Question

我在下面有4张桌子：

曲目

+----+-----+----------------+-------+
| ID | TID |     TITLE      | ALBUM |
+----+-----+----------------+-------+
|  1 | AAA | Yesterday      |     1 |
|  2 | BBB | Happy          |     2 |
|  3 | CCC | Gangname Style |     3 |
+----+-----+----------------+-------+

专辑

+----+-----+---------+-------+
| ID | AID |  TITLE  | COVER |
+----+-----+---------+-------+
|  1 | AAA | Album A | 1.jpg |
|  2 | BBB | Album B | 2.jpg |
|  3 | CCC | Album C | 3.jpg |
+----+-----+---------+-------+

track_artist

+----+-----+-----------+
| ID | TID | ARTIST_ID |
+----+-----+-----------+
|  1 |   1 |         1 |
|  2 |   2 |         2 |
|  3 |   3 |         3 |
+----+-----+-----------+

艺术家

+----+--------+--------+
| ID |  NAME  | AVATAR |
+----+--------+--------+
|  1 | Taylor | 1.JPG  |
|  2 | T-ara  | 2.JPG  |
|  3 | M2M    | 3.JPG  |
+----+--------+--------+

我想基于上面4表中的track.TITLE（album.TITLE）获得artist.NAME，track.TID和where TID = $XXX。当我使用原始的INNER JOIN MySQL代码时很容易，但是我想在Laravel中使用Eloquent ORM。我该怎么做？谢谢。

我已经分析了：

一条曲目 belongsTo一条专辑

一条曲目 hasMany track_artist

一位 track_artist hasOne 艺术家

一条曲目 hasMany 艺术家

Answer 1

首先，您的关系（我认为）需要一些更改。

一首曲目属于一张专辑，而一本专辑具有许多曲目（一对多）
一首曲目有一位歌手，而一位歌手有多首曲目（一对多）
一张专辑有多位歌手，而一位歌手有多张专辑（多对多）

您将需要一个与album id和artist id的多对多关系的表

这是考虑以上几点的示例

<?php
namespace App;
use Illuminate\Database\Eloquent\Model;

class track extends Model{
    function album(){
        $this->belongsTo(album::class);
    }
    function artist(){
        $this->belongsTo(artist::class);
    }
}

class album extends Model{
    function track(){
        $this->hasMany(track::class);
    }
    function artist(){
        $this->belongsToMany(artist::class, 'my_many_to_many_table');
    }
}

class artist extends Model{
    function track(){
        $this->hasMany(track::class);
    }
    function album(){
        $this->belongsToMany(album::class, 'my_many_to_many_table');
    }
}

Answer 2

您可以尝试

桌子应该是这样，您需要4张桌子

public IEnumerable<DataGridRow> GetDataGridRows(DataGrid grid)
    {
        var itemsSource = grid.ItemsSource as IEnumerable;
        if (null == itemsSource) yield return null;
        foreach (var item in itemsSource)
        {
            var row = grid.ItemContainerGenerator.ContainerFromItem(item) as DataGridRow;
            if (null != row) yield return row;
        }
    }

    private void DgCustomerInfo_SelectionChanged(object sender, SelectionChangedEventArgs e)
    {

        try
        {
            var row_list = GetDataGridRows(customersPage.dgCustomerInfo);
            foreach (DataGridRow single_row in row_list)
            {
                if (single_row.IsSelected == true)
                {

                    customersPage.txtCustomerName.Text=single_row.Item.ToString();

                }
            }

        }

模型

albums(id, title, cover)
artist(id, name, avatar)
tracks(id, album_id, title)
track_artists (id, artist_id, track_id) ManyToMany with tracks

获取数据

class Album extends Model{

    function tracks(){
        $this->hasMany(Track::class);
    }
}

class Artist extends Model{

   function tracks(){
      $this->belongsToMany(Track::class, 'track_artists'); //here track_artists table is as pivot table
   }   
}

class Track extends Model{

   function album(){
        $this->belongsTo(Album::class);
   }

   function artist(){
      $this->belongsToMany(Artist::class, 'track_artists'); //here track_artists table is as pivot table
   }   
}

注意：您无需为$track = Track::with('album','artists')->find($trackId); echo $track->title." - ". $track->album->title; //now print all artist of this track foreach($track->artists as $artist){ echo $artist->name; }创建模型，因为它是track_artists和tracks表的数据透视表。您可以使用artists或Track laravel雄辩的模型在数据透视表中插入和更新。不过，如果您愿意，可以创建扩展Artist的文件。

将数据保存在数据透视表中

Pivot

有关详细信息，https://laravel.com/docs/5.6/eloquent-relationships#many-to-many

Laravel中的多联接表

2 个答案: