Data of a Form sent in 2 different pages - php

I have a form where the data is sent through an email when I click on a button.
The issue I have is that I want the same data to be inserted into a table.
I found a solution on here: send the form values to multiple pages
but I don't know how to insert that portion of code in mine.
This is the code I put in my page:
<script language="Javascript">
<!--
$('form').submit(function() {
$.ajax({
async: false,
method: 'POST',
url: 'submit.php',
data: $( this ).serialize(),
success: function() {
window.location.href = "http://stackoverflow.com"; //redirect to this page
}
});
$.ajax({
async: false,
method: 'POST',
url: 'appuntamento.php',
data: $( this ).serialize(),
success: function() {
window.open("http://www.stackoverflow.com"); //this page will be open in new tab
}
});
});
-->
</script>
The fact is that I don't know if I inserted it right and I don't know how to call the function on the button.

You don't exactly have to call the function on the button. If you include the js in your html or php file which contains the form tag <form> once the button with type="submit" is clicked the event will be triggered.
However, if you must use a button that doesn't automatically trigger form submit on click then you can trigger manually by adding an id to the form and submit onclick of the button
if you have
<form id="target">
<input type="text" value="Hello there">
<input type="submit" value="Go">
</form>
You can do this
$('form').submit(function() {
$.ajax({
async: false,
method: 'POST',
url: 'submit.php',
data: $( this ).serialize(),
success: function() {
window.location.href = "http://stackoverflow.com"; //redirect to this page
}
});
$.ajax({
async: false,
method: 'POST',
url: 'appuntamento.php',
data: $( this ).serialize(),
success: function() {
window.open("http://www.stackoverflow.com"); //this page will be open in new tab
}
});
});
If you have
<form id="target">
<input type="text" value="Hello there">
<button id="submit-form">Submit</button
</form>
you can have this
$( "#submit-form" ).click(function() {
$( "#target" ).submit();
});
$('form').submit(function() {
$.ajax({
async: false,
method: 'POST',
url: 'submit.php',
data: $( this ).serialize(),
success: function() {
window.location.href = "http://stackoverflow.com"; //redirect to this page
}
});
$.ajax({
async: false,
method: 'POST',
url: 'appuntamento.php',
data: $( this ).serialize(),
success: function() {
window.open("http://www.stackoverflow.com"); //this page will be open in new tab
}
});
});

Related

Multiple forms with file input on same page

I'm newby with jquery and have a problem with dealing multiple multipart forms on same page. I'm trying to add some data to mysql via php also uploading mp3 files at same time. Each form uses samename+PHPID. There is no problem with first form but im not getting file data when i use other forms. Can anyone help me?
JS:
$(".msc_send_button").click(function(e) { // changed
var a = this.id; // Button id
var form = $('#'+a).parents('form').attr('id'); // Get buttons parent form id
e.preventDefault();
var formData = new FormData($('#'+form)[0]); // Form Data
$.ajax({
url: '/formposts',
type: 'POST',
cache: false,
contentType: false,
processData: false,
data: formData,
beforeSend: function(){
// change submit button value text
$('#'+a).html('Sending...');
},
success: function(data) {
if(data) {
// Message
$('#info').html(data);
//Button Reset
$('#'+a).html('Send');
}
},
error: function(e){
alert(e);
}
});
return false;
});
PHP Form:
<form name="music-form" id="music-form<?php echo $cont['id']; ?>" enctype="multipart/form-data" novalidate>
<input type="text" name="songno" id="songno" value="<?php echo $cont['song_no']; ?>">
<input type="file" id="mp3" name="mp3" class="inputfile" accept="audio/*" multiple>
<button type="submit" class="msc_send_button" id="msc_send_button<?php echo $cont['id']; ?>">Send</button>
</form>
I think you should change your jquery selector. I didn't see your html codes but you may have created nested forms. Maybe you can use closest('from') instead of parents('form')
$(".msc_send_button").click(function(e) { // changed
var buttonEl = $(this);
var form = buttonEl.closest('form');
e.preventDefault();
var formData = new FormData(form[0]); // Form Data
$.ajax({
url: '/formposts',
type: 'POST',
cache: false,
contentType: false,
processData: false,
data: formData,
beforeSend: function(){
// change submit button value text
buttonEl.html('Sending...');
},
success: function(data) {
if(data) {
// Message
$('#info').html(data);
//Button Reset
buttonEl.html('Send');
}
},
error: function(e){
alert(e);
}
});
return false;
});

