i'm trying to add data to database with jquery. My code is here, but its not working.
Using bootstrap form, with that form i'm trying to send input datas to specific page (in this situation, adduser.php) in that page, i'm trying to check this values to database for be sure there is no same data (i'm checking email adresses)
Can you help me guys?
<script>
$(document).ready(function(){
$('#adduser').click(function(){
var add_name = $('#add_name').val();
var add_surname = $('#add_surname').val();
var add_email = $('#add_email').val();
var add_password = $('#add_password').val();
if(add_name == '' || add_surname == '' || add_email == '' || add_password == '' ){
$('#add_user_error').html("<strong class='text-danger'>*** Please enter all details</strong>");
}else{
$.ajax({
url: "adduser.php",
method: "post",
data:{add_name:add_name,add_surname:add_surname,add_email:add_email,add_password:add_password},
success: function(data){
if (data == 1) {
$('#add_user_error').html("<strong class='text-danger'>This email have in database</strong>");
}else{
$('#add_user_error').html("<strong class='text-success'>Success</strong>");
}
}
}); return false;
}
});
});
</script>
<?php
include ('setup.php'); //Database connection dbc
if(isset($_POST['adduser'])){
$add_name = $_POST['add_name'];
$add_surname = $_POST['add_surname'];
$add_email = $_POST['add_email'];
$add_password = $_POST['password'];
$q = "SELECT * FROM users WHERE email = '$add_email'";
$r = mysqli_query($dbc, $q);
$adduser = mysqli_fetch_assoc($r);
if($adduser['email'] !== $add_email){
$q = "INSERT INTO users (name,surname,email,password) VALUES ('$add_name','$add_surname','$add_email','$add_password')";
$r = mysqli_query($dbc, $q);
}
}
?>
There could be any other mistake too but I think you need to send adduser from ajax. Then only $_POST['adduser'] will be true (it is false now, as it isn't set)
Replace your data line with line below and try
data:{add_name:add_name,add_surname:add_surname,add_email:add_email,add_password:add_password,adduser:1}
In your ajax code pass your form data using serialize method
$.ajax({
url: "adduser.php",
method: "post",
data:$('#yourformid').serialize(),
/* your remaining code as it is */
Related
I have created the login and register page using ajax and php. Some user send me an email "we can not register". But i see they are already registered from database. Users are not redirected after registration. This problem is sometimes said that they had seen on the computer, but mobile in multiple times.
I wonder if there's a mistake I missed, would you check for me?
The Ajax Code:
<script type="text/javascript">
$(document).ready(function(){
var form = $('#registerform');
var submit = $('#register');
var firstname = $('#register_username');
var fullnamereg = $("#reg_fullname");
var regemail = $('#reg_email');
var regpassword = $('#reg_password');
var validated = true;
form.on('submit', function(e) {
// prevent default action
e.preventDefault();
jQuery.ajax({
type: "POST",
url: URL+"register.php",
data: form.serialize(),
beforeSend: function(){},
success:function(data){
if($.trim(data) === 'ok'){
setTimeout(function() {
$('.login_form').html("Registered!");
setTimeout(function() {
window.location.href = siteurl;
}, 1000);
}, 1000);
}
},
error: function(data){
$('#form_error').html('Error, intenta de nuevo.');
},
})
});
});
</script>
and PHP code Here:
if(isset($_POST["reusername"]) && isset($_POST["fullname"]) && isset($_POST["password"]) && isset($_POST["regemail"])){
$firstname = mysqli_real_escape_string($db, $_POST["reusername"]);
$fullname = mysqli_real_escape_string($db, ucfirst(strtolower($_POST["fullname"])));
$password = mysqli_real_escape_string($db, sha1(md5(trim($_POST["password"]))));
