php Laravel eloquent 获取关系数
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Laravel eloquent get relation count
提问by Vincent Decaux
I use Laravel 5.3.
我使用 Laravel 5.3。
I have 2 tables :
我有 2 张桌子：
Articles
---------
id
cat_id
title
And
和
Category
---------
id
parent_id
title
I have defined my relations in my models :
我已经在我的模型中定义了我的关系：
// Article model
public function category()
{
return $this->belongsTo(Category::class);
}
// Category model
public function children()
{
return $this->hasMany(Category::class, 'parent_id', 'id');
}
Is there an easy way using Eloquent to have a list a categories with count of articles. The difficulty is that I want to group categories where id_parent = 0, i.e. I want to display only parent categories with count of articles in children.
有没有一种简单的方法使用 Eloquent 来列出一个包含文章计数的类别。困难在于我想将类别分组 where id_parent = 0，即我只想显示父类别以及子项中的文章数。
I tried something like that :
我试过这样的事情：
$category = new \App\Models\Category();
$categoryTable = $category->getTable();
return $category->leftJoin('article', 'article.cat_id', '=', 'category.id')
->whereIn('article.cat_id', function($query)
{
$query->select('cat_id')
->from('categories')
->where('categories.parent_id', ???)
->orWhere($this->tableName .'.cat_id', $id);
})
->groupBy('cat_id');
But I am lost...
但是我迷路了...
回答by kapil.dev
you can use withCount(). It is available from 5.3 version
你可以使用withCount(). 从 5.3 版本开始可用
for more info about eloquent visit : https://laravel.com/docs/5.3/eloquent-relationships
有关雄辩访问的更多信息：https: //laravel.com/docs/5.3/eloquent-relationships
回答by Carter Fort
You can use the hasManyThrough()Eloquent method to fetch all of the childrens' Articles, then add the article counts together in a nice little getter. I added the getter to the $appendsarray on the model to help illustrate it in the Tinker output.
您可以使用hasManyThrough()Eloquent 方法获取所有孩子的文章，然后将文章计数添加到一个漂亮的小 getter 中。我将 getter 添加到$appends模型上的数组中，以帮助在 Tinker 输出中进行说明。
class Category extends Model
{
protected $appends = [
'articleCount'
];
public function articles()
{
return $this->hasMany(Article::class);
}
public function children()
{
return $this->hasMany(Category::class, 'parent_id');
}
public function childrenArticles()
{
return $this->hasManyThrough(Article::class, Category::class, 'parent_id');
}
public function getArticleCountAttribute()
{
return $this->articles()->count() + $this->childrenArticles()->count();
}
}
Here's the Tinker output:
这是 Tinker 的输出：
Psy Shell v0.8.0 (PHP 7.0.6 — cli) by Justin Hileman
>>> $cat = App\Category::first();
=> App\Category {#677
id: "1",
name: "Cooking",
parent_id: null,
created_at: "2016-12-15 18:31:57",
updated_at: "2016-12-15 18:31:57",
}
>>> $cat->toArray();
=> [
"id" => 1,
"name" => "Cooking",
"parent_id" => null,
"created_at" => "2016-12-15 18:31:57",
"updated_at" => "2016-12-15 18:31:57",
"articleCount" => 79,
]
>>>
If you want to restrict your Category query to ones that have children that have articles, you could do that using the has()method:
如果要将 Category 查询限制为具有文章的子项的查询，则可以使用以下has()方法：
Category::has('children.articles')->get();
Here's more on the has()method:
以下是该has()方法的更多信息：
https://laravel.com/docs/5.3/eloquent-relationships#querying-relationship-existence
https://laravel.com/docs/5.3/eloquent-relationships#querying-relationship-existence
And the hasManyThrough()method:
和hasManyThrough()方法：
https://laravel.com/docs/5.3/eloquent-relationships#has-many-through
https://laravel.com/docs/5.3/eloquent-relationships#has-many-through
回答by Amit Gupta
Define a articles()relation in your Categorymodel as:
将模型中的articles()关系定义Category为：
public function articles()
{
return $this->hasMany(Article::class, 'cat_id');
}
Then you can try it as:
然后你可以尝试它：
Category::where('parent_id', 0)->withCount('articles')->get();
回答by DevK
This should work:
这应该有效：
$category
->where('categories.parent_id', 0)
->leftJoin('article', 'article.cat_id', '=', 'categories.id')
->select('categories.id', \DB::raw('COUNT(article.id)'))
->groupBy('categories.id')
->get();
The above query will get you category IDs and count of all articles that belong to the category.
上面的查询将为您提供类别 ID 和属于该类别的所有文章的数量。
After reading your question and comments again, if I understand correctly you want to get the count of all articles that belong to those categories (with parent_id = 0) + the count of articles that belong to sub categories (those with parent_id = (id of some category)).
再次阅读您的问题和评论后，如果我理解正确，您想获得属于这些类别的所有文章的计数（parent_id = 0）+ 属于子类别的文章计数（那些具有 parent_id = (id of某些类别））。
Now I have no way of testing this easily, but I think something along these lines should work for that.
现在我没有办法轻松地测试这个，但我认为这些方面的东西应该适用于此。
$category
->where('categories.parent_id', 0)
->leftJoin('article', 'article.cat_id', '=', 'categories.id')
->leftJoin('categories as c2', 'c2.parent_id', '=', 'categories.id')
->leftJoin('article as a2', 'a2.cat_id', '=', 'c2.id')
->select('categories.id', \DB::raw('(COUNT(article.id)) + (COUNT(a2.id)) as count'))
->groupBy('categories.id')
->get();
That beign said, I think you're better of having a column named count in categories and update it each time a new article gets added. For performance.
也就是说，我认为您最好在类别中创建一个名为 count 的列，并在每次添加新文章时对其进行更新。为了性能。
回答by Carlos_E.
I am sure somebody is still going through this, I was able to solve it the following way, suppose I have an Agent model and a Schedule model, i.e. one agent may have many schedules:
我确定有人还在经历这个，我能够通过以下方式解决它，假设我有一个代理模型和一个调度模型，即一个代理可能有多个调度：
class Schedule extends Model {
public function agent() {
return $this->belongsTo(Agent::class, 'agent_id');
}
}
class Agent extends Model {
public function user(){
return $this->belongsTo(User::class);
}
public function schedules(){
return $this->hasMany(Schedule::class);
}
}
Well some agents may not necessarily have schedules assigned, thus, I filtered those before calling the with()method, like this:
好吧，一些代理可能不一定分配了时间表，因此，我在调用该with()方法之前过滤了这些，如下所示：
$agents = Agents::whereIn(
'id',
Schedule::distinct()->pluck('agent_id')
)->with('schedules')->get();
Hope this helps!.
希望这可以帮助！。