How to check id of button clicked in php

I have 2 buttons "removeD" and "updateRecord"(by id) and I have written same ajax for them as follows:
$.ajax({
url: 'DB/tableDisplay.php',
type: 'POST',
data: 'id='+uid,
dataType: 'html'
})
But in tableDisplay.php I want to have different functionality for both the buttons.How to check the id of the button clicked in php?
I've tried using :
if(isset($_POST['removeD'])){
}else{
}
But this is not working.
Try this:
$(document).ready(function(){
$('button').click(function(){
var id = $(this).attr('id');
$.ajax({
url: 'DB/tableDisplay.php',
type: 'POST',
data: {id: id},
dataType: 'html'
})
});
});
$('body').on('click','#id1',function(){
$.ajax({
url : 'url',
data : {variable:values},
type : 'html',
dataType : 'GET/POST',
success : function(data){
console.log('Message after Success');
},
error : function(){
console.log('Error Message')
}
});
});
In your url page you can find whether ajax request is posted or not
if(isset($_REQUEST['variable'])){
write your php code here........
}
Try the following code to detect the button clicked:
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<script>
$(document).ready(function () {
$('.click-button').on('click',function () {
var button_id = $(this).attr('id');
if(button_id == 'removeD'){
alert('removeD button clicked');
}else if(button_id == 'updateRecord'){
alert('updateRecord button clicked');
}
});
});
</script>
</head>
<input type="button" class="click-button" id="removeD" value="removeD">
<input type="button" class="click-button" id="updateRecord" value="updateRecord">
You can use target it return which DOM element triggered the event. see below code its too easy and short to get id of clicked element.
$('button').click(function(e){
alert(e.target.id);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<button id="remove">Remove</button><br>
<button id="update">Update</button><br>

How to upload image through jQuery Steps

I know how to upload images by ajax but I want to upload images via jQuery steps. I've tried multiple ways but its not not working. If anyone has ever done that please help me. Thanks.
HTML
<input type="file" style="background-color: #2184b3; color: #fff;" class="btn btn-default" name="upload_doc" id="upload_doc" title="Search for a file to add">
jQuery
if(currentIndex == 0)
{
var upload_doc = $("#upload_doc").val();
event.preventDefault();
$.ajax({
async: false,
url: myUrl,
dataType: 'json',
type: 'POST',
cache: false,
contentType: false,
processData: false,
data : { upload_doc : upload_doc, step1 : step1},
success: function(response) {
console.log(response);
}
});
}
Follow this way for upload an image, In this way you don't want HTML form.
Add this code to your mainpage.php
<input type="file" name="upload_doc" id="upload_doc" title="Search for a file to add"/>
<input id="uploadImage" type="button" value="Upload Image" name="upload"/>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.4/jquery.min.js"></script>
<script type="text/javascript">
jQuery.noConflict();
jQuery('#uploadImage').on("click", function (e) {
var uploadedFile = new FormData();
uploadedFile.append('upload_doc', upload_doc.files[0]);
jQuery.ajax({
url: 'lab1.php',
type: 'POST',
processData: false, // important
contentType: false, // important
dataType : 'json',
data: uploadedFile
});
});
</script>
Then add this for upload.php
<?php
// check record array
print_r($_FILES);
move_uploaded_file($_FILES['upload_doc']['tmp_name'], $_FILES['upload_doc']['name']);
?>
first in your ajax call include success & error function and then check if it gives you error or what?You can use jquery.form.js plugin to upload image via ajax to the server.
<form action="" name="imageUploadForm" id="imageUploadForm" method="post" enctype="multipart/form-data">
<input type="file" style="background-color: #2184b3; color: #fff;" class="btn btn-default" name="upload_doc" id="upload_doc" title="Search for a file to add">
</form>
<script type="text/javascript">
$(document).ready(function (e) {
$('#imageUploadForm').on('submit',(function(e) {
e.preventDefault();
var formData = new FormData(this);
$.ajax({
type:'POST',
url: $(this).attr('action'),
data:formData,
cache:false,
contentType: false,
processData: false,
success:function(data){
console.log("success");
console.log(data);
},
error: function(data){
console.log("error");
console.log(data);
}
});
}));
$("#upload_doc").on("change", function() {
$("#imageUploadForm").submit();
});
});
</script>
From your comment,
actually the thing is that I'm submitting many values also when uploading the image. so one click of next i sends so many data including image. rest data goes well except image only.
If you're uploading file, along with other input values, through AJAX, use FormData object. But keep in mind that old browsers don't support FormData object, FormData support starts from the following desktop browsers versions: IE 10+, Firefox 4.0+, Chrome 7+, Safari 5+, Opera 12+.
So your jQuery should be like this:
if(currentIndex == 0){
event.preventDefault();
var formData = new FormData($('form')[0]);
$.ajax({
async: false,
url: myUrl,
dataType: 'json',
type: 'POST',
cache: false,
contentType: false,
processData: false,
data : formData,
success: function(response) {
console.log(response);
}
});
}
first in your ajax call include success & error function and then check if it gives you error or what?You can use jquery.form.js plugin to upload image via ajax to the server.
$(document).ready(function (e) {
$('#imageUploadForm').on('submit',(function(e) {
e.preventDefault();
var formData = new FormData(this);
$.ajax({
type:'POST',
url: $(this).attr('action'),
data:formData,
cache:false,
contentType: false,
processData: false,
success:function(data){
console.log("success");
console.log(data);
},
error: function(data){
console.log("error");
console.log(data);
}
});
}));
$("#ImageBrowse").on("change", function() {
$("#imageUploadForm").submit();
});
});