$thisEmail = mysqli_real_escape_string($db, $_POST['regemail']);
$registerTime = time();
$ip=$_SERVER['REMOTE_ADDR']; // user ip
$saveUser = mysqli_query($db,"INSERT INTO `users`
(email,username,fullname ,password,registered,ip)
VALUES
('$thisEmail','$firstname','$fullname','$password','$registerTime','$ip')") or die(mysqli_error($db));
if($saveUser){
$getUserID = mysqli_query($db,"SELECT user_id,username FROM users WHERE email = '$thisEmail' AND username = '$firstname'") or die(mysqli_error($db));
$row=mysqli_fetch_array($getUserID,MYSQLI_ASSOC);
$uid=$row['user_id']; // Get User ID
$dbunm = $row['username']; // Get User Username
$_SESSION['user_id'] = $uid; // Session User ID
$hash = sha1($dbunm).$registerTime;
if($hash){
setcookie($cookiename,$hash,time()+31556926 ,'/');
mysqli_query($db, "INSERT INTO `session` (userid, auth) VALUES ('$uid', '$hash')");
echo 'ok';
}
}
}
If the insert fails because the user already exists, then the PHP code will return a 200 success status but it won't echo ok. So, your Javascript code will fall into the success block, but then skip the if statement. Thus, if a user tries to re-register, they will get no error message and no redirect. You could have the PHP code issue a explicit error status:
if($saveUser){
...
} else {
http_response_code(500);
exit;
}
Then, if a registration fails, the AJAX call will fall into the error block.
I have read all the related questions that reference to this topic, but still cannot find answer here. So, php and ajax works great. The problem starts when i try to include json, between php and ajax, to passing data.
here is my ajax:
function likeButton(commentId, userId, sessionUserId) {
// check if the comment belong to the session userId
if(sessionUserId == userId) {
alert("You cannot like your own comment.");
}
else if(sessionUserId != userId) {
var like_upgrade = false;
$.ajax({
url: "requests.php",
type: "POST",
dataType: "json",
data: {
keyLike: "like",
commentId: commentId,
userId: userId,
sessionUserId: sessionUserId,
like_upgrade: like_upgrade
},
success: function(data) {
var data = $.parseJSON(data);
$("#comment_body td").find("#updRow #updComLike[data-id='" +commentId+ "']").html(data.gaming_comment_like);
if(data.like_upgrade == true) {
upgradeReputation(userId);
}
}
});
}
}
Note, that i try not to include this:
var data = $.parseJSON(data);
Also i tried with diferent variable like so:
var response = $.parseJSON(data);
and also tried this format:
var data = jQuery.parseJSON(data);
None of these worked.
here is requests.php file:
if(isset($_POST['keyLike'])) {
if($_POST['keyLike'] == "like") {
$commentId = $_POST['commentId'];
$userId = $_POST['userId'];
$sessionUserId = $_POST['sessionUserId'];
$sql_upgrade_like = "SELECT * FROM gaming_comments WHERE gaming_comment_id='$commentId'";
$result_upgrade_like = mysqli_query($conn, $sql_upgrade_like);
if($row_upgrade_like = mysqli_fetch_assoc($result_upgrade_like)) {
$gaming_comment_like = $row_upgrade_like['gaming_comment_like'];
}
$gaming_comment_like = $gaming_comment_like + 1;
$sql_update_like = "UPDATE gaming_comments SET gaming_comment_like='$gaming_comment_like' WHERE gaming_comment_id='$commentId'";
$result_update_like = mysqli_query($conn, $sql_update_like);
$sql_insert_like = "INSERT INTO gaming_comment_likes (gaming_comment_id, user_id, user_id_like) VALUES ('$commentId', '$userId', '$sessionUserId')";
$result_insert_like = mysqli_query($conn, $sql_insert_like);
$like_upgrade = true;
//json format
$data = array("gaming_comment_like" => $gaming_comment_like,
"like_upgrade" => $like_upgrade);
echo json_encode($data);
exit();
}
}
Note: i also try to include this to the top of my php file:
header('Content-type: json/application');
but still not worked.
