Submit button intermittently not submitting form information - javascript

So I've got this form for adding comments under a post. The methods utilized here are MYSQL(holds the submitted form data in a database) PHP(communicating with the database) and JavaScript, more specifically AJAX (for hooking up the submit button and handling events).
Typing in your comment into the form and pressing submit is supposed to print the comment onto the screen.
When I click submit, it doesn't print anything. Then, when I type another comment and click submit once more, it prints the contents of that comment. Other times, it successfully prints the contents of the comment instead of failing to submit.
I checked it out in inspect element and in the console log, whenever it misses, it still sends some blank <p> tags through with the class of the comment that should be submitted.
The PHP page for the comment form:
<head>
<script src="https://code.jquery.com/jquery-3.3.1.js"></script>
<script src="https://ajax.aspnetcdn.com/ajax/jQuery/jquery-3.3.1.min.js"></script>
<link rel="stylesheet" href="//netdna.bootstrapcdn.com/bootstrap/3.1.1/css/bootstrap.min.css">
<link rel="stylesheet" href="Forums.css">
</head>
<body>
<?php
$result = mysqli_query($link, $displayPost); ?>
<?php $row = mysqli_fetch_assoc($result);?>
<p> <?php echo $row["title"];?> </p>
<br>
<p> <?php echo $row["body"];?> </p>
<form action="<?php echo $url ?>" method="post" id="form-group">
<div class="forum col-md-12">
<textarea type="text" style="overflow: auto; resize: none;" name="body" class="txtBody"></textarea>
<input type="submit" name="submit" class="btnCreate" style="margin-bottom: 4px;">
</div>
</form>
</body>
<script>
function refreshData() {
$.ajax({
type:'GET',
url: 'getcomments.php?id=<?php echo $id ?>',
dataType: 'html',
success: function(result){
console.log(result);
$('#comments').html(result);
}
});
}
$(document).ready(function () {
refreshData();
$("#form-group").submit(function (event) {
var $form = $(this);
console.log($form.attr('action'));
var serializedData = $form.serialize();
$.ajax({
url: $form.attr('action'),
type: 'POST',
data: serializedData
});
refreshData();
event.preventDefault();
});
});
</script>
<div id="comments"></div>
The PHP page for getting previously submitted comments and printing them on the screen
<?php
$link = mysqli_connect("localhost", "root", "WassPord64", "forum");
$id = $_GET["id"];
$displayPost = "SELECT * FROM comments WHERE post_id='$id'";
$link->query($displayPost);
$result = mysqli_query($link, $displayPost);
if (mysqli_num_rows($result) > 0) :
// output data of each row
while($row = mysqli_fetch_assoc($result)) :
$row = mysqli_fetch_assoc($result);?>
<p class="postBody"><?php echo $row['body'];?></p>
<?php endwhile; ?>
<?php endif; ?>

You are calling refreshData() when the Ajax is not done. You can make a callback function by using $.ajax.success
Try this:
$(document).ready(function () {
refreshData();
$("#form-group").submit(function (event) {
var $form = $(this);
console.log($form.attr('action'));
var serializedData = $form.serialize();
$.ajax({
url: $form.attr('action'),
type: 'POST',
data: serializedData,
success: function(){
refreshData();
}
});
event.preventDefault();
});
});

Related

PHP: multiple button with different name to update MySQL database with AJAX

I have an problem, I have multiple button generate with while, with different names (button[$nostation]).
Now, I want to update MySQL database (table: smt, column: no) with the same id ($nostation).
How I can generate AJAX function for that?
This is my code:
<?php
$query1 = mysqli_query($connect,"SELECT * FROM smt WHERE no <= 15");
while ( $data=mysqli_fetch_array($query1)){
$nostation = $data['no'];
$namastation = $data['name'];
echo "
<div class='col-xs-2-2'>
<form action='coba.php' method='post'>
<button name='button[$nostation]' value='2' style='background-color:#02780d; width:140px; height:75px; margin : 2px; border-radius:10%;'>
<center>
<b style='font-size:15px; color: #fff; font-family:Calibri;'>$namastation</b>
</center>
</button>
</form>
</div>
";}?>
And this is my code for update database with PHP:
<?php
include 'connect.php';
$array=$_POST['button'];
foreach ($array as $nostation => $value) {
$updch=mysqli_query($connect,"UPDATE smt SET status='$value' WHERE no='$nostation'");
}?>
How I can update with AJAX without refreshing the page?
View Part :-
<?php
$query1 = mysqli_query($connect,"SELECT * FROM smt WHERE no <= 15");
while ( $data=mysqli_fetch_array($query1)){
$nostation = $data['no'];
$namastation = $data['name'];
?>
<div class='col-xs-2-2'>
<form method='post'>
<input type="hidden" value="<?php echo $nostation;?>" id="name_<?=$nostation;?>" name="name">
<button type="submit" id="button_<?=$nostation;?>" data-id="<?=$nostation;?>">SAVE</button>
<center>
<b style='font-size:15px; color: #fff; font-family:Calibri;'>$namastation</b>
</center>
</button>
</form>
</div>
<?php } ?>
jQuery / AJAX Part:-
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script>
$(document).ready(function(){ //when DOM is Ready.
$("[id^=button_]").click(function () { //when Button is Clicked.
var id = $(this).data('id'); // Get the ID of the button that was clicked on.
var name = $("#name_"+id).val(); // value from `input` which is connected the clicked button.
// console.log(id+"---"+name);
$.ajax({ // AJAX request
url: 'update.php', // send request to server.
method: 'POST', // method is POST.
data: { //data which is sent to server.
id: id,name: name
},
success: function (data) { //success function called.
alert(data); // alert success data.
}
});
});
});
</script>
update.php:-
And in the php-side We catch it by:-
echo $id = $_POST['id'];
echo $name = $_POST['name'];
//use update query.
Note:- For more info regarding click()
https://api.jquery.com/click

jQuery on("click",....) function doesn't execute for my dynamically created div

I'm creