Sending two forms after button click and run php in background - javascript

I have created this site and integrated with payfast, it works well. but now i want to execute another php script on place order button which gets details into the db
here is my code quite long.... I no nothing about Ajax or javascript!! Please help
<form id="form1" action="payment.php" method="POST">
some stuff...
</form>
<?php
$htmlForm = '<form action="https://'.$pfHost.'/eng/process" method="post" id="form2">';
$htmlForm .= '<input type="submit" id="submit" name="submit" class="btn btn-primary btn-lg btn-flat" value="PLace Order" onclick="submitForms()"></form>';
?>
<?php
echo"
".$htmlForm."
</div>";
}
else{
echo '...';
}
?>
<script>
submitForms = function() {
document.getElementById("form1").submit();
document.getElementById("form2").submit();
}
</script>
$(function() {
$(".submit").click(function() {
var uid = $("#uid").val();
var prodtls = $("#prodtls").val();
var fname = $("#fname").val();
var amnt = $("#amnt").val();
var mail = $("#mail").val();
var dataString = 'uid='+ uid + '&prodtls=' + prodtls + '&fname' +fname + '&amnt' + amnt + '&mail' + mail;
if(time=='' || date=='')
{
$('.success').fadeOut(200).hide();
$('.error').fadeOut(200).show();
}
else
{
$.ajax({
type: "POST",
url: "payment.php",
data: dataString,
success: function(){
$('.success').fadeIn(200).show();
$('.error').fadeOut(200).hide();
}
});
}
return false;
});
});
I tried the ajax i found but, it never worked. Check at the end of my code!
What i need is payment.php will execute in the background while it directs user to payfast
I want to submit only the first one via AJAX.

Your attempt doesn't make much sense because it doesn't appear to target the first form (which you say you want) and may not prevent the default postback behaviour.
As I understand it, you want that whichever form the user clicks on to submit, the code will actually then submit the first form via AJAX, and then the second one via standard postback.
This should do the job:
HTML:
<form id="form1" class="doubleForm" action="payment.php" method="post">
<input type="submit" id="submit1" name="submit" class="btn btn-primary btn-lg btn-flat" value="Submit">
</form>
<form id="form2" class="doubleForm" action="https://example.com/eng/process" method="post">';
<input type="submit" id="submit2" name="submit" class="btn btn-primary btn-lg btn-flat" value="Place Order">
</form>
JavaScript:
//handle submission of both forms
$(".doubleForm").submit(function(event) {
event.preventDefault(); //stop standard postback
var uid = $("#uid").val();
var prodtls = $("#prodtls").val();
var fname = $("#fname").val();
var amnt = $("#amnt").val();
var mail = $("#mail").val();
var dataString = 'uid='+ uid + '&prodtls=' + prodtls + '&fname' +fname + '&amnt' + amnt + '&mail' + mail;
if(time == '' || date == '')
{
$('.success').fadeOut(200).hide();
$('.error').fadeOut(200).show();
}
else
{
//submit first form via AJAX
var request = $.ajax({
type: "POST",
url: "payment.php",
data: dataString
});
request.done(function(response) {
$('.success').fadeIn(200).show();
$('.error').fadeOut(200).hide();
//now the first form is submitted and server has responded, we can trigger "normal" (non-AJAX) submission of second form
document.querySelector("#form2").submit();
});
}
});

So this actually worked well using .serialize(). Thanks #ADyson for the headsup just had to tweak it a little bit, though i'm not experienced in js
<script type="text/javascript">
$(function(){
$("#submit2").click(function(){
var dataString = $("#form1").serialize();
$.ajax({
type: "POST",
url: "payment.php",
data: dataString,
success: function(data)
{
alert('Success!');
$("#form1")[0].reset();
}
});
});
});
</script>

Related

Get value from document.querySelector and insert into localhost database

I want to get the value from document.querySelector and save the data into the database via form with ajax and php, but the code I wrote did not run well / could not display the result, if there was an error in writing my code
<script>
//on the click of the submit button
$(document).ready(function () {
$("#submit").click(function () {
//get the form value
let no = document.querySelector('i[data-name="no"]').value;
let question = document.querySelector('i[data-name="question"]').value;
let countdown = document.querySelector('i[data-name="countdown"]').value;
let answer = document.querySelector('i[data-name="answer"]').value;
//make the postdata
var postData = 'no=' + no + '&question=' + question + '&countdown=' + countdown +
'&answer=' + answer;
if (no == '' || question == '' || countdown == '' || answer == '') {
alert("Please Fill All Fields");
} else {
// AJAX Code To Submit Form.
$.ajax({
type: "POST",
url: "post-answer.php",
data: postData,
cache: false,
success: function (result) {
alert(result);
}
});
}
return false;
});
});
</script>
this is the form I created
<form class="glass userform" method="post" data-ajax="false">
<i data-name='countdown'></i><i data-name='answer'></i>
<i data-name='no'></i>
<a href=""
style="float:right;font-size: 30px;width: 50px;margin-right:50px;margin-top:30px;display:
inline-block;"><i>Next</i></a>
<i data-name='question'></i>
<br>
<a href="#" onclick="checkboxtrue();"
style="width:100%;text-align: center;font-size: 30px;display: inline-block;"><i>Benar
</i></a><br>
<a href="#" onclick="checkboxfalse();"
style="width:100%;text-align: center;font-size: 30px;display: inline-block;"><i>Salah</i></a>
<br><br>
<input type="hidden" id="checkboxtotext" name="checkboxtotext" onkeyup="valueoftrueanswer()" />
<input id="submit" type="button" value="Submit">
</form>