What am i missing here?
Don't call $.parseJSON. jQuery does that automatically when you specify dataType: 'json', so data contains the object already.
You should also learn to use parametrized queries instead of substituting variables into the SQL. Your code is vulnerable to SQL injection.
I'm creating a follow button, more or less like the twitter one.
You click the button, and you follow the user.
You click again, and you unfollow the user.
I have done this code
HTML
<div data-following="false" class='heart canal'><i class='fa fa-heart awesome'></i></div>
AJAX
$(document).ready(function() {
$(".heart.canal").click(function() {
if($(".heart").attr("data-following") == '0'){
$(".heart").attr('data-following', '1');
} else if($(".heart").attr("data-following") == '1'){
$(".heart").attr('data-following', '0');
}
var usuario = $(".left h4").attr("data-id");
var seguidor = $("#user_account_info .profile_ball").attr("data-id");
var seguir = $(".heart").attr("data-following");
$.ajax({
type: "POST",
url: "./include/php/follow.php",
data: { user: usuario, follower: seguidor, follow: seguir },
success: function(response) {
if(response == '0'){
$(".heart").addClass("like");
} else if(response == '1'){
$(".heart").removeClass("like");
}
}
});
return false;
});
});
PHP
<?php
$dsn = "mysql:host=localhost;dbname=tapehd;charset=utf8";
$usuario = "root";
$contraseña = "";
$conexion = new PDO($dsn, $usuario, $contraseña);
$resultado = null;
$sql = "";
$user = $_POST["user"];
$seguidor = $_POST["follower"];
$follow = $_POST["follow"];
if($follow == '0'){
$sql = "INSERT INTO seguidores(id_canal, id_seguidor) VALUES('$user', '$seguidor')";
} else if($follow == '1'){
$sql = "DELETE FROM seguidores WHERE id_canal = '$user' AND id_seguidor= '$seguidor'";
}
if($conexion){ $resultado = $conexion->query($sql); }
return $follow;
?>
The problem is, everytime I click the button, I only insert data in the database. I mean, I only create follows.
When I click twice, it doesnt remove the follow.
Is there anyway to insert data when data-following = true and remove it when data-following = false ?
UPDATED
I have changed the boolean false and true for 2 strings, 0 and 1. But it doesn't work anyway.
There are numerous problems here. For one, like #Mark said, you need to understand that when sending ajax requests to PHP, you are sending strings. Also, in your JS, you are binding a click function to the .heart.canal, but then the function changes all elements with that class rather than the actual clicked element. Lastly, once you send the right information to PHP you need to print your results in order to see it in ajax.
Try the following:
JS:
$(document).ready(function () {
$(".heart.canal").click(function () {
var $heart = $(this);
if ($heart.data("following")) {
$heart.data("following", false)
} else {
$heart.data("following", true);
}
var usuario = $(".left").find("h4").data("id");
var seguidor = $("#user_account_info").find(".profile_ball").data("id");
$.ajax({
type: "POST",
url: "follow.php",
data: {user: usuario, follower: seguidor, follow: $heart.data("following")},
success: function (result) {
if (result) {
console.log("true");
} else {
console.log("false");
}
}
});
return false;
});
});
PHP:
$user = (int)$_POST["user"];
$seguidor = (int)$_POST["follower"];
$follow = ($_POST["follow"] === 'true') ? true : false;
if ($follow) {
// insert
} else {
// delete
}
print $follow;
I have built a login script that uses AJAX to submit form data.
The PHP part works fine without AJAX. But the system doesnt work with AJAX Implementation.
It always Displays the below message even though the PHP file returns true[correct username & password] ... Seems like the if condition in Jquery is not working.
