Pop-up message after submitting a form with php - php

I'm trying to get a pop-up message saying if it was successfully submitted or not without having to go to a different page.
Now chrome gives me the pop-up message but it redirects me to a blank page after.
Here is my current code.
<?php
include "header.php";
include "conexao.php";
echo "<h1 align='center'>Pagina para alterar produtos</h1><div class='container'><hr>";
$referencia=$_GET['id'];
$sql = "SELECT * ";
$sql = $sql . " FROM tb_produto ";
$sql = $sql . " WHERE pr_codigo='".$referencia."'";
$produtos = $db->query($sql);
foreach ($produtos as $produto) {
$referencia = $produto["pr_codigo"];
$nome = $produto["pr_descricao"];
$preco = $produto["pr_preco"];
$disponivel = $produto["disponivel"];
}
echo "<h2>Referencia: ".$referencia."</h2>";
echo "<h2>Nome: ".$nome."</h2><hr>";
?>
<form action="confirmaAlterar.php">
<div class="form-group">
<label>Referencia</label>
<input class="form-control" type="text" name="referencia" value="<?php echo $referencia?>">
</div>
<div class="form-group">
<label>Nome</label>
<input class="form-control" type="text" name="nome" value="<?php echo $nome?>">
</div>
<div class="form-group">
<label>Preço</label>
<input class="form-control" type="text" name="preco" value="<?php echo $preco?>">
</div>
<button class="btn btn-primary">Alterar</button>
</form>
Here is where it submits the information of the form.
<?php
include "header.php";
include "conexao.php";
$nome=$_GET['nome'];
$referencia=$_GET['referencia'];
$preco=$_GET['preco'];
$sql="UPDATE tb_produto SET pr_descricao='".$nome;
$sql.="', pr_preco=".$preco;
$sql.= " WHERE pr_codigo='".$
try{
$comando=$db->prepare($sql);
$comando->execute();
echo "<script type='text/javascript'>alert('submitted successfully!')</script>";
header( "refresh2;Location:index.php" );
}
catch (PDOException $e){
echo "A";
}

To pass values using ajax. Form:
<form id="form">
<input type="text" value="test" name="akcija">
</form>
All inputs fields values in your form will be passed.
Ajax:
jQuery(function(){
$('#form').on('submit', function (e) { //on submit function
e.preventDefault();
$.ajax({
type: 'post', //method POST
url: 'yoururl.php', //URL of page where u place query and passing values
data: $('#form').serialize(), //seriallize is passing all inputs values of form
success: function(){ //on success function
$("#input").attr("disabled", true); //example
$("#input").removeClass('btn-primary').addClass('btn-success');//example
},
});
}
});
And on the ajax page you can get values by
$akcija = $_POST['akcija']

for this Problem you must use ajax method .
1- create html form and all input Required .
<form id="contactForm2" action="/your_url" method="post">
...
</form>
2- add jQuery library file in the head of html page .
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js">
</script>
...
3- add this method Under the jQuery library
<script type="text/javascript">
var frm = $('#contactForm2');
frm.submit(function (ev) {
$.ajax({
type: frm.attr('method'),
url: frm.attr('action'),
data: frm.serialize(),
success: function (data) {
if(data == 'pass')
alert('ok');
else(data == 'fail')
alert('no');
}
});
ev.preventDefault();
});
</script>
4- in your_url .php file
<?php
$a = ok ;
if( $a == 'ok' ){
echo 'pass';
}else{
echo 'fail';
}
?>
this top answer is easy management form with jquery , but you need managment Complex form better use this library http://jquery.malsup.com/form/

Related

Submit form using Ajax/Jquery and check result?

