如何將php，sql和Javascript連接在一起，

Question

我正在登錄頁面，我有javascript做驗證（檢查字段是否為空）sql存儲數據和php做什么php做（idk）....無論如何當我按提交它告訴我不能POST /登錄.PHP

是在那里測試它在網站上，看看它是否真的有效或代碼完全錯誤。

<?php

$server = 'localhost';
$username = 'root';
$passowrd = 'cosc_453';
$dbname = 'login'


if(!empty($_POST['user']))


{ $query = mysql_query("SELECT * FROM UserName where userName ='$_POST[user]' AND pass = '$_POST[pass]'") or die(mysql_error()); 

$row = mysql_fetch_array($query) or die(mysql_error());


{ $_SESSION['userName'] = $row['pass']; echo "SUCCESSFULLY LOGIN TO USER PROFILE PAGE..."; } 



else { echo "SORRY... YOU ENTERD WRONG ID AND PASSWORD... PLEASE RETRY..."; 
    }
  }
 }

if(isset($_POST['submit'])) 


{ SignIn();

 } ?>

php外部

     function validate(){

   if ( document.getElementById (user).value=="")
      {
       alert ("Please enter your user name");
   }
   else if ( document.getElementById(pass).value=="")
alert("Please enter you password");
 else {
alert("Processing Login........");
}
      }

javscript外部

  CREATE TABLE UserName ( 
 UserNameID int(9) NOT NULL auto_increment,
userName VARCHAR(40) NOT NULL, 
 pass VARCHAR(40) NOT NULL, 
PRIMARY KEY(UserNameID) );

 INSERT INTO 
UserName (userName, pass) 
VALUES
 ("cosc" , "453");

sql外部

<!DOCTYPE HTML> 
 <html>
<head>
<title>Sign-In</title>
<link rel="stylesheet" type="text/css" href="home.css">
<script src ="login.js"></script> 
</head> 
<body id="body-color"> 

<div id="Sign-In">
<fieldset style="width:30%">
<legend>LOG-IN HERE</legend>

<form method="Post" action="login.php" submit =" validate()"> 

User:<br><input type="text" name="user" size="40"><br> 

Password:<br><input type="password" name="pass" size="40"><br>

<input id="button" type="submit" name="submit" value="Log-In">

   </form> 
  < /fieldset>
   </div> 
  </body> 
        </html>

Answer 1

您的mysql沒有與數據庫的連接。 請停止使用mysql，改為使用mysqli

<?php

    $con = mysqli_connect("localhost","root","cosc_453","login");

    // Check connection
    if (mysqli_connect_errno())
    {
       echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $sql = "SELECT * FROM UserName WHERE userName ='".$_POST[user]."' AND pass = '".$_POST[pass]."'";
    $result = mysqli_query($conn,$sql);
    $count_result = mysqli_num_rows($result);

    // Login Success URL

    if($count_result  == 1)
    {
        // If you validate the user you may set the user cookies/sessions here
        #setcookie("logged_in", "user_id");
        #$_SESSION["logged_user"] = "user_id";

        $_SESSION["secret_id"] = $row['secret_id'];

        if($row['level'] == 1)
        {
            // Set the redirect url after successful login for admin
            $resp['redirect_url'] = 'admin/';
        }

        else if($row['level'] == 2)
        {
            // Set the redirect url after successful login for user
            $resp['redirect_url'] = 'user/';
        }

    }
    else
    {
         echo "Invalid username or pass";
    }


?>

Answer 2

要添加Eh Ezani所說的內容，您的HTML中存在問題。 當我相信你的意思是onsubmit時，你的表單屬性會讀取提交。 可能想嘗試類似的東西。

<form method="Post" action="login.php" onsubmit ="return validate()"> 

     User:<br><input type="text" name="user" size="40"><br> 

     Password:<br><input type="password" name="pass" size="40"><br>

     <input id="button" type="submit" name="submit" value="Log-In">

</form>

此外，“使用MySQLi超過舊的MySQL功能。”i“代表”改進“。改進列表可以在文檔中找到。

- 來自

mysqli和mysql之間的區別？

如何將php，sql和Javascript連接在一起，

問題描述

2 個解決方案

解決方案1
0 2016-05-12 04:17:10

解決方案2
0 2016-05-12 05:08:15

如何將php，sql和Javascript連接在一起，

問題描述

2 個解決方案

解決方案1 0 2016-05-12 04:17:10

解決方案2 0 2016-05-12 05:08:15

解決方案1
0 2016-05-12 04:17:10

解決方案2
0 2016-05-12 05:08:15