用phpmyadmin连接php
用phpmyadmin连接php
提问于 2018-06-21 05:13:03
我正在建立一个简单的用户登录页面，当我输入我的用户名和密码时，没有任何反应，它似乎没有连接到数据库
下面是我的html代码和一些CSS，以及下面的php代码：
<!DOCTYPE html>
<html lang="en">
<head>
<!-- Required meta tags -->
<meta charset="utf-8">
<meta name="viewport"
content="width=device-width, initial-scale=1, shrink-to-fit=no">
<!-- Bootstrap CSS -->
<link rel="stylesheet" href="/css/bootstrap.min.css">
<title>Intacs Login</title>
<!-- inner css -->
<style>
#login_logo {
margin: 15px 10px 10px 0;
display: block;
margin-left: auto;
margin-right: auto;
}
.container{
background-color: white;
border-radius: 30px;
border-color: lightgray;
border-style:outset;
width: 320px;
margin-top: 30px;
margin:auto;
padding-top: 50px;
padding-bottom: 50px;
}
#login_table{
margin-top: auto;
margin-left: auto;
margin-right: auto;
}
#login_remember{
text-align: center;
}
</style>
</head>
<body>
<!-- Optional JavaScript -->
<!-- jQuery first, then Popper.js, then Bootstrap JS -->
<script src="/js/jquery-3.3.1.slim.min.js"></script>
<script src="/js/popper.min.js"></script>
<script src="/js/bootstrap.min.js"></script>
<!-- HTML code -->
<div class="container">
<div id="login_form">
<div id="login_block">
<img id="login_logo" src="img/Intacs Master Logo.jpg"
width="187" height="63"> <!--div container-->
</div>
</div>
<table id="login_table" cellspacing="1" cellpadding="3" border="0">
<tbody>
<tr>
<form method="post" action="connect.php"> <!--form to connect to php file called connect.php -->
<th>Username: </th>
<td>
<input type="text" name="username" id="usernamefield" class="loginfield" validate="/^[a-z0-9_]{2,}$/i" valmsg="Please enter a valid username." value="" size="14" maxlength="32">
</td>
</tr>
<tr>
<th>Password: </th>
<td nowrap="nowrap">
<input type="password" name="password"
id="passwordfield" class="loginfield" validate="/^[^\s]{4,}$/"
valmsg="Please enter a valid password." value="" size="14"
maxlength="32">
</td>
</form>
</tr>
</tbody>
</table>
<div id="login_remember"><input type="checkbox" name="remember" value="1"> Remember me</div>
<br />
<center><input type="submit" value="Submit"></center>
</body>
</html>
<?php
function Db() {
$host = "localhost:8888";
$username = "root";
$password = "";
$db = "intacslogin";
$conn = new mysqli($host, $username, $password, $db);
if(!$conn){
die("Could not connect");
}
}
if(isset($_POST['login'])){
$uid = trim ($_POST['username']);
$pwd = trim($_POST['password']);
if($uid ==""){
$err[] = "Username is missing";
} else if($pwd == ""){
$err[] = "Password is missing";
} else{
$db = Db();
$uid = $db->real_escape_string($uid);
$pwd = $db->real_escape_string($pwd);
$sql = "SELECT * FROM users
WHERE username = '$uid'
and password = '$pwd'";
$result = $db->query($sql);
}
}
?>请帮助，谢谢和快乐的编码，谢谢你的帮助。这是一个很小的个人项目
回答 1
Stack Overflow用户
发布于 2018-06-21 07:20:34
<?php
//THIS FUNCTION SHOULD RETURN A CONNECTION TO BE USED IN A QUERY
function Db() {
$host = "localhost:8888";
$username = "root";
$password = "";
$db = "intacslogin";
//MAKE SURE U CONFIRM A SUCCESSFULL CONNECTION LIKE SO
if ($conn = new mysqli($host, $username, $password, $db)) {
return $conn;
}else {
return false;
}
//if(!$conn){
// die("Could not connect");
//}
}
if(isset($_POST['login'])){
$uid = trim ($_POST['username']);
$pwd = trim($_POST['password']);
if($uid ==""){
$err[] = "Username is missing";
} else if($pwd == ""){
$err[] = "Password is missing";
} else{
if (!$conn) {
//you can die() becouse it returned false
}else {
//proced when we a valid connection
//dont use $db use $conn
$uid = $conn->real_escape_string($uid);
$pwd = $conn->real_escape_string($pwd);
$sql = "SELECT * FROM users
WHERE username = '$uid'
and password = '$pwd'";
$result = $conn->query($sql);
}
}
}
?>页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50957105
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