I am wanting to submit a form using ajax to go to my page 'do_signup_check.php'.
There it will check the email address the user entered against the database to see if there is a match.
If there is a match I want my ajax form to redirect the user to the login.php page.
If there isn't a match I want it to load my page 'do_signup.php'
for some reason the code seems to be doing nothing. Please can someone show me where I am going wrong?
My Ajax form:
<html lang="en">
<head>
<script src="https://www.google.com/recaptcha/api.js" async defer></script>
<script src="https://www.google.com/recaptcha/api.js?onload=onloadCallback&render=explicit"
async defer>
</script>
<?php include 'assets/config.php'; ?>
<script>
$(function () {
$('form').on('submit', function (e) {
e.preventDefault();
$.ajax({
type: 'post',
url: 'do_signup_check.php',
data:{"name":name,"email":email},
success: function () {
if(result == 0){
$('.signup_side').fadeOut(500).promise().done(function() {
$('.signup_side').load('do_signup.php',function(){}).hide().fadeIn(500);
});
}else{
$('.signup_side').fadeOut(500).promise().done(function() {
$('.signup_side').load('login.php',function(){}).hide().fadeIn(500);
}
});
});
});
</script>
</head>
<body>
<div class="sign_up_contain">
<div class="container">
<div class="signup_side">
<h3>Get Onboard.</h3>
<h6>Join the directory.</h6>
<form id="signup" action="" method="POST" autocomplete="off" autocomplete="false">
<div class="signup_row action">
<input type="text" placeholder="What's your Name?" name="name" id="name" class="signup" autocomplete="new-password" autocomplete="off" autocomplete="false" required />
<input type="text" placeholder="Got an Email?" name="email" id="email" class="signup" autocomplete="new-password" autocomplete="off" autocomplete="false" required />
<div class="g-recaptcha" style="margin-top:30px;" data-sitekey="6LeCkZkUAAAAAOeokX86JWQxuS6E7jWHEC61tS9T"></div>
<input type="submit" class="signup_bt" name="submit" id="submt" value="Create My Account">
</div>
</form>
</div>
</div>
</div>
</body>
</html>
My Do_Signup_Check.php page:
<?php
session_start();
require 'assets/connect.php';
$myName = $_POST["name"];
$myEmail = $_POST["email"];
$check = mysqli_query($conn, "SELECT * FROM user_verification WHERE email='".$myEmail."'");
if (!$check) {
die('Error: ' . mysqli_error($conn));
}
if (mysqli_num_rows($check) > 0) {
echo '1';
} else {
echo '0';
}
?>
Try to use input id 'submt' to trigger the form as such :
$("#submt").click(function(){
e.preventDefault();
//your logic
You have not included the jquery in your page, the function is also missing the closing parentheses.
Include the jquery at the top
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
Replace your function with the following code:-
<script>
$(function () {
$('form').on('submit', function (e) {alert(123);
e.preventDefault();
$.ajax({
type: 'post',
url: 'receive-callback.php',
data:{"name":name,"email":email},
success: function () {
if(result == 0){
$('.signup_side').fadeOut(500).promise().done(function() {
$('.signup_side').load('do_signup.php',function(){}).hide().fadeIn(500);
});
}else{
$('.signup_side').fadeOut(500).promise().done(function() {
$('.signup_side').load('login.php',function(){}).hide().fadeIn(500);
});
}
}
});
});
});
</script>
After receiving success in Ajax and doing all necessary logic (fadeOut etc) just relocate user to login.php:
window.location.href = 'path/to/login.php';
Otherwise, in error function (it is goes just after success function) do the next:
window.location.href = 'path/to/do_signup.php';
Upd:
In case you want to have some piece of code from another file, you can use jQuery method load.
$('#some-selector').load('path/to/course/login.php');
It's a very basic example of using 'load', go google around it to explore more.
Please Try this code.
$.ajax({
type: 'post',
url: 'do_signup_check.php',
data:{"name":name,"email":email},
success: function (result) {
if(result == 0){
window.location.href = 'do_signup.php';
}else{
window.location.href = 'login.php';
}
});
My Do_Signup_Check.php page:
<?php
session_start();
require 'assets/connect.php';
$myName=$_POST["name"];
$myEmail=$_POST["email"];
$check = mysqli_query($conn, "SELECT * FROM user_verification WHERE email='".$myEmail."'");
if (!$check) {
die('Error: ' . mysqli_error($conn));
}
if(mysqli_num_rows($check) > 0){
echo json_encode(array('result'=>'1'));
exit;
}else{
echo json_encode(array('result'=>'0'));
exit;
}
?>
I hope this will work for you.

