php mysql选择和更新错误

Question

我有两个数据库表，“用户”，它有3列（id，[自动增量]用户名和夹点） 另一个表是只有一列（划痕）的“pins” 我的桌面

USER 
Id  username  Pinc 
1   Josh    
2   Angela   
3   Chika   

PINS 
scratches 
123456 
234515 
124564 

我想如果用户通过表单提交他的用户名和密码，它会检查是否存在其他用户的密码。如果是这样，它会给出一个错误“对不起，该代码只有一个人有权”，如果它存在于他的柱子中，它会记录他。如果它不存在或他或另一个用户，它将检查是否密码存在于引脚表中。如果是，它会将其更新到用户的pinc柱。然后如果它不存在于引脚表中，则将其登录，它将显示“对不起，该代码不存在”的错误。 我的代码

//if the password already exist for another user if yes give error 

$sql = "SELECT * FROM user WHERE pinc = '" . $user_password . "';"; 
$query = $this->db_connection->query($sql); 


if ($query->num_rows== 1) { 
    $this->errors[] = "Sorry, that code is entitled to one person only";} 

//if no, check if it exist in the pins table,   
elseif ($query->num_rows== 0) { 


    $sql = "SELECT * FROM pins WHERE scratches = '" . $user_password . "';"; 
    $query = $this->db_connection->query($sql); 

//if no give error   
elseif ($query->num_rows== 0) { 
    $this->errors[] = "Sorry, that code does not exist.";} 

//if yes update the user table   
elseif ($query->num_rows== 1) { 


$sql = "UPDATE user ". 
    "SET pinc = '$user_password' ". 
    "WHERE user_name = '$user_name'" ; 


    $query_new_user_insert = $this->db_connection->query($sql); 

    $sql = "SELECT user_name, pinc 
        FROM user 
        WHERE user_name = '" . $user_name . "' ;"; 
      $result_of_login_check = $this->db_connection->query($sql); 

      // if this user exists 
      if ($result_of_login_check->num_rows == 1) { 

       // get result row (as an object) 
       $result_row = $result_of_login_check->fetch_object(); 



        // write user data into PHP SESSION 
        $_SESSION['user_name'] = $result_row->user_name; 


        $_SESSION['user_login_status'] = 1; 

       } else { 
        $this->errors[] = "Wrong PIN. please check the PIN at the back of your scrathch and Try again."; 
       } 
      } else { 
       $this->errors[] = "A student with this Reg_No does not exist."; 
      } 
      }  else { 
       $this->errors[] = "Database connection problem."; 
     } 
    } 
} 

当我运行上面的代码，它提供了解析错误：语法错误，意外 'elseif的'

Answer 1

你缺少}这行之前：

elseif ($query->num_rows== 0) { 
    $this->errors[] = "Sorry, that code does not exist.";} 

这是你的原代码：

if ($query->num_rows== 1) { 
    $this->errors[] = "Sorry, that code is entitled to one person only";} 

//if no, check if it exist in the pins table,   
elseif ($query->num_rows== 0) { 

$sql = "SELECT * FROM pins WHERE scratches = '" . $user_password . "';"; 
    $query = $this->db_connection->query($sql); 

//if no give error   
elseif ($query->num_rows== 0) { 
    $this->errors[] = "Sorry, that code does not exist.";} 

并没有失踪}在此之前//if no give error