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php – Laravel：选择Eloquent Eager Loading的关系

2020-07-15 Laravel 前端之家

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@H_404_0@ 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

$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,用户可以选择国家/地区名称,而laravel应仅使用具有指定名称的国家/地区的帖子来回答请求.

如何将此条件添加到我现有的分页查询中？我试过这个：

$query = app(Post::class)->with('country')->newQuery(); 
// ...
if ($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)

whereHas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);
}

所以改变代码一点,

$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();
}

顺便说一句,上面的代码可以稍微简化如下;

$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();

}