我做了一個PhoneGap應用程序,當記錄日誌記錄是否成功,或者即使不成功時,我也遇到了在else部分中處理的相同錯誤。 這是我的Javascript Ajax登錄代碼。PhoneGap登錄Error
$(document).ready(function() {
$("#login").click(function() {
var email = $("#email").val();
var password = $("#password").val();
var dataString = "email=" + email + "&password=" + password + "&login=";
if ($.trim(email).length > 0 & $.trim(password).length > 0) {
$.ajax({
type: "POST",
url: "http://192.168.43.173/viamobile/login.php",
data: dataString,
crossDomain: true,
cache: false,
beforeSend: function() {
$("#login").html('Connecting...');
},
success: function(data) {
if (data == "success") {
localStorage.login = "true";
localStorage.email = email;
window.location.href = "index.html";
} else if (data = "failed") {
alert("Login error");
$("#login").html('Login');
}
}
});
}
return false;
});
});
和我的PHP腳本
<?php
include "db.php";
if(isset($_POST['login']))
{
$email = $_POST['email'];
$password = $_POST['password'];
$login = mysqli_num_rows(mysqli_query($con,"select * from `admin` where `username`='$email' and `password`='$password'"));
if($login!=0)
{
echo "success";
}
else
{
echo "failed";
}
}
?>
<?php
header("Access-Control-Allow-Origin: *");
$con= mysqli_connect("localhost","root","","phonegap");
?>
我真的需要驗證用戶,並根據用戶的訪問重定向到不同的頁面。但是,既然這個錯誤我不能這樣做。可以有人告訴我我錯了哪裏?
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