The response received in AJAX call is shown in another HTML page

I have ajax request call which sends an ID to the server, then the server sends a JSON response. I want to update the innerHTML of the pre tag using the value in that JSON Response.
Form HTML
<form id="AssociateForm" class="form form-inline" style="float:right" action="{% url 'Project:MyView' TR.id %}" method="POST" target="_blank">
<div class="form-group">
<input type="text" name="JIRA_ID" style="width:150px" placeholder="ID" class="form-control has-success" id="{{TR.id}}">
<button name="button" type="submit" id='Submit_{{TR.id}}' class="btn btn-primary">Associate</button>
</div>
</form>
AJAX
<script>
$("#AssociateForm").submit(function(e) {
e.preventDefault();
var form = $(this);
var url = form.attr('action');
var local_id = $('input[name=J_ID]').attr('id');
var formData = {
'J_ID' : $('input[name=J_ID]').val()
};
console.log(formData)
$.ajax({
url: url,
data: formData,
dataType: 'json',
success: function (datas) {
var data = JSON.parse(datas);
if(datas.status){
alert(datas);
//$('#Failure_'+local_id).innerHTML = data.category + ' issue: '+data.j_id +' ('+data.j_status+')'
}
},
error: function(jqXHR, textStatus){
alert("In error")
}
})
.done(function(data){
alert(data)
});
});
</script>
for some reason, the above code is not printing the console log as well.
But,
When the response comes, the success section is not triggered. Instead, the complete JSON string is printed on a different page.
JSON Response
{"category": "known", "j_id": "AU298", "j_status": "Confirmed"}
below is from View-Page-source
<html>
<head></head>
<body data-gr-c-s-loaded="true">
<pre style="word-wrap: break-word; white-space: pre-wrap;">
{"category": "known", "j_id": "AU298", "j_status": "Confirmed"}
</pre>
</body>
</html>
This is possibly because you are submitting a form, and after submitting it will open a new tab, as Form is submitted.
To resolve this, you can probably use the below code:
<form action="..." method="POST" target="_blank">
<input type="submit" id="btn-form-submit"/>
</form>
<script>
$('#btn-submit').click( function(){ $('#btn-form-submit').click(); } );
</script>
success: function (datas) {
if (datas.status) {
alert(datas);
$('pre#<ID>').html(datas.category + ' issue: ' + datas.j_id + ' (' + datas.j_status + ')');
}
}
This worked for me, I removed the form completely.
Code in-place of Form
<div class="form-group AssociateForm" style="float:right">
<input type="text" name="J_ID" style="width:150px;float:left" class="form-control has-success">
<button name="button" type="submit" id="{{TR.id}}" class="Associater btn btn-primary">Associate</button>
</div>
AJAX
<script>
$('.Associater').on('click', function () {
var local_id = $(this).attr('id');
var j_id = $(this).closest("div.AssociateForm").find('input[name=J_ID]').val();
if (j_id === "") {
alert("JID cannot be empty")
return false
}
var url = "{% url 'Project:View' 0 %}".replace('0', local_id);
var formData = {
'J_ID' : j_id,
'csrfmiddlewaretoken': '{{ csrf_token }}'
};
console.log(local_id);
console.log(j_id);
console.log(url);
console.log(formData);
$.ajax({
type: 'POST',
url: url,
data: formData,
dataType: 'json',
success: function (data) {
if (data.status) {
ele = 'Failure_'+local_id;
document.getElementById(ele).innerHTML = data.category + ' issue: '+data.j_id +' ('+data.j_status+')';
}
},
error: function (jqXHR, textStatus ) {
alert("For some reason im here");
}
});
});
</script>

