如果数据存在，laravel 连接表
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laravel join table if data exists
提问by Gregor Voinov
What is the best way to join table A with table B, if table B has data and if not just give me data from table A? because if I do it in this way, and there are no photos in table B I don't get data from that row from table A.
将表 A 与表 B 连接的最佳方法是什么，如果表 B 有数据，如果不只是给我来自表 A 的数据？因为如果我这样做，并且表 BI 中没有照片，则不会从表 A 的该行中获取数据。
$data = Category::join('photos', 'categories.cover_id', '=', 'photos.id')
->get(['categories.id',
'categories.position',
'categories.visible',
'categories.created_at',
'categories.updated_at',
'categories.title',
'photos.filename']);
return $data;
my idea is just to make another request to get all data from table A where categories.cover_id is 0 (without join)
我的想法只是提出另一个请求，从表 A 中获取所有数据，其中 category.cover_id 为 0（无连接）
my tables are just
我的桌子只是
table A (categories)
-------------------------------
| id | title | cover_id | ... |
-------------------------------
| 1 | lorem | 1 | ... |
-------------------------------
| 2 | ipsum | 12 | ... |
-------------------------------
| 3 | dolor | 0 | ... |
-------------------------------
table B (Photos, there is no data for dolor, because i created dolor recently in table A)
---------------------------------
| id | title | filename | ... |
---------------------------------
| 1 | lorem | lorem.jpg | ... |
---------------------------------
| .. | ..... | ...jpg | ... |
---------------------------------
| 12 | ipsum | ipsum.jpg | ... |
---------------------------------
回答by lukasgeiter
You should be fine by just using a leftJoin(). A normal ("inner join") will only return results from both tables. But a left joinreturns all results from the lefttable (in this case categories) and everything that exists from the other table.
只需使用leftJoin(). 正常（“内连接”）只会返回两个表的结果。但是左连接返回左表中的所有结果（在本例中categories）以及其他表中存在的所有内容。
$data = Category::leftJoin('photos', 'categories.cover_id', '=', 'photos.id')
->get(['categories.id',
'categories.position',
'categories.visible',
'categories.created_at',
'categories.updated_at',
'categories.title',
'photos.filename']);
Or you could...
或者你可以...
Use the power of Eloquent
使用 Eloquent 的强大功能
You only need to define the relationship (I assume you already have a Photomodel) and this gets a lot easier
你只需要定义关系（我假设你已经有了一个Photo模型），这变得容易多了
class Category extends Eloquent {
public function photos(){
return $this->hasMany('Photo', 'cover_id');
}
}
And then...
进而...
$data = Category::with('photos')->get();
And you will have the photos model nested inside the category models. Accessible like this:
您将拥有嵌套在类别模型中的照片模型。可以这样访问：
foreach($data as $category){
foreach($category->photos as $photo){
echo $photo->filename;
}
}
回答by Chintan Parekh
I would rather do it this way:
我宁愿这样做：
// Just assuming a few variables for better understanding
$allCategories = Category::all();
foreach ($allCategories as $category) {
$photo = Photo::find($category->cover_id);
if ($photo) { // $photo!=null or isset($photo) - you can use anything
// photo is found do the additional processing
}
//proceed with your normal processing
}
