Question

我在这样的原始SQL中搜索功能

//test condition variable
 if ($param_type != "") {
    if($param_type =='SAS'){
        $str .= " AND pos.type IN('SAS1','SAS2') ";
    }
    else if($param_type =='SME'){
        $str .= " AND pos.type IN('SME1','SME2') ";
    }
    else if($param_type =='PT'){
        $str .= " AND pos.type IN('PT','PT0') ";
    }else{
        $str .= " AND pos.type ='$param_type'";
    }
}

//SQL

$query =" SELECT * FROM tbl_user AS u INNER JOIN tbl_position AS pos 
         ON u.user_id = pos.user_id where u.levels IN ('7','8','9') .$str ;" ;

如何在laravel中编写这样的SQL？

Answer 1

您可以使用原始查询。

$query = DB::select(SELECT * FROM tbl_user AS u INNER JOIN tbl_position AS pos 
     ON u.user_id = pos.user_id where u.levels IN ('7','8','9') .$str ;");

当然，您需要在控制器的顶部添加use DB;

无论如何Raw查询可能是危险的，并且使用查询构建器会更好简单的例子：

$query = DB::table('tbl_user')
     ->join('tbl_position','tbl_user.user_id','=','tbl_position.user_id')
     ->where('something','=','something')
     ->get();

查询构建器：query builder

Answer 2

回答我的问题是谁遇到像我这样的问题：

像这样的例子

$query = DB::table('user');

if ($param_type =='PT')
    $query->where('pos.type', '=', 1);
}
 $result = $query->get();

Laravel5.2 SQL构建器测试条件到哪里查询

2 个答案: