我想使用php Web服务完成这个简单的登录任务。我只是试图根据我在 php 中回显的文本结果来验证用户名和密码。
.PHP：
<?php
// Include confi.php
include_once('confi.php');
$found = false;
$email = isset($_POST['email']) ? mysql_real_escape_string($_POST['email']) : "";
$password = isset($_POST['password']) ? mysql_real_escape_string($_POST['password']) : "";
if(!empty($email) && !empty($password)){
$login=mysql_num_rows(mysql_query("select *
from `login`
where `email`='$email'
and `password`='$password'"));
$result =array();
if($login!=0)
{
echo "success";
}
else
{
echo "failed";
}
}
@mysql_close($conn);
/* Output header */
header('Content-type: text/plain');
?>
如果用户名和密码匹配;则显示成功。
Jquery
<script>
$(function () {
$("#logon").click(function () {
var email = $("#username").val();
var password = $("#pass").val();
var dataString = "email=" + email + "&password=" + password;
if ($.trim(email).length > 0 & $.trim(password).length > 0) {
$.ajax({
type: "POST",
url: "http://*****/login.php",
data:dataString,
crossDomain: true,
cache: false,
beforeSend: function () { $("#logon").html('Connecting...'); },
success: function (data) {
if (data == "success") {
alert(result+"You are in");
localStorage.login = "true";
localStorage.email = email;
window.location.href = "test.html";
}
else if (data == "failed") {
alert("Login error");
$("#logon").html('Login');
}
}
});
}
});
});
</script>
您缺少 JSON 函数，您必须对要发送回请求的内容进行编码
<?php
/*output the header here*/
header("Content-Type: application/json");
// Include confi.php
include_once('confi.php');
$response['Status'];// declare a variable to store msg
$found = false;
$email = isset($_POST['email']) ?
mysql_real_escape_string($_POST['email']) : "";
$password = isset($_POST['password']) ?
mysql_real_escape_string($_POST['password']) : "";
if(!empty($email) && !empty($password)){
$login=mysql_num_rows(mysql_query("select *
from `login`
where `email`='$email'
and `password`='$password'"));
$result =array();
if($login!=0)
{
$response['Status']=""success";
}
else
{
$response['Status']="failed";
}
}
@mysql_close($conn);
在这里进行更改
$result= json_encode($message);
echo $result;
？>
在你 jQuery 数据到 AJAX 函数添加
dataType:"json",
success: function (data) {
if (data.Status == "success") { //here check the condition
alert(result+"You are in");
localStorage.login = "true";
localStorage.email = email;
window.location.href = "test.html";
}
else if (data.Status== "failed") { //here check the condition
alert("Login error");
$("#logon").html('Login');
}
}