$response["success"] = 1;
}
期望的输出
[
{
"album_name": "Startboy",
"artist_name": "Weeknd",
"genres_name":"R&B",
"album_poster": "www.yyy.storage.x.png",
"songs_details": [{
"title": "False Alarm",
"song_location": "www././../yw.mp3"
},
{
"title": "Reminder",
"song_location": "www././../x.mp3"
}
]
},
{
"album_name": "25",
"artist_name": "Adele",
"genres_name":"",
"album_poster": "www.yyy.storage.a.png",
"songs_details": [{
"title": "Hello",
"song_location": "www././../yy.mp3"
},
{
"title": "I Miss You",
"song_location": "www././../yx.mp3"
}
]
}
]
实际输出
[
{
"album_name": "Startboy",
"artist_name": "Weeknd",
"genres_name":"R&B",
"album_poster": "www.yyy.storage.x.png",
"songs_details": [{
"title": "False Alarm",
"song_location": "www././../yw.mp3"
},
{
"title": "Reminder",
"song_location": "www././../x.mp3"
},
{
"title": "Hello",
"song_location": "www././../yy.mp3"
},
{
"title": "I Miss You",
"song_location": "www././../yx.mp3"
}]
},
{
"album_name": "25",
"artist_name": "Adele",
"genres_name":"",
"album_poster": "www.yyy.storage.a.png",
"songs_details": [{
"title": "False Alarm",
"song_location": "www././../y.mp3"
},
{
"title": "Reminder",
"song_location": "www././../x.mp3"
},
{
"title": "Hello",
"song_location": "www././../yy.mp3"
},
{
"title": "I Miss You",
"song_location": "www././../yx.mp3"
}
]
}
]
解决方案
当您选择歌曲时,您的 SQL 不会限制曲目来自哪个专辑(尽管它确实确保它们与 ALBUM 匹配)...
$sql2 = "SELECT*"
. " FROM SONG,ALBUM where SONG.ALBUM_ID = ALBUM.ID";
您只需根据当前专辑选择 SONG 记录...
$sql2 = "SELECT * FROM SONG where ID =".$row["ALBUM_ID"];
您可以在此处使用准备好的语句,但由于这是使用另一个表中的字段,因此可以使用它 - 但不是必需的。
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