Question

我是SQL的新手，我正在尝试建立一个登录系统。我已经按照指南进行了操作，但是当我尝试登录时，我收到了这两条消息：

警告：mysqli_fetch_array（）要求参数1为mysqli_result，第13行的C：\ xampp \ htdocs \ loginguide / login.php中给出布尔值

警告：mysqli_num_rows（）要求参数1为mysqli_result，第16行的C：\ xampp \ htdocs \ loginguide \ login.php中给出布尔值

我在MyPHPAdmin上创建了一个数据库但是，我不知道如何正确地做到这一点，我是不是想在这个数据库中添加表？

这是login.php代码：

<?php
   include("config.php");
   session_start();

   if($_SERVER["REQUEST_METHOD"] == "POST") {
      // username and password sent from form 

      $myusername = mysqli_real_escape_string($db,$_POST['username']);
      $mypassword = mysqli_real_escape_string($db,$_POST['password']); 

      $sql = "SELECT id FROM admin WHERE username = '$myusername' and passcode = '$mypassword'";
      $result = mysqli_query($db,$sql);
      $row = mysqli_fetch_array($result,MYSQLI_ASSOC);
      $active = $row['active'];

      $count = mysqli_num_rows($result);

      // If result matched $myusername and $mypassword, table row must be 1 row

      if($count == 1) {
         session_register("myusername");
         $_SESSION['login_user'] = $myusername;

         header("location: welcome.php");
      }else {
         $error = "Your Login Name or Password is invalid";
      }
   }
?>
<html>

   <head>
      <title>Login Page</title>

      <style type = "text/css">
         body {
            font-family:Arial, Helvetica, sans-serif;
            font-size:14px;
         }

         label {
            font-weight:bold;
            width:100px;
            font-size:14px;
         }

         .box {
            border:#666666 solid 1px;
         }
      </style>

   </head>

   <body bgcolor = "#FFFFFF">

      <div align = "center">
         <div style = "width:300px; border: solid 1px #333333; " align = "left">
            <div style = "background-color:#333333; color:#FFFFFF; padding:3px;"><b>Login</b></div>

            <div style = "margin:30px">

               <form action = "" method = "post">
                  <label>UserName  :</label><input type = "text" name = "username" class = "box"/><br /><br />
                  <label>Password  :</label><input type = "password" name = "password" class = "box" /><br/><br />
                  <input type = "submit" value = " Submit "/><br />
               </form>

               <div style="font-size:11px;<?php echo $error; ?></div>

            </div>

         </div>

      </div>

   </body>
</html>

Answer 1

1）您需要在SQL查询中包含“active”列，或者只需将'id'替换为*

select * from admin where ....
(or)
select id, active from admin where ....

2）检查您在给定示例中包含的config.php文件，如果DB_USERNAME，DB_PASSWORD，DB_DATABASE名称错误（语法错误），您可能会收到该警告。

<?php
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', 'root');
define('DB_DATABASE', 'testDB');
$db = mysqli_connect(DB_SERVER,DB_USERNAME,DB_PASSWORD,DB_DATABASE);
?>

Answer 2

我认为您的查询失败并返回false值。检查您的查询，如果它正常工作，请将其放在您的代码中：

$result = mysqli_query($db,$sql);

if (!$result) {
    printf("Error: %s\n", mysqli_error($db));
    exit();
}

了解更多信息。

http://www.php.net/manual/en/mysqli.error.php

Answer 3

您只选择ID，然后使用＆＃34;有效＆＃34;

所以你应该先解决这个问题：

SELECT * FROM admin WHERE username = '$myusername' and passcode = '$mypassword'

它应该现在可以使用

mysqli_fetch_array＆amp; row期望参数1为mysqli_result

3 个答案: