Updating the query on the other page and set the button disabled - php

I have a question regarding ajax and modal. What I need to do is, when user click submit of the button, it should set the button disabled and sending the values to the other page to update the database. But my approach is using modal.
My modal code is like this,
<div class="modal fade" id="qcRejectModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title" id="myModalLabel">QC <font color="CC0000"><b>REJECT</b></font> Confirmation Window</h4>
</div>
<div class="modal-body">
<div class="row">
<div class="col-md-12">
<div class="form-group">
This is the confirmation window to PASS all the Quality Control for, <br/><br/>
PROJECT : <b>'.$row['PROJECT_NAME'].'</b>
HEADMARK : <b>'.$row['HEAD_MARK'].'</b>
ID : <b>'.$row['ID'].'
</div>
</div>
</div><br/>
<div class="row">
<div class="col-md-12">
<div class="form-group">
<textarea class="form-control" placeholder="Reason for rejection" rows="2" name="rejectionreason" id="rejectionreason" required></textarea>
</div>
</div>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Cancel</button>
<button type="button" name="rejectbutton" id="rejectbutton" class="btn btn-danger">Confirm REJECT!</button>
</div>
</div>
</div>
</div>
</div>
So in this code when user click the button,
echo "<input type='hidden' data_project='$row[PROJECT_NAME]' data_headmark='$row[HEAD_MARK]' data_id='$row[ID]'></input>";
echo '<button type="button" name="qcreject" id="qcreject" class="btn btn-primary btn-danger btn- lg btn-block" data-toggle="modal" data-target="#qcRejectModal">
<span style="font-size:35px"> '.$row['PROJECT_NAME'].' / <b>'.$_POST ["hm"].'</b>/'.$row['ID'].' ~ FAIL</span></button>';
modal code is executed and showed and waiting for the user to input the comment in the textfield and when user click REJECT button, it should pass
$row['PROJECT_NAME'], $row['HEAD_MARK'], $row['ID'] and the textfield value (#rejectionreason)into the processclass.php.
And on the processclass.php, those variables will be used to update the database with SQL query.
This is the snippet of my ajax.
$('.rejectbutton').click(function() {
$.ajax({
url: 'processclass.php',
type: 'POST',
data: ,
success: function (result) {
alert("Your data has been submitted");
}
});
});
I am having problem with executing on the process class. that means passing those values to the processclass.php.

$('.rejectbutton').click(function() {
$(this).attr('disabled','disabled');
$.ajax({
url: 'processclass.php',
type: 'POST',
data: {param_name: '<?php echo $param_value ?>'},
success: function (result) {
alert("Your data has been submitted");
},
complete: function(){
$(this).removeAttr('disabled');
}
});
});

Related

How do I do a reset of the Bootstrap modal window without reloading the page ? Using PHP , jQuery, AJAX

I have a comment system. The user can report a bad comment (jquery and AJAX). When there are several comments for the same post the Bootstrap modal window retains the message from the previous report.
How do I do a reset of the modal window without reloading the page?
post.php
<button type="button" id="showModal" class="btn btn-primary btn-sm reporting" data-comment-id = "<?= $comment->getId() ?>" data-toggle="modal" data-target="#reportModal"></button>
<div class="modal fade" id="reportModal" tabindex="-1" role="dialog" aria-labelledby="title-report" aria-hidden="true">
<div class="modal-dialog" role="document">
<div class="modal-content">
<div class="modal-header">
<h5 class="modal-title" id="title-report">Confirm reporting</h5>
<button type="button" class="close" data-dismiss="modal" aria-label="Close">
<span aria-hidden="true">×</span>
</button>
</div>
<div class="modal-body">
</div>
<div class="modal-footer">
<button type="button" class="btn btn-primary submit-reporting">Validate</button>
<button type="button" class="btn btn-secondary" data-dismiss="modal">Close</button>
</div>
</div>
<input type="hidden" name="id" id="commentId" value="">
</div>
</div>
report.js
$(document).ready(function(){
$('.reporting').click(function(e) {
e.preventDefault();
$('#commentId').val($(this).data('comment-id'));
});
$('.submit-reporting').click(function(e) {
e.preventDefault();
var commentId = $('#commentId').val();
$.ajax({
url: 'post/moderateComment/' + commentId,
success: function(){
$(".modal-body").html("<p>The report has been forwarded!</p>");
$('.submit-reporting').attr("disabled", "disabled");
$('[data-comment-id = "'+commentId+'"]').hide();
}
});
});
});
I think what you are looking for is to have a modal with it's initial state? If that is so, you can do it by tweaking your modal when it is opened and resetting the elements that you want. So in your javascript, you should have something like:
$(document).ready(function(){
$('.reporting').click(function(e) {
e.preventDefault();
//Clean your message here*** with something like
//$('.elementWHereMessageIs').empty(); //Is faster than text('') or html('')
$(".modal-body").empty(); // I don't know if this is the message you want to empty?
$('#commentId').val($(this).data('comment-id'));
});
$('.submit-reporting').click(function(e) {
e.preventDefault();
var commentId = $('#commentId').val();
$.ajax({
url: 'post/moderateComment/' + commentId,
success: function(){
$(".modal-body").html("<p>The report has been forwarded!</p>");
$('.submit-reporting').attr("disabled", "disabled");
$('[data-comment-id = "'+commentId+'"]').hide();
}
});
});
});
Hope this helps!
Leo.

