[英]login form with ajax and javascript
I tried to use ajax to show error message if username or password are incorrect, and to go to tabeeb.html or secc.php if they are correct. 如果用户名或密码不正确,我尝试使用ajax显示错误消息,如果正确,则尝试使用tabeeb.html或secc.php。 but instead of going to the correct page, the page is presented in the dialog, like this img . 但是,而不是转到正确的页面,该页面显示在对话框中,如img所示 。 what is the wrong? 怎么了
this is the form 这是表格
<form method="post" class="bounceInDown animated">
<input type="text" name="u" id="u" placeholder="User Name" class="bbb"><br>
<input type="password" name="p" id="p" placeholder="Password" class="bbb">
<input type="button" onClick="myFunction()" name="loginbtn" class=" btn
btn-primary " value="Log in">
</form>
this is myfunction() function 这是myfunction()函数
function myFunction(){
var u = document.getElementById("u").value;
var p = document.getElementById("p").value;
var dataString = 'u1=' + u + '&p1=' + p ;
if (u == '' || p == '') {
alert("Please Fill All Fieldsssssss");}
else {
// AJAX code to submit form.
$.ajax({
type: "POST",
url: "ajaxjss.php",
data: dataString,
cache: false,
success: function(html) {
alert(html);
}
});
}
return false;
and this is ajaxjss.php 这是ajaxjss.php
<?php
// Fetching Values From URL
$u2 = $_POST['u1'];
$p2 = $_POST['p1'];
$connection = mysql_connect("localhost", "root", "password");
// Establishing Connection with Server..
$db = mysql_select_db("project", $connection); // Selecting Database
if (isset($_POST['u1'])) {
// من تحت هو التجربة
$result = mysql_query("select * from test where student_name = '$u2' ") or
die("faild");
$row = mysql_fetch_array($result);
if($row['student_name'] == $u2 && $row['student_password']==$p2)
{ $NoOFLetter = strlen($u2);
$result = substr($u2, $NoOFLetter-3, $NoOFLetter);
if($result == "doc"){header('Location: tabeeb.html'); exit; }
if($result == "sec"){header('Location: secc.php'); exit;}
exit; }
else{
$error_message = "Error in username or password!";}
// كل اللى فوق هو التجربة
if(!$query){
echo "error";
}else{echo "Form Submitted succesfully";}}
mysql_close($connection); // Connection Closed
?>
Now check it 现在检查一下
function myFunction(){ var u = document.getElementById("u").value; var p = document.getElementById("p").value; var dataString = 'u1=' + u + '&p1=' + p ; if (u == '' || p == '') { alert("Please Fill All Fieldsssssss");} else { // AJAX code to submit form. $.ajax({ type: "POST", url: "ajaxjss.php", data: dataString, cache: false, success: function(html) { if(html=="error"){ alert(html); } else{ window.location.href="secc.php"; } } }); } return false;
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