Ajax code not working even if it is (i think) correct

The function that this do is when the user clicked the button, it will execute the Ajax codes and then get the value of the input and send it to the PHP file and then send it back to the Ajax code to display the message from the MySQL table.
I tried changing my codes, changing div ids, changing syntax, clearing block of codes but none seems to work.
AJAX
<script>
$(document).ready(function() {
$("#snd").click(function() {
var msgg = $('input[name=message]').val();
$.ajax({
type: "POST",
url: 'automatedchat_func.php',
data: {newmsg: msgg},
success: function(data) {
$("#conversation").html(data);
}
});
});
});
</script>
HTML UPDATED
<div class="convo">
<div class="convo_field" id="conversation">
</div>
<div class="obj">
<div class="txtbox">
<form method="POST">
<input type="input" id="msg" name="message" placeholder="Type exact or related word(s) of your question"/>
</form>
</div>
<div class="but_send"><button id="snd" name="send">SEND</button></div>
</div>
</div>
PHP UPDATED
<?php
include 'database/connect.php';
session_start();
$sql = "SELECT * FROM ai WHERE keywords LIKE '%$_POST[message]%' OR '$_POST[message]%_' OR '$_POST[message]_'";
$result = $conn->query($sql);
if ($row = $result->fetch_assoc()) {
echo "Hi ". $_SESSION['name'] .".<br> " . $row['message'];
}
?>
Changes with suggestion(in comment):-
<div class="convo">
<div class="convo_field" id="conversation">
</div>
<div class="obj">
<div class="txtbox">
<input type="input" id="msg" name="message" placeholder="Type exact or related word(s) of your question"/>
</div><!-- form not requird -->
<div class="but_send"><button id="snd" name="send">SEND</button></div>
</div>
</div>
<!-- Add jquery library so that jquery code wiil work -->
<script>
$(document).ready(function() {
$("#snd").click(function() {
var msgg = $('#msg').val(); // id is given so use that, its more easy
$.ajax({
type: "POST",
url: 'automatedchat_func.php',
data: {newmsg: msgg},
success: function(data) {
$("#conversation").html(data);
}
});
});
});
</script>
automatedchat_func.php(must be in the same working directory where your above html file exist)
<?php
error_reporting(E_ALL);// check all type of error
ini_set('display_errors',1);// display all errors
include 'database/connect.php';
session_start();
$final_result='';//a variable
if(isset($_POST['newmsg']) && !empty($_POST['newmsg'])){// its newmsg not message
$message = $_POST['newmsg'];
$sql = "SELECT * FROM ai WHERE keywords LIKE '%$message%' OR '$message%' OR '$message'"; // check the change ere
$result = $conn->query($sql);
while($row = $result->fetch_assoc()) { // while needed
$final_result .="Hi ". $_SESSION['name'] .".<br> " . $row['message']."<br>";
}
}else{
$final_result .="Hi please fill the input box first";
}
echo $final_result; // send final result as response to ajax
?>
Note:- your query is still vulnerable to SQL Injection. So read for prepared statements and use them

Displaying a success message after form submission

I'm trying to understand jquery and particularly inserting and displaying data in a mysql table using ajax.
I have been experimenting with this code which inserts and displays records from a mysql database. I'm now trying to get it to display the success message in the div with the id "info" whilst also displaying all the records. I seem to only do one but never both. Many thanks.
form.php
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js">
</script>
<script src="https://netdna.bootstrapcdn.com/bootstrap/3.0.0/js/bootstrap.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
function showComment(){
$.ajax({
type:"post",
url:"process_ajax.php",
data:"action=showcomment",
success:function(data){
$("#comment").html(data);
}
});
}
showComment();
$("#button").click(function(){
var username=$("#username").val();
var review=$("#review").val();
$.ajax({
type:"post",
url:"process_ajax.php",
data:"username="+username+"&review="+review+"&action=addcomment",
success:function(data){
$("#info").html(data);
}
});
});
});
</script>
</head>
<body>
<form>
Username : <input type="text" name="username" id="username"/>
</br>
Review : <input type="text" name="review" id="review" />
</br>
<input type="button" value="Send Comment" id="button">
</form>
<div id="info" />
<ul id="comment"></ul>
</body>
</html>
Process_ajax.php
<?php
include_once("db_conx.php");
$action=$_POST["action"];
if($action=="showcomment"){
$show="Select * from user_reviews ORDER by date desc";
$result = $db_conx->query($show);
while($row=mysqli_fetch_array($result)){
echo "<li><b>$row[username]</b> : $row[review]</li>";
}
}
else if($action=="addcomment"){
$username= ($_POST['username']);
$review= ($_POST['review']);
$stmt = $db_conx->prepare('INSERT user_reviews SET username = ?, review=?');
$stmt->bind_param('ss', $username, $review);
$stmt->execute();
if ($stmt->errno) {
echo "There was an error in saving your review. Please try again." . $stmt->error;
}else{
echo "Your review has been saved";
}
}
?>
Save the record data as an array and the message as a normal var and create a json:
$list = '';
while($row=mysqli_fetch_array($result)){
$list .= "\n<li><b>$row[username]</b> : $row[review]</li>";
}
if ($stmt->errno) {
$msg = "There was an error in saving your review. Please try again." . $stmt->error;
}else{
$msg = "Your review has been saved";
}
echo json_encode(["list"=>$list, "msg"=>$msg]);
Remember to put in the dataType into the Ajax:
type:"post",
dataType: "json"
Then in your success function, you can access the vars: data.list and data.msg