How to submit a remote modal form via ajax - jQuery

This is a little different from the method of submitting a form via static form value in modal. Using the same method is not working because AJAX not submitting the form value from the shown modal. So,
html
load form
modal
<div class="modal fade" id="xModal" tabindex="-1" role="dialog">
<div class="modal-dialog" role="document">
<div class="modal-content">
<div class="modal-header">
<h4 class="modal-title" id="xModalLabel">loading...</h4>
</div>
<div class="modal-body">...</div>
<div class="modal-footer">
<button type="button" class="btn btn-link waves-effect">SAVE CHANGES</button>
<button type="button" class="btn btn-link waves-effect" data-dismiss="modal">CLOSE</button>
</div>
</div>
</div>
</div>
jquery
//clear modal
$('body').on('hidden.bs.modal', '.modal', function () {
$(this).removeData('bs.modal');
});
//submit form
$('#xModal').on('shown.bs.modal',function (e) {
e.preventDefault();
var form=$(this).find('form').serialize();
$('#save_btn').on('click',function(){
$.ajax({
url:'inc/data.php',
type:'POST',
data:form,
success : function(data){
console.log(data);
}
});
});
});
data.php
<?
print_r($_POST);
?>
remote_form.html
<div class="modal-content">
<div class="modal-header">
<h4 class="modal-title" id="xModalLabel">Reg Form</h4>
</div>
<div class="modal-body">
<form>
<input type="text" name="fname" /><br />
<input type="text" name="lname" /><br />
<input type="text" name="email" />
</form>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-link waves-effect" id="save_btn">SAVE</button>
</div>
</div>
Thanks, any comment is welcome!
Delegate your click event, move the event outside the on('shown.bs.modal') event
$(document).on('click', '#save_btn',function(){
var form=$(this).closest('form').serialize();
$.ajax({
url:'inc/data.php',
type:'POST',
data:form,
success : function(data){
console.log(data);
}
});
});
You can set ids to input fields and then getting data from that id as follows.
<input type="text" name="fname" id="fname"/>
and on the script do like this.
//submit form
$('#xModal').on('shown.bs.modal',function (e) {
e.preventDefault();
$('#save_btn').on('click',function(){
var fname = $("#fname");
$.ajax({
url:'inc/data.php',
type:'POST',
data:{
fname = fname
},
success : function(data){
console.log(data);
}
});
});
});

How to update data using modal with ajax, php and mysql?

I want to update data using modal with ajax in PHP. I am beginer, plz tell me where is my mistake .
HTML Code
<button type="submit" class="btn btn-success" data-toggle="modal" data-cphone='.$row['country_phon'].' data-cname='.$row['country_name'].' >Update</button>';
<div id="myModal" class="modal fade" tabindex="-1" role="dialog">
<div class="modal-dialog">
<!-- Modal content-->
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">×</button>
<h4 class="modal-title">Update Country</h4>
</div>
<form action="" method="post">
<div class="modal-body">
<input type="text" id="cn" name="pcountry">
</div>
<div class="modal-body">
<input type="text" id="cph" name="pphone">
</div>
<div class="modal-footer">
<button type="submit" class="btn btn-primary updt" name="updatecountry" >Save</button>
JS Code: Here when I click on Button, modal appears with the values fetched from database with the help of jquery.
<script>
$(document).ready(function(){
$(".btn").click(function(){
var cphone =$(this).data('cphone');
var cname = $(this).data('cname');
$("#cph").val(cphone);
$("#cn").val(cname);
$("#myModal").modal('show');
});
});</script>
JS code : This is ajax code . I want to update the text of textbox.
<script>
$(".updt").click(function(e){
e.preventDefault();
var cnt = $("#myModal").find("input[name='pcountry']").val();
var cp = $("#myModal").find("input[name='pphone']").val();
$.ajax({
dataType: 'json',
type:'POST',
url: 'test.php',
data:{pcountry:cnt, pphone:cp},
})
)};
</script></div>
PHP Code :
<?php
if(isset($_POST['updatecountry']))
{$country1 = $_POST['pcountry'];
$phone1 = $_POST['pphone'];
echo $country1;
echo $phone1;
echo "Updated Successfully";
mysqli_query($conn,"update country set country_name='$country1' , country_phon='$phone1' where country_id=18");
mysqli_close($conn);
}?>
On view btn class use for update and save. I suggest you please use different class name or id.
<button type="submit" class="form_update btn btn-success" data-toggle="modal" data-cphone='.$row['country_phon'].' data-cname='.$row['country_name'].' >Update</button>';
JS changes
<script>
$(document).ready(function(){
$(".form_update").click(function(){
var cphone = $(this).attr("data-phone");
var cname = $(this).attr("data-cname");
$("#cph").val(cphone);
$("#cn").val(cname);
$("#myModal").modal('show');
});
});
</script>
Hope this code is workinf fine.

