在网页中创建小程序

Question

I want to have a simple web page with fields for userName and password.Once a user enters the correct userName and password and press logIn button then it should display an applet . 我想要一个简单的网页，其中包含userName和password的字段。一旦用户输入正确的userName和password并按logIn按钮， 则它应该显示一个applet 。 Here's my code. 这是我的代码。

<?php  
$un=$_POST["username"];  
$pw=$_POST["password"];  
$log=$_POST["Login"];  
$con=mysqli_connect("localhost","root","","");  
if(mysqli_connect_errno($con))  
    {  
    echo "Failed to connect".mysqli_connect_error();  
    }  
mysqli_select_db($con,student);  
$query="SELECT * FROM studentinfo WHERE stName=$un AND stP=$pw";  

$result=mysqli_query($con,$query);  
$num_rows = mysqli_num_rows($result);  
if($log){  
    if($num_rows==1){  
    $isLogged=true;  
    }  
    else{  
    echo "Error log In.Invalid username or password";     
    $isLogged=false;  
    }  
}  
?>  
<head>  
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />  
<title>User LogIn</title>  
</head>  

<body>  
<form method="post">  
            <label>Username<input name="username" type="text" /></label> <label>Password<input name="password" type="password" /></label>  
            <input name="cmd" type="submit" value="Login" />  
        </form>  
<?php if($isLogged) {?>  
<applet code="studentWeb.html" width="32" height="32" alt="Couldn't launch applet" title="Student Details">  
</applet>  
<?php }?>  
</body>  
</html>

Right now it gives an error as 现在它给出了一个错误

mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in and it doesn't open up a applet even when correct username and password has been entered. mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in即使输入正确的用户名和密码也不会打开applet。

Edit: After I changed to action="logindata.php" in the form, here's the logindata.php 编辑：在表单中更改为action =“ logindata.php”后，这是logindata.php

<?php
$connect=mysqli_connect("localhost","root","");
if(mysqli_connect_errno($connect))
{
    echo "Failed to connect".mysqli_connect_error();
}
mysqli_select_db($connect,"student") or die("couldn't connect to db");
$un=$_POST["username"];
$pw=$_POST["Password"];

$sql="SELECT * FROM studentinfo WHERE stUserName='$un' AND stPassword='pw'";

$query=mysqli_query($connect,$sql) or die("couldn't find values");

if($query){
    include_once("studentWeb.html");
}
else{
echo ("Invalid username or password");

}


?>

Why does this allow incorrect password and username ? 为什么这允许使用错误的密码和用户名 ？

Answer 1

Quote these WHERE stName='$un' AND stP='$pw' since we're dealing with strings, which is why you are getting a boolean error. 引用这些WHERE stName='$un' AND stP='$pw'因为我们正在处理字符串，这就是为什么会出现布尔错误的原因。

Plus, use isset() around your executed code since you're using your entire code inside the same file. 另外，由于在同一文件中使用了整个代码，因此请在执行的代码周围使用isset() 。

You stand at getting an undefined index warning upon page entry. 您站在页面进入时收到未定义的索引警告。

Using error reporting will give you that: 使用错误报告将为您提供：

Add error reporting to the top of your file(s) which will help find errors. 将错误报告添加到文件顶部，这将有助于发现错误。

<?php 
error_reporting(E_ALL);
ini_set('display_errors', 1);

// rest of your code

Sidenote: Error reporting should only be done in staging, and never production. 旁注：错误报告仅应在登台进行，而不应在生产过程中进行。

Also add or die(mysqli_error($con)) to mysqli_query() 还要添加or die(mysqli_error($con))到mysqli_query()

Footnotes: 脚注：

Your present code is open to SQL injection . 您当前的代码可以进行SQL注入 。

Use prepared statements , or PDO with prepared statements , they're much safer . 使用预处理语句或将PDO与预处理语句结合 使用起来更加安全 。

Edit: 编辑：

If you wish to include a file (show contents of), do the following, which I am under the impression you wish to do: 如果您希望包括一个文件（显示其内容），请执行以下操作，我对您的印象是：

<?php if($isLogged) { 

    include 'studentWeb.html';
}
?>

This is to be replaced by your present code: 这将由您当前的代码代替：

<?php if($isLogged) {?>  
<applet code="studentWeb.html" width="32" height="32" alt="Couldn't launch applet" title="Student Details">  
</applet>  
<?php }?>

You can also try: 您也可以尝试：

<?php $file = file_get_contents('studentWeb.html', true);
  echo $file;
?>

or '../foldername/studentWeb.html' depending on where the file is located. 或'../foldername/studentWeb.html'具体取决于文件的位置。

The <applet> tag is not supported in HTML5. HTML5不支持<applet>标记。 Use the <object> tag instead if you are having problems with your present code. 如果当前代码有问题，请使用<object>标记。 Yet, applets usually have the .class extension. 但是，小程序通常具有.class扩展名。

Consult http://www.tutorialspoint.com/html/html_applet_tag.htm 咨询http://www.tutorialspoint.com/html/html_applet_tag.htm

Edit #2: 编辑＃2：

As per your edit, change: 根据您的修改，更改：

if($query){
    include_once("studentWeb.html");
}
else{
echo ("Invalid username or password");

}

to: 至：

$numrows = mysqli_num_rows($query);

if($numrows > 0){
   include_once("studentWeb.html");
}

Answer 2

mysqli_query() is returning false because of failure. 由于失败， mysqli_query()返回false 。 Use $query="SELECT * FROM studentinfo WHERE stName=" . $un . " AND stP= " . $pw; 使用$query="SELECT * FROM studentinfo WHERE stName=" . $un . " AND stP= " . $pw; $query="SELECT * FROM studentinfo WHERE stName=" . $un . " AND stP= " . $pw; Moreover, this is not a good way because if your query fails, it will return boolean false which will again generate error for mysqli_num_rows() 而且，这不是一个好方法，因为如果query失败，它将返回布尔值false ，这将再次为mysqli_num_rows()生成错误。

You may look into this for further information and this for better ways to query your database. 你可以看看到这个进一步的信息和这个更好的方式来查询数据库。

在网页中创建小程序

问题描述

2 个解决方案

解决方案1
2 已采纳 2014-11-15 17:32:32

解决方案2
0 2014-11-15 17:37:41

在网页中创建小程序

问题描述

2 个解决方案

解决方案1 2 已采纳 2014-11-15 17:32:32

解决方案2 0 2014-11-15 17:37:41

解决方案1
2 已采纳 2014-11-15 17:32:32

解决方案2
0 2014-11-15 17:37:41