Php 在Laravel'内搜索;他急着装货
我有这样写的可选路线Php 在Laravel'内搜索;他急着装货,php,laravel-5,eager-loading,Php,Laravel 5,Eager Loading,我有这样写的可选路线 Route::get('{id}/{inid?}/{task?}', 'WelcomeController@station')->where(['id' => '[0-9]+', 'inid' => '[0-9]+']); public function station($id, $inid = null, $task = null) { $obj = Station::findOrFail($id); $template = 'stat
Route::get('{id}/{inid?}/{task?}', 'WelcomeController@station')->where(['id' => '[0-9]+', 'inid' => '[0-9]+']);
public function station($id, $inid = null, $task = null)
{
$obj = Station::findOrFail($id);
$template = 'station';
$app = array(
'docTitle' => $obj->st_name,
'lat' => $obj->st_lat,
'lng' => $obj->st_lng,
'appName' => $obj->setting->set_name,
'linkId' => $id
);
if ($inid) {
$obj = $obj->instruments()->where("in_id", $inid);
$template = 'form.index';
dd($obj);
}
return view($template, compact('obj', 'app'));
}
控制器是这样的
Route::get('{id}/{inid?}/{task?}', 'WelcomeController@station')->where(['id' => '[0-9]+', 'inid' => '[0-9]+']);
public function station($id, $inid = null, $task = null)
{
$obj = Station::findOrFail($id);
$template = 'station';
$app = array(
'docTitle' => $obj->st_name,
'lat' => $obj->st_lat,
'lng' => $obj->st_lng,
'appName' => $obj->setting->set_name,
'linkId' => $id
);
if ($inid) {
$obj = $obj->instruments()->where("in_id", $inid);
$template = 'form.index';
dd($obj);
}
return view($template, compact('obj', 'app'));
}
现在,我失败的是,实际上,我如何使用站的$obj
内指定的$inid
提取仪器
下面是我如何设置模型的
class Station extends Model {
public function instruments()
{
return $this->hasMany('App\Instrument', 'in_stid');
}
}
至于仪器呢
class Instrument extends Model {
protected $primaryKey = 'in_id';
public function station()
{
return $this->belongsTo('App\Station', 'in_stid');
}
}
希望有人能告诉我在哪里可以找到官方文件来解决这个问题
谢谢也许你在找这样的东西
替换
$obj = $obj->instruments()->where("in_id", $inid);
与