Bootstrap modal form submit not working

please i need someone to help me check if i'm missing something. The form in the Bootstrap modal wont submit.
my HTML codes for the modal (sidebar.php)
<!-- start Joel's modal -->
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title">E-Logbook Entry</h4>
</div>
<div class="modal-body">
<form id="modal-form" method="post" action="notes_functions.php">
<fieldset>
<label>Log Entry</label>
<textarea rows="4" cols="50" class="form-control" name="note_content" placeholder="What have you learnt today?..."></textarea>
</form>
</div>
<div class = "modal-footer">
<button type = "button" class = "btn btn-default" data-dismiss = "modal">
Close
</button>
<input type="submit" name="submit" class="btn btn-default" id="submitnote" value="Submit" />
</div>
</div>
</div>
</div>
<!-- end Joel's modal -->
codes for the PHP file (notes_functions.php)
<?php
include_once 'database-config.php';
if (isset($_POST['submitnote'])) {
$noteContent = strip_tags($_POST['note_content']);
$sql = "INSERT INTO account_notes (note_contents) VALUES ('$noteContent')";
$dbh->exec($sql);
echo "New record created successfully";
echo "Log details = ".$note_contents;
}
?>
my AJAX codes for submitting the form
<script type="text/javascript">
var frm = $('#modal-form');
frm.submit(function (ev) {
$.ajax({
type: frm.attr('method'),
url: frm.attr('action'),
dataType: "JSON",
data: frm.serialize(),
success: function (data) {
alert('ok');
}
});
ev.preventDefault();
});
</script>
I can't seem to spot the error :-(
your submit button is not in the form. place it in the form or add this to the button
form="modal-form"
like so:
<input type="submit" name="submit" class="btn btn-default" id="submitnote" value="Submit" form="modal-form" />
Two things:-
1.Instead of if (isset($_POST['submitnote'])) { use if (isset($_POST['submit'])) {
2.Put ev.preventDefault(); before $.ajax({

modal Pop Up window not waiting for Ajax Call

i have tried all my possible best and searched every where but to no avail that is why i have decided to do this after about 4years of learning from this website.
i noticed that my modal pop up window is not waiting for my ajax that sends the value retrived from the anchor tag which is then sent to the PHP file here by casuing the php to give an index undefined error. but if i just use the alert() on success: function() i see the values been retrived from the anchor tag. now this is my code the html link <a href="#" class="btn btn-info btn-sm" id="editr" data-id="<?php echo $postID; ?>" data-owner="<?php echo $op; ?>" data-toggle="modal" data-target="#modal-form" >Edit Page</a>
$(function(){
$("#editr").click(function(e) {
e.preventDefault();
var id = $(this).attr("data-id");
var owner = $(this).attr("data-owner");
var dataString = 'id=' + id + '&user=' + owner;
$.ajax({
url: "../config/edit.php",
type: "POST",
data : dataString,
cache: false,
success: function(data){
alert(data);
}
});
});
});
this is the modal part
<div class="modal fade" id="modal-form" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal"><span aria-hidden="true">×</span><span class="sr-only">Close</span></button>
<h4 class="modal-title" id="myModalLabel">Form in moadl</h4>
</div>
<div class="modal-body">
<form role="form">
<div class="form-group">
<label for="exampleInputEmail1"> Email address</label>
<input type="email" class="form-control" id="exampleInputEmail1" placeholder="Enter email">
</div>
<div class="form-group">
<label for="exampleInputPassword1">Password</label>
<input type="password" class="form-control" id="exampleInputPassword1" placeholder="Password">
</div>
<div class="checkbox">
<label>
<input type="checkbox"> Check me out
</label>
</div>
</form>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button type="button" class="btn btn-primary">Save changes</button>
</div>
</div>
</div>
</div>
You need to wait for the promise to return and then continue on with what you were trying to do.
$(function(){
$("#editr").click(function(e) {
e.preventDefault();
var id = $(this).attr("data-id");
var owner = $(this).attr("data-owner");
var dataString = 'id=' + id + '&user=' + owner;
var promise = $.ajax({
url: "../config/edit.php",
type: "POST",
data : dataString,
cache: false
});
promise.then(function(data){
//show modal with data (in place of alert)
alert(data);
});
});
});

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