我使用Laravel雄辩的查询构建器，我有一个查询，我想在多个条件上有一个where子句。它能起作用，但并不优雅。
例子:
$results = User::where('this', '=', 1)
->where('that', '=', 1)
->where('this_too', '=', 1)
->where('that_too', '=', 1)
->where('this_as_well', '=', 1)
->where('that_as_well', '=', 1)
->where('this_one_too', '=', 1)
->where('that_one_too', '=', 1)
->where('this_one_as_well', '=', 1)
->where('that_one_as_well', '=', 1)
->get();
有没有更好的方法，或者我应该坚持这个方法?
当前回答
DB::table('users')
->where('name', '=', 'John')
->orWhere(function ($query) {
$query->where('votes', '>', 100)
->where('title', '<>', 'Admin');
})
->get();
其他回答
public function search()
{
if (isset($_GET) && !empty($_GET))
{
$prepareQuery = '';
foreach ($_GET as $key => $data)
{
if ($data)
{
$prepareQuery.=$key . ' = "' . $data . '" OR ';
}
}
$query = substr($prepareQuery, 0, -3);
if ($query)
$model = Businesses::whereRaw($query)->get();
else
$model = Businesses::get();
return view('pages.search', compact('model', 'model'));
}
}
使用纯Eloquent，像这样实现它。这段代码返回所有帐户处于活动状态的已登录用户。 $用户= \ App \用户::(“状态”,“积极”)——> (logged_in,真的)- > ();
Model::where('column_1','=','value_1')
->where('column_2 ','=','value_2')
->get();
OR
// If you are looking for equal value then no need to add =
Model::where('column_1','value_1')
->where('column_2','value_2')
->get();
OR
Model::where(['column_1' => 'value_1',
'column_2' => 'value_2'])->get();
查询作用域可以帮助您提高代码的可读性。
http://laravel.com/docs/eloquent#query-scopes
用一些例子更新这个答案:
在你的模型中，像这样创建作用域方法:
public function scopeActive($query)
{
return $query->where('active', '=', 1);
}
public function scopeThat($query)
{
return $query->where('that', '=', 1);
}
然后，你可以在构建查询时调用这个作用域:
$users = User::active()->that()->get();
根据我的建议，如果你正在做筛选或搜索
那么你应该选择:
$results = User::query();
$results->when($request->that, function ($q) use ($request) {
$q->where('that', $request->that);
});
$results->when($request->this, function ($q) use ($request) {
$q->where('this', $request->that);
});
$results->when($request->this_too, function ($q) use ($request) {
$q->where('this_too', $request->that);
});
$results->get();