Submit form using ajax and pass the value submitted to new page

i have form that need the previous value inserted....i'm using ajax,but the success: function(data) wont let me moved to the next page...
here is my code
HTML
<form>
<input type="text" name="id_1" id="id_1>
<input type="text" name="id_2" id="id_2>
<input type="text" name="id_3" id="id_3>
<button type="button" onclick="next();">
</form>
<div id="tabelna"></div>
JQuery
var id_1 = $('#id_1').val();
var id_2= $('#id_2').val();
var id_3= $('#id_3').val();
var datana = 'id_1='+id_1+'&id_2='+id_2+'&id_3='+id_3;
var urlna="<?=base_url()?>something/something/something";
$.ajax({
type: 'POST',
url: urlna,
data: datana,
beforeSend:function(data){
},
message:"<center>><h3>Loading Data. . .</h3></center>"
});
},
error: function(data) {
jAlert('Failed');
},
success: function(data) {
load();
}
})
return false;
}
function load()
{
$('#tabelna').load('<?=base_url()?>something/something/something') (This is my mistake)
}
CONTROLLER
function set_value()
{
extract($_POST);
$d['id1'] = $this-db->query('SELECT * FROM TBL1 where id='.$id_1);
$d['id2'] = $this-db->query('SELECT * FROM TBL2 where id='.$id_2);
$d['id3'] = $this-db->query('SELECT * FROM TBL3 where id='.$id_3);
$this->load->view('something/v_add',$d); (this is my mistake)
}
How can i pass the submitted value to the controller and shows new form ?
we can call controller function using window.location
function load()
{
window.location.href = "<?php echo site_url('controller_d/login/admin_link_delete_user');?>";
}

Why doesn't jquery submit work properly?

This code works fine in its current state where im using a click event on a button. But if i use a form tag around the input tags in my html code and use jquery's submit method it doesnt give any results. why is that happening? i have to use the form element because i want to be able to search with an enter key insted of clicking on a button
Html:
<body>
Title: <input type="text" id="title"><br/>
<input type="submit" value="Submit" id="btn">
<p id="results"></p>
<body>
jQuery:
$(document).ready(function(){
$("#btn").on("click", function(){
var textval = $('#title').val();
var playListURL = 'https://en.wikipedia.org/w/api.php?action=query&list=search&srsearch=' + textval + '&format=json&callback=?';
$.ajax({
type: "GET",
url: playListURL,
contentType: "application/json; charset=utf-8",
async: false,
dataType: "json",
success: function (data, textStatus, jqXHR) {
for (var i = 0; i < 10; i++)
{
var url = "https://en.wikipedia.org/wiki/" + data.query.search[i].title;
$("#results").append("<b> <a href='" + url + "' > " + data.query.search[i].title + "</a></b></br> ");
//console.log(url);
}
},
error: function (errorMessage) {
}
});
//alert($('#title').val());
});
});
Try to change this line
$("#btn").on("click", function(){
to this:
$("#btn").on("click", function(e){
e.preventDefault();
Then the form will not submit default way on click

Make javascript validation happen sooner on click?

So I have a form as such:
<form action="URL" name=myform method=post onsubmit="return validateFormOnSubmit(this)" autocomplete=off>
When the form is submitted, it checks for validation.
Earlier though, I have code like this:
<script type="text/javascript">
$(function() {
$("#submitbutton").click(function() {
var fname = $("input#first_name").val();
var lname = $("input#last_name").val();
var email = $("input#email").val();
var leadsource = $("input#leadsource").val();
var country = $("input#country").val();
var phone = $("input#phone").val();
var oid = $("input#oid").val();
var retURL = $("input#retURL").val();
var crazy
= $("input#00N40000001mCkP").val();
var dataString = '&email=' + email + '&phone=' + phone + '&oid=' + oid + '&retURL=' + retURL + '&leadsource=' + leadsource + '&country=' + country; //alert (dataString);return false; $.ajax({
type: "POST",
url: "MYURL.php?first_name="
+ fname + "&last_name=" + lname,
data: dataString,
success: function(response) { $(myform).submit(); }
});
return false; });
}); </script>
Essentially whats happening is, the form data is initially sent to a different url through an AJAX submit prior to completing the real form. On success of the ajax push - the form is submitted, and then the form is validated.
I need the form validation to have right when #submitbutton is clicked though - prior to the ajax request. Any ideas?
That code should work just fine. If you want to add it to a onclick= property on the element (I wouldn't, but if you insist), you can use this syntax to reference the form:
<input type="submit" onclick="return validateFormOnSubmit(this.parentNode)" />
And you'd have to do an e.preventDefault() and return false (maybe just one?) from the validateFormOnSubmit() function.
Take a look at preventDefault . Also, you can call validateFormOnSubmit in the anon function above and return false if it validation doesn't pass.
you should put some " " around your form tags attributes, and give it an id!
just call your validate function as the first line in your button click handler :)
$("#submitbutton").click(function(e) {
if( validateFormOnSubmit($('#myform'))){
//do all the posting stuff
}
else{
e.preventDefault();
return false;
}

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