Ajax POST and php query

Been looking at some tutorials, since I'm not quite sure how this works (which is the reason to why I'm here: my script is not working as it should). Anyway, what I'm trying to do is to insert data into my database using a PHP file called shoutboxform.php BUT since I plan to use it as some sort of a chat/shoutbox, I don't want it to reload the page when it submits the form.
jQuery:
$(document).ready(function() {
$(document).on('submit', 'form#shoutboxform', function () {
$.ajax({
type: 'POST',
url: 'shoutboxform.php',
data: form.serialize(),
dataType:'html',
success: function(data) {alert('yes');},
error: function(data) {
alert('no');
}
});
return false;
});
});
PHP:
<?php
require_once("core/global.php");
if(isset($_POST["subsbox"])) {
$sboxmsg = $kunaiDB->real_escape_string($_POST["shtbox_msg"]);
if(!empty($sboxmsg)) {
$addmsg = $kunaiDB->query("INSERT INTO kunai_shoutbox (poster, message, date) VALUES('".$_SESSION['username']."', '".$sboxmsg."'. '".date('Y-m-d H:i:s')."')");
}
}
And HTML:
<form method="post" id="shoutboxform" action="">
<input type="text" class="as-input" style="width: 100%;margin-bottom:-10px;" id="shbox_field" name="shtbox_msg" placeholder="Insert a message here..." maxlength="155">
<input type="submit" name="subsbox" id="shbox_button" value="Post">
</form>
When I submit anything, it just reloads the page and nothing is added to the database.
Prevent the default submit behavior
$(document).on('submit', 'form#shoutboxform', function(e) {
e.preventDefault();
$.ajax({
type: 'POST',
url: 'shoutboxform.php',
data: $(this).serialize(),
dataType: 'html',
success: function(data) {
alert('yes');
},
error: function(data) {
alert('no');
}
});
return false;
});
Use the following structure:
$('form#shoutboxform').on('submit', function(e) {
e.preventDefault();
// your ajax
}
Or https://api.jquery.com/submit/ :
$("form#shoutboxform").submit(function(e) {
e.preventDefault();
// your ajax
});

Ajax post form working 2 times then reloads page with a get URL

I am working on a login form that gets loaded inside a div (parent of .messageboxcontent) with .load on a button press. It all works till the 3rd time I press submit where the div disappears again (I guess by reload of the page and the div CSS is hidden). The URL has the $_POST data added after the 3rd submit (?username=<whatever_I_Fill_In_As_3rd>).
<div class="messageboxcontent">
<form id="ajaxform">
<table>
<tr>
<td>Gebruikersnaam: </td><td><input type="text" name="username" /></td><td>
</tr>
</table>
<input type="submit" value="Registreer" id="submit" />
</form>
</div>
<script>
$('form').on('submit', function( event )
{
var dataString = $(this).serialize();
event.stopPropagation();
//event.preventDefault();
$.ajax(
{
type: "POST",
url: "register.php",
data: dataString,
success: function(response)
{
$('.messageboxcontent').html(response);//FIXED by changing .messageboxcontent to parent.
}
});
return false;
});
</script>
I tried different kind of approaches like:
$('form').submit(function(event) {
//..
}
//
$('#ajaxform').submit(function(event) {
//..
}
//
$(document).ready(function()
{
$("#ajaxform").on("submit", function( event )
{
var dataString = $(this).serialize();
//event.stopPropagation();
event.preventDefault();
$.ajax(
{
type: "POST",
url: "register.php",
data: dataString,
success: function(response)
{
$("div.messageboxcontent").html(response);
}
});
return false; //with and without this.
});
});
Be consistent with quotes. Also close your div (<div class="messageboxcontent"></div>)
Try this:
$(document).ready(function(){
$("#ajaxform").on("submit", function( event ){
var dataString = $(this).serialize();
event.preventDefault();
$.ajax({
type: "POST",
url: "register.php",
data: dataString,
success: function(response)
{
$("div.messageboxcontent").html(response);
}
});
return false;
});
});
Hope that helps.
return false; or event.preventDefault(); inside your ajax function would stop the page from reloading.
Secondly, jQuery works with selector methods via class or id - so, in your case, you will want to use $('#ajaxform').
Lastly, the possible reason why you are facing with unexpected result like after 3rd time is because your form is wrapped inside a div that you want to manipulate the result. So, try rewrapping your DIV element to this: <div class="messageboxcontent"></div> and have your <form> stand on its own separately from messageboxcontent div.

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