这些获取用户输入的用户名和密码,并且应该发送到validate.php文件以检查记录是否存在于数据库中。并收到回复。
<script type="text/javascript">
function validatelogin() {
var user=document.forms["login"]["username"].value;
var pass=document.forms["login"]["password"].value;
if(user==null || user=="") {
alert("Please enter username");
$('#username').focus();
return false;
} else if(pass==null || pass=="") {
alert("Please enter password");
$('#password').focus();
return false;
} else if(user!==null || user!=="" || pass!==null || pass!=="") {
// These gets the username and password inputted by the user and is
// expected to be sent to the validate.php file to check whether the
// records is present in the database or not. and receives a response.
$.ajax({
url : "validate.php",
type : 'POST',
data : $('#login').serialize(),
success : function(msg) {
$('#login-box').html(msg);
}
});
return false;
} else {
alert("form submitted");
}
}
</script>
php文件 - 用于查询数据库以检查现有记录,并期望触发回到ajax的真或假响应 - 请问我该如何做到这一点。然后代码现在可以决定是否登录
<?php
error_reporting(0);
mysql_pconnect("localhost","root","");
mysql_select_db("sim_tracker");
$username=$_POST['username'];
$password=$_POST['password'];
mysql_query("select * from tbl_user where
username='$username' and password='$password'");
$row=mysql_affected_rows();
if($row>0) {
return false;
echo"$username,$password";
}
?>
HTML
<div id="login-box" class="login-popup">
<a href="#" class="close">
<img src="images/close_pop.png" class="btn_close"
title="Close Window" alt="Close" />
</a>
<form method="post" id="login" class="signin"
action="#" onSubmit="return validatelogin()">
<fieldset class="textbox">
<label class="username">
<span>Username or email</span>
<input id="username" name="username" value=""
type="text" autocomplete="on" placeholder="Username">
</label>
<label class="password">
<span>Password</span>
<input id="password" name="password" value=""
type="password" placeholder="Password">
</label>
<button class="submit button" type="submit">Sign in</button>
<p>
<a class="forgot" href="#">Forgot your password?</a>
</p>
</fieldset>
</form>
</div>
答案 0 :(得分:0)
我在您的代码中看到至少2个问题:
$result = mysql_query("select * from tbl_user where username='$username' and password='$password'"); $row_count = mysql_num_rows(); if ($row_count > 0) { // login correct, add some code to actually log the user in (i. e. some session stuff) echo "You were logged in!"; // if you want just true or false say echo 1 instead } else { // login incorrect echo "Wrong username or password!" // or again echo 0 }