我有两个数据库表:
帖子
from pynput import mouse
class MyException(Exception):pass
NumberOfMouseClicks = 0
def on_click(x, y, button, pressed):
global NumberOfMouseClicks
print(x, y)
NumberOfMouseClicks = NumberOfMouseClicks + 1
if (NumberOfMouseClicks==10):
raise MyException(button)
with mouse.Listener(on_click=on_click) as listener:
try:
listener.join()
except MyException as e:
pass
国家/地区
$table->increments('id');
$table->integer('country_id')->unsigned();
$table->foreign('country_id')->references('id')->on('countries');
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:
$table->increments('id');
$table->string('name', 70);
...导致以下错误:
$query = app(Post::class)->with('country')->newQuery();
// ...
if ($request->exists('country')) {
$query->where('country.name', $request->country);
}
// ...
答案 0 :(得分:13)
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);
}
所以稍微改变代码,
$query = ""
if ($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();
}
顺便说一句,上面的代码可以稍微简化一下;
$query = ""
if ($request->has('country'){
$query = Post::with(["country" => function($q) use($request){
$q->where("name","=",$request->country);
}])->first()
}else{
$query = Post::with("country")->get();
}
答案 1 :(得分:0)
$query = ""
if ($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();
}