Incorrect Username/Password
HTML Result Div
<div id="user-result" align="center"></div>
Jquery
<script type="text/javascript">
$(document).ready(function () {
var form = $('#loginform');
form.submit(function (ev) {
ev.preventDefault();
$.ajax({
type: form.attr('method'),
url: form.attr('action'),
cache: false,
data: form.serialize(),
success: function (data) {
if (data == "true") {
$("#user-result").html("<font color ='#006600'> Logged in | Redirecting..</font>").show("fast");
setTimeout(
function () {
window.location.replace("index.php");
}, 1500);
} else {
$("#user-result").html("<font color ='red'> Incorrect Username/Password</font>").show("fast");
}
}
});
});
});
</script>
fn_login.php
<?php
{
session_start();
include_once 'db_connect.php';
if (isset($_POST))
{
$email = filter_input(INPUT_POST, 'email', FILTER_SANITIZE_STRING);
$logpwd = filter_input(INPUT_POST, 'password', FILTER_SANITIZE_STRING);
$stmt = $conn->prepare("SELECT password FROM manager WHERE email = ? LIMIT 1");
$stmt->bind_param("s", $email);
$stmt->execute();
$stmt->store_result();
// get variables from result.
$stmt->bind_result($password);
$stmt->fetch();
// Check if a user has provided the correct password by comparing what they typed with our hash
if (password_verify($logpwd, $password))
{
$sql = "SELECT * from manager WHERE email LIKE '{$email}' LIMIT 1";
$result = $conn->query($sql);
$row=mysqli_fetch_array($result);
$id = $row['id'];
$conn->query("UPDATE manager SET lastlogin = NOW() WHERE id = $id");
$_SESSION['manager_check'] = 1;
$_SESSION['email'] = $row['email'];
$_SESSION['fullname'] = $row['fullname'];
$_SESSION['designation'] = $row['designation'];
$_SESSION['id'] = $row['id'];
echo "true";
}
else {
die();
}
}
}
?>
Can someone please point out the mistake in the code/practice.
EDIT
Just Tried disabling AJAX, the PHP file works correctly echoing true when username/pass is correct
You have spaces after ?>
So, the AJAX response is having spaces after true.
Solution:
Remove ?> from the end of PHP file.
It will not affect any PHP functionality.
And you AJAX response will be without spaces.
Excluding closing tag ?> from the end of PHP file is standard practice for modern PHP frameworks and CMSs.
Tips for debugging AJAX:
1) Always use Firefox (with Firebug Add) on Or Chrome.
2) Use Console tab of Firebug, to check which AJAX requests are going.
3) Here, you can see input parameters, headers and most important response.
4) So, in short you can debug a whole AJAX request life cycle.
You can echo json_encode(array('success'=>true)) from php code and modify your if condition in jquery with if(data.success){} Your modified code becomes
<?php
{
session_start();
include_once 'db_connect.php';
if (isset($_POST))
{
$email = filter_input(INPUT_POST, 'email', FILTER_SANITIZE_STRING);
$logpwd = filter_input(INPUT_POST, 'password', FILTER_SANITIZE_STRING);
$stmt = $conn->prepare("SELECT password FROM manager WHERE email = ? LIMIT 1");
$stmt->bind_param("s", $email);
$stmt->execute();
$stmt->store_result();
// get variables from result.
$stmt->bind_result($password);
$stmt->fetch();
// Check if a user has provided the correct password by comparing what they typed with our hash
if (password_verify($logpwd, $password))
{
$sql = "SELECT * from manager WHERE email LIKE '{$email}' LIMIT 1";
$result = $conn->query($sql);
$row=mysqli_fetch_array($result);
$id = $row['id'];
$conn->query("UPDATE manager SET lastlogin = NOW() WHERE id = $id");
$_SESSION['manager_check'] = 1;
$_SESSION['email'] = $row['email'];
$_SESSION['fullname'] = $row['fullname'];
$_SESSION['designation'] = $row['designation'];
$_SESSION['id'] = $row['id'];
echo json_encode(array('success'=>true));
}
else {
die();
}
}
}
AND JQuery becomes
<script type="text/javascript">
$(document).ready(function () {
var form = $('#loginform');
form.submit(function (ev) {
ev.preventDefault();
$.ajax({
type: form.attr('method'),
url: form.attr('action'),
cache: false,
data: form.serialize(),
success: function (data) {
if (data.success) {
$("#user-result").html("<font color ='#006600'> Logged in | Redirecting..</font>").show("fast");
setTimeout(
function () {
window.location.replace("index.php");
}, 1500);
} else {
$("#user-result").html("<font color ='red'> Incorrect Username/Password</font>").show("fast");
}
}
});
});
});
</script>
I apologise if this comes across as really stupid. I have searched but can't seem to find an answer. I hope I can explain what it is I am trying to do.
