Question

我有两个数据库表，＆＃34; user＆＃34;它有3列（id，[自动增量]用户名和捏）另一张桌子是＆＃34;针脚＆＃34;只有一列（划痕）我的表格

USER
Id       username      Pinc


1         Josh             

2         Angela         

3          Chika 


PINS
scratches


123456


234515

124564

我想要一个用户通过表单提交密码的情况，它会检查引脚表中是否存在此类数据，如果存在，它将使用表单发布更新我的用户表的pinc列，用它登录。如果引脚表中没有它，则会出现错误＆＃34;抱歉引脚不存在。＆＃34; 我的代码

$sql = "SELECT * FROM    pins WHERE scratches = '" .' $user_password '. "';";
$query = $this->db_connection->query($sql);
if ($query->num_rows== 0){
   $this->errors[] = "Sorry, that PIN does not exist.";
} elseif ($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();
        $_SESSION['user_name'] = $result_row->user_name;
        $_SESSION['user_login_status'] = 1;
    } else {
       $this->errors[] = "Wrong password. Try again.";
    }
} else {
    $this->errors[] = "This user does not exist.";
}
} else {
   $this->errors[] = "Database connection problem.";
}
}
}

当我运行代码时，我得到了'＃34;抱歉引脚不存在。＆＃34;有人能告诉我它有什么问题吗？

Answer 1

奇怪的字符串结构：

$sql = "SELECT * FROM pins WHERE scratches = '" .' $user_password '. "';";

它将从引脚WHERE scrathes =＆＃34; $ user_password＆＃34;中选择所有数据。它将返回未找到的数据。

像这样修改：

$sql = "SELECT * FROM    pins WHERE scratches = '" . $user_password . "';";

告诉我它是如何运作的。

Answer 2

你在''附近使用singe quete $userpassword，所以你传递的是字符串像这样更改你的第一个查询

$sql = "SELECT * FROM    pins WHERE scratches = '".$user_password."';";

你的第二个查询也错了。因为我假设$username是一个字符串，你需要将它包裹在像这样的单''

中

$sql = "UPDATE user ".
   "SET pinc = '$user_password' ".
   "WHERE user_name = '$user_name'" ;//You are missing single quete here if username is a string

我不知道这也是问题，但在第三个查询中有额外的空间。对于紧随其后的$username变量（点. .之后）。如果这不起作用就像这样删除它

$sql = "SELECT  user_name, pinc 
              FROM user
              WHERE user_name = '".$user_name."' ;";

**

这是您的完整代码应该是

**

$sql = "SELECT * FROM    pins WHERE scratches = '".$user_password ."';";
$query = $this->db_connection->query($sql);
if ($query->num_rows== 0){
   $this->errors[] = "Sorry, that PIN does not exist.";
} elseif ($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();
        $_SESSION['user_name'] = $result_row->user_name;
        $_SESSION['user_login_status'] = 1;
    } else {
       $this->errors[] = "Wrong password. Try again.";
    }
} else {
    $this->errors[] = "This user does not exist.";
}
} 
}
}

Answer 3

变化：

$sql = "SELECT * FROM    pins WHERE scratches = '" .' $user_password '. "';";

到

 $sql = "SELECT * FROM    pins WHERE scratches = '".$user_password."'";

更新并从数据库表中选择

3 个答案:

这是您的完整代码应该是