how to insert value in database using php, jquery and ajax

I am struggling very hard to get this to work and I don't know what I'm doing wrong. I have a register page that I want to take the data inserted into the form and INSERT it to the database with jQuery and AJAX. I'm not very experienced with AJAX AND jQuery so be gentle! :P I will show you the files that I have...this is a msg.php page when i have submit data sometimes post submit in database`
mostly not so i want to know that why it s happening i am new in this field
<?
php $id=$_GET['id'];
$id1=$_SESSION['id'];
?>
<form method="post" class="msgfrm" id="msgfrm">
<input type="hidden" value="<?php echo $_GET['id']; ?>" name="rcvrid" id="rcvrid">
<input type="hidden" name="senderid" id="senderid" value="<?php echo $id1;?>" >
<div class="msgdiv" id="chatbox"></div>
<div class="textdiv">
<input type="text" name="msg" id="msg" class="textmsg">
<input type="submit" value="Send" onClick="sendChat()">
</div>
</form>
function sendChat()
{
$.ajax({
type: "POST",
url: "msg_save.php",
data: {
senderid:$('#senderid').val(),
rcvrid:$('#rcvrid').val(),
msg: $('#msg').val(),
},
dataType: "json",
success: function(data){
},
});
}
msg_save.php file
<?php
require_once('include/util.php');
$rcvrid=$_POST['rcvrid'];
$senderid=$_POST['senderid'];
$msg=$_POST['msg'];
$sql="insert into message(rcvrid,senderid,msg) values($rcvrid,$senderid,'$msg')";
mysql_query($sql);
?>
$.ajax({
type: "POST",
url: "msg_save.php",
data: " senderid="+$('#senderid').val()+"rcvrid="+$('#rcvrid').val()+"msg="+$('#msg').val(),
dataType: "json",
success: function(data){
},
});
please try this code and send data ,and use post method in php to get data,it will work
if you are trying chat application check this, it is old but just for idea:
http://www.codeproject.com/Articles/649771/Chat-Application-in-PHP
use mysqli_query instead of mysql_query recommended
<?php
$id=$_GET['id'];
//$id1=$_SESSION['id']; COMMENTED THIS AS I AM NOT IN SESSION. HARDCODED IT IN THE FORM AS VALUE 5
?>
<html>
<head>
<script src="//code.jquery.com/jquery-1.10.2.min.js"></script>
</head>
<body>
<form method="post" class="msgfrm" id="msgfrm">
<input type="hidden" value="<?php echo $_GET['id']; ?>" name="rcvrid" id="rcvrid">
<input type="hidden" name="senderid" id="senderid" value="5" >
<div class="msgdiv" id="chatbox"></div>
<div class="textdiv">
<input type="text" name="msg" id="msg" class="textmsg">
<input type="submit" value="Send" >
</div>
</form>
<script>
$("#msgfrm").on("submit", function(event) {
event.preventDefault();
$.ajax({
type: "POST",
url: "msg_save.php",
data: $(this).serialize(),
success: function(data) {
$("#chatbox").append(data+"<br/>");//instead this line here you can call some function to read database values and display
},
});
});
</script>
</body>
</html>
msg_save.php
<?php
//require_once('include/util.php');
$rcvrid = $_POST['rcvrid'];
$senderid = $_POST['senderid'];
$msg = $_POST['msg'];
echo $rcvrid.$senderid.$msg;
$con = mysqli_connect("localhost", "root", "", "dummy");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql = "insert into message(rcvrid,senderid,msg) values($rcvrid,$senderid,'$msg')";
mysqli_query($con,$sql);
mysqli_close($con);
echo "successful"
?>
check that whether you have inserted jquery file or not.
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
Then include your function sendChat() inside <script> tags.
On submit button
<button type="submit" id="button">SAVE</button>
<script>
$(document).ready(function(){
$("#button").click(function(){
var firstname=$("#firstname").val();
var lastname=$("#lastname").val();
var email=$("#email").val();
$.ajax({
url:'dbConfigAndInsertionQuery.php',
method:'POST',
data:{
firstname:firstname,
lastname:lastname,
email:email
},
success:function(data){
alert(data);
}
});
});
});
</script>

