错误:创建简单的php&时出错mysql登录:( [英] Error: Getting error while creating simple php & mysql login :(

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问题描述

这是我第一次尝试做一个简单的PHP& MySql登录但收到此错误 - >不推荐使用:mysql_connect():不推荐使用mysql扩展,将来会删除它:在第11行的C:\ wamp \ www \ _login_form \ dbConfig.php中使用mysqli或PDO



和mysql和mysqli有什么区别?



这里分别是我的login.php代码和dbConfig.php代码:

login.php

This is first time I am trying to do a simple PHP & MySql login but am getting this error --> Deprecated: mysql_connect(): The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead in C:\wamp\www\login_form\dbConfig.php on line 11

and what is the difference between mysql and mysqli ?

here is my login.php code and dbConfig.php code respectively:
login.php

<?php include "dbConfig.php";



$msg = "";

if ($_SERVER['REQUEST_METHOD'] == 'POST') {

    $name = $_POST["name"];

    $password = md5($_POST["password"]);

     if ($name == '' || $password == '') {

        $msg = "You must enter all fields";

    } else {

        $sql = "SELECT * FROM members WHERE name = '$name' AND password = '$password'";

        $query = mysql_query($sql);



        if ($query === false) {

            echo "Could not successfully run query ($sql) from DB: " . mysql_error();

            exit;

        }



        if (mysql_num_rows($query) > 0) {

            header('Location: www.google.com');
            exit;
        }

        $msg = "Username and password do not match";
    }
}
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>discussdesk.com - Login form in PHP mysql</title>
<meta name="description" content=""/>
<meta name="keywords" content=""/>
<link href="style.css" rel="stylesheet" type="text/css">

</head>
<body>

    <form name="frmregister"action="<?= $_SERVER['PHP_SELF'] ?>" method="post" >
        <table class="form" border="0">

            <tr>
            <td></td>
                <td style="color:red;">
                <?php echo $msg; ?></td>
            </tr>

            <tr>
                <th><label for="name"><strong>Name:</strong></label></th>
                <td><input class="inp-text" name="name" id="name" type="text" size="30" /></td>
            </tr>
            <tr>
                <th><label for="name"><strong>Password:</strong></label></th>
                <td><input class="inp-text" name="password" id="password" type="password" size="30" /></td>
            </tr>
            <tr>
            <td></td>
                <td class="submit-button-right">
                <input class="send_btn" type="submit" value="Submit" alt="Submit" title="Submit" />

                <input class="send_btn" type="reset" value="Reset" alt="Reset" title="Reset" /></td>

            </tr>
        </table>
    </form>

<div style="line-height: 30px; margin-left: 307px;"><b>Name:</b> discussdesk <br/>  <b>Password:</b> discussdesk</div>
<div style="line-height: 30px; margin-left: 207px;">For More Info: <a href="http://www.discussdesk.com">Visit our Website</a></div>

</body>
</html>







dbConfig.php






dbConfig.php

<?php

define ("DB_HOST", "localhost"); // set database host

define ("DB_USER", "root"); // set database user

define ("DB_PASS",""); // set database password

define ("DB_NAME","innodb"); // set database name

$link = mysql_connect(DB_HOST, DB_USER, DB_PASS) or die("Couldn't make connection.");

$db = mysql_select_db(DB_NAME, $link) or die("Couldn't select database");

?>

推荐答案

msg = < span class =code-summarycomment>;

if
msg = ""; if (


_SERVER ['REQUEST_METHOD'] = = POST) {

_SERVER['REQUEST_METHOD'] == 'POST') {


name =
name =


这篇关于错误:创建简单的php&amp;时出错mysql登录:(的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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