I want to be able to query a database and if there is a record in it to show the record in the span/div or show a not found error message if there isn't.
I have a jquery check up and running to check if a username is in the database, what I want to know is how easy it would be to ammend this to pull all the data and show it in the span/div on the original page.
This is the jquery I have:
$(document).ready(function () {
$('#username').keyup(username_check);
});
function username_check() {
var username = $('#username').val();
if (username == "" || username.length < 2) {
$('#username').css('border', '1px #D5D5D5');
$('#cross').hide();
$('#tick').hide();
} else {
jQuery.ajax({
type: "POST",
url: "check.php",
data: 'username=' + username,
cache: false,
success: function (response) {
if (response == 1) {
$('#username').css('border', '2px #C33 solid');
$('#tick').hide();
$('#cross').fadeIn();
} else {
$('#username').css('border', '2px #090 solid');
$('#cross').hide();
$('#tick').fadeIn();
}
}
});
}
}
Can I do all this on the one page and query the db from the same page, instead of posting it to another page as I don't know how to get the results back to the calling page?
I hope I have explained what I want to do. Apologies if I haven't
Here is the PHP code:
$username = trim(strtolower($_POST['username'])); $username = mysql_escape_string($username); $query = "SELECT adbkid FROM person WHERE adbkid = '$username' LIMIT 1"; $result = mysql_query($query); $num = mysql_num_rows($result); echo $num; mysql_close()
You will normally send ajax requests to pages hosted on your server. So you can't directly access your database without going through your server. You'll need to write a function on your server that queries the database, and then call that function from javascript using ajax.
You can output a string in PHP and then set that text value to an element with jQuery ( $('#element').val(responseFromServer);
or $('#element').html(responseFromServer);
Instead of sending back "1" send back a json response something like:
/* record exists */
{status:1, html:'server generated message about record'}
/* doesn't exist */
{status:0}
This will allow you to still change css based on response data status value
Can use $.post ajax shorthand method:
$.post('check.php',{username: username}, function(response){
var upDateElement=$('#spanID');
if(response.status && response.status== 1){
$('#username').css('border', '2px #C33 solid');
$('#tick').hide();
$('#cross').fadeIn();
upDateElement.html( response.html)
}else{
$('#username').css('border', '2px #090 solid');
$('#cross').hide();
$('#tick').fadeIn();
upDateElement.html('Message html for no record found')
}
},'json')
check.php
$data = array();
$data['exists'] = false;
if(!isset($_POST['username'])) {
echo json_encode($data);
exit();
}
$username = mysql_escape_string($_POST['username']);
$query = "SELECT adbkid FROM person WHERE adbkid = '$username' LIMIT 1";
$result = mysql_query($query);
$row = mysql_fetch_assoc($result);
if(count($row) == 1) {
$data = $row;
$data['exists'] = true;
}
return json_encode($data);
from your jQuery:
success: function(response) {
/**
* For instance, for a table with id, username, password and email you have
* data.exists = true/false;
* data.id = 1;
* data.username = 'foo';
* data.password = 'sample data password';
* data.email = 'foo#bar.com';
*/
if(response.exists === true) {
$('#username').val(response.username);
$('#email').val(response.email);
}
}