Ajax won't post form data to PHP script

I got a form with some data that needs to be sent and I want to do it with ajax. I got a function that respond on an onclick event of a button. When I click the button I got some post data in firebug but it just doesn't reach my PHP script. Does anyone know what's wrong?
JS:
function newItem() {
var dataSet = $("#createItem :input").serialize();
confirm(dataSet); //Below this code box is the output of this variable to check whether it is filled or not
var request = $.ajax({
type: "POST",
url: "/earnings.php",
data: dataSet,
dataType: "json"
});
request.done(function(){
$('.create_item').children(".row").slideUp('100', 'swing');
$('.create_item').children("h2").slideUp('100', 'swing');
confirm("succes");
});
request.fail(function(jqXHR, textStatus) {
confirm( "Request failed:" + textStatus );
});
}
dataSet result when the form is completly filled in:
id=123&date=13-09-2013&amount=5&total=6%2C05&customer=HP&invoicenumber=0232159&quarter=1&description=Test
The PHP:
<?php
include('includes/dbconn.php');
function clean_up($string){
$html = htmlspecialchars($string);
$string = mysql_real_escape_string($html);
return $string;
}
if($_POST){
$date = clean_up($_POST['date']);
$amount = clean_up($_POST['amount']);
$total = clean_up($_POST['total']);
$customer = clean_up($_POST['customer']);
$invoicenumber = clean_up($_POST['invoicenumber']);
$quarter = clean_up($_POST['quarter']);
$description = clean_up($_POST['description']);
$sql = ("INSERT INTO earnings (e_date, e_amount, e_total, e_customer, e_invoicenumber, e_quarter, e_description)
VALUES ($date, '$amount', '$total', '$customer', $invoicenumber, $quarter, '$description')");
echo $sql;
if($mysqli->query($sql) === true){
echo("Successfully added");
}else{
echo "<br /> \n" . $mysqli->error;
}
}
?>
The form works fine without the ajax but with it it just doesn't work.
Your help is appreciated!
Try this snippet code bro...
<form id="F_login">
<input type="text" name="email" placeholder="Email">
<input type="password" name="password" placeholder="Password">
<button id="btn_login" type="submit">Login</button>
</form>
$("#btn_login").click(function(){
var parm = $("#F_login").serializeArray();
$.ajax({
type: 'POST',
url: '/earnings.php',
data: parm,
success: function (data,status,xhr) {
console.info("sukses");
},
error: function (error) {
console.info("Error post : "+error);
}
});
});
Reply me if you try this...
prevent your form submitting and use ajax like this:
<form id="createItem">
<input id="foo"/>
<input id="bar"/>
<input type="submit" value="New Item"/>
</form>
$('#createItem).on("submit",function(e){
e.preventDefault;
newItem();
});
Try this
<input type="text" id="foo"/>
<input type="text" id="bar"/>
<input type="button" id="btnSubmit" value="New Item"/>
<script type="text/javascript">
$(function() {
$("#btnSubmit").click(function(){
try
{
$.post("my php page address",
{
'foo':$("#foo").val().trim(),
'bar':$("#bar").val().trim()
}, function(data){
data=data.trim();
// alert(data);
// this data is data that the server sends back in case of ajax call you
//can send any type of data whether json or json array or any other type
//of data
});
}
catch(ex)
{
alert(ex);
}
});
});
</script>

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