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
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)- > ();
其他回答
一定要对子查询应用任何其他过滤器，否则or可能会收集所有记录。
$query = Activity::whereNotNull('id');
$count = 0;
foreach ($this->Reporter()->get() as $service) {
$condition = ($count == 0) ? "where" : "orWhere";
$query->$condition(function ($query) use ($service) {
$query->where('branch_id', '=', $service->branch_id)
->where('activity_type_id', '=', $service->activity_type_id)
->whereBetween('activity_date_time', [$this->start_date, $this->end_date]);
});
$count++;
}
return $query->get();
$variable = array('this' => 1,
'that' => 1
'that' => 1,
'this_too' => 1,
'that_too' => 1,
'this_as_well' => 1,
'that_as_well' => 1,
'this_one_too' => 1,
'that_one_too' => 1,
'this_one_as_well' => 1,
'that_one_as_well' => 1);
foreach ($variable as $key => $value) {
User::where($key, '=', $value);
}
您可以在where子句中使用数组，如下所示。
$result=DB::table('users')->where(array(
'column1' => value1,
'column2' => value2,
'column3' => value3))
->get();
使用这个
$users = DB::table('users')
->where('votes', '>', 100)
->orWhere('name', 'John')
->get();
在这种情况下，你可以使用这样的代码:
User::where('this', '=', 1)
->whereNotNull('created_at')
->whereNotNull('updated_at')
->where(function($query){
return $query
->whereNull('alias')
->orWhere('alias', '=', 'admin');
});
它应该为你提供这样的查询:
SELECT * FROM `user`
WHERE `user`.`this` = 1
AND `user`.`created_at` IS NOT NULL
AND `user`.`updated_at` IS NOT NULL
AND (`alias` IS NULL OR `alias` = 'admin')
