如何在laravel 5.2中传递连接数组以选择查询

时间:2017-06-17 11:21:25

标签: php laravel

我正在尝试从多个表中获取数据。我已将连接语句存储在$ Joins数组中,并将该数组传递给模型
这是我到目前为止所尝试的内容 -

$tbl_name="user_master";
$select= ['user_master.*', 'country.country_name', 'city.city_name', 'login_master.email_id', 'login_master.password'];
$joins=[
         "'country', 'user_master.country_id', '=', 'country.country_id'",
         "'city', 'user_master.city_id', '=', 'city.city_id'",
         "'login_master', 'user_master.user_id', '=', 'login_master.user_id'"
       ]; 
$users=$obj->getdata($tbl_name,$select,$joins);  

模型 -

public function getdata($tbl_name,$select,$joins)
    {
        $users = DB::table($tbl_name)
                ->join($joins)
                ->select($select)
                ->paginate(5);

                return $users;
    }  

我正在尝试将$ joins数组传递给查询,但它显示错误 -
缺少Illuminate \ Database \ Query \ Builder :: join()的参数2,在第61行的/opt/lampp/htdocs/demo_laravel/app/Models/Userdata.php中调用并定义

如果我尝试 -

$users = DB::table($tbl_name)
                ->join('country', 'user_master.country_id', '=', 'country.country_id')
                ->join('city', 'user_master.city_id', '=', 'city.city_id')
                ->join('login_master', 'user_master.user_id', '=', 'login_master.user_id')
                ->select($select)
                ->paginate(5);

                return $users;  

它工作正常但是当我将数组传递给连接时它显示错误 怎么解决?
请帮帮我。

1 个答案:

答案 0 :(得分:0)

试试这个:

$joins采用以下格式:

$joins = [
    ['country', 'user_master.country_id', '=', 'country.country_id'],
    ['city', 'user_master.city_id', '=', 'city.city_id'],
    ['login_master', 'user_master.user_id', '=', 'login_master.user_id'],
];

和型号:

public function getdata($tbl_name, $select, $joins)
{
    $users = DB::table($tbl_name);

    foreach ($joins as $join) {
        $users = $users->join(...$join);
    }

    $users = $users
        ->select($select)
        ->paginate(5);

    return $users;
}