Laravel：雄辩渴望加载关系的选择

Question

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');

国别

$table->increments('id');
$table->string('name', 70);

我使用laravel作为后端。 现在，我想为前端实现过滤数据。 因此，用户可以选择一个国家名称，而laravel仅应使用具有指定名称国家的帖子来回答请求。

如何将此条件添加到现有的分页查询中？ 我尝试了这个：

$query = app(Post::class)->with('country')->newQuery(); 
// ...
if ($request->exists('country')) {
        $query->where('country.name', $request->country);
}
// ...

...导致以下错误：

Column not found: 1054 Unknown column 'country.name' in 'where clause' (SQL: select count(*) as aggregate from `posts` where `country`.`name` = Albania)

Answer 1

其中，方法根据Laravel代码库接受参数，

 /**
 * 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);
}

所以稍微修改一下代码，

$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();
}

顺便说一下，上面的代码可以简化如下：

$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();

}

Answer 2

$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();
}