Laravel:雄辩渴望加载关系的选择
问题描述:
我有两个数据库表:
帖子
$table->increments('id');
$table->integer('country_id')->unsigned();
$table->foreign('country_id')->references('id')->on('countries');
国家/地区
$table->increments('id');
$table->string('name', 70);
我使用laravel作为后端.现在,我想为我的前端实现过滤数据.因此,用户可以选择一个国家/地区名称,而laravel仅应使用具有指定国家/地区名称的帖子来回答请求.
I use laravel as back-end. Now I want to implement filtering data for my front-end. So the user can select a country name and laravel should answer the request only with posts that have a country with the specified name.
如何将此条件添加到我现有的分页查询中?我试过了:
How could I add this condition to my existing pagination query? I tried this:
$query = app(Post::class)->with('country')->newQuery();
// ...
if ($request->exists('country')) {
$query->where('country.name', $request->country);
}
// ...
...导致以下错误:
... resulting in the following error:
Column not found: 1054 Unknown column 'country.name' in 'where clause' (SQL: select count(*) as aggregate from `posts` where `country`.`name` = Albania)
答
其中,Has方法根据Laravel代码库接受参数,
whereHas method accepts parameter as per Laravel code base,
/**
* Add a relationship count / exists condition to the query with where clauses.
*
* @param string $relation
* @param \Closure|null $callback
* @param string $operator
* @param int $count
* @return \Illuminate\Database\Eloquent\Builder|static
*/
public function whereHas($relation, Closure $callback = null, $operator = '>=', $count = 1)
{
return $this->has($relation, $operator, $count, 'and', $callback);
}
所以稍微更改一下代码,
so Changing the code a little as,
$query = ""
if ($request->has('country'){
$query = Post::with("country")->whereHas("country",function($q) use($request){
$q->where("name","=",$request->country);
})->get()
}else{
$query = Post::with("country")->get();
}
通过上面的代码可以稍微简化如下;
By the way above code can be a little simplified as follow;
$query = ""
if ($request->has('country'){
$query = Post::with(["country" => function($q) use($request){
$q->where("name","=",$request->country);
}])->first()
}else{
$query = Post::with("country")->get();
}