php - 无法从给定的插入查询中将值插入mysql数据库..是否正确

Question

我试图从另一页发布的变量中将值插入mysql数据库中，并将其保存在数据库中，但是代码用于将值插入mysql数据库中，但查询不起作用，它在编辑时显示错误。

解析错误：语法错误，意外的''。$ Type_of_leave。''（T_CONSTANT_ENCAPSED_STRING）在53行的C：\ wamp \ www \ saveetha \ staff \ process_leave.php中

如何使用查询进行插入
                
                
                
                新假期
                
                
                

            $con=mysqli_connect("localhost","root","","saveetha");
            // Check connection
            if (mysqli_connect_errno())
              {
              echo "Failed to connect to MySQL: " . mysqli_connect_error();
              }

            mysqli_query($con,"INSERT INTO `leave_details`(`leave_id`, `type`, `no_of_days`, `date_from`, `date_till`, `availed_leave`, `leave_reason`, `alternate_date`, `period`, `semester`, `fac_name`, `fac_desg`, `dept`, `approved_authority`) VALUES (
            NULL,
            "'.$Type_of_leave.'",
            "'.$No_of_days.'",
            "'.$from.'",
            "'.$upto.'",
            "'.$Leave_Availed.'",
            "'.$Reason_for_Leave.'",
            "'.$Alternate_leave.'",
            "'.$Period.'",
            "'.$Semester.'",
            "'.$Fac_name.'",
            "'.$Fac_desg.'",
            "'.$dept.'",
            "'.$Approved_by.'")");

             echo 'leave from updated !! ';
            mysqli_close($con);
            ?>

Answer 1

您的报价倒退：

mysqli_query($con,"INSERT INTO `leave_details`(`leave_id`, `type`, `no_of_days`, `date_from`, `date_till`, `availed_leave`, `leave_reason`, `alternate_date`, `period`, `semester`, `fac_name`, `fac_desg`, `dept`, `approved_authority`) VALUES (
        NULL,
        '".$Type_of_leave."',
        '".$No_of_days."',
        '".$from."',
        '".$upto."',
        '".$Leave_Availed."',
        '".$Reason_for_Leave."',
        '".$Alternate_leave."',
        '".$Period."',
        '".$Semester."',
        '".$Fac_name."',
        '".$Fac_desg."',
        '".$dept."',
        '".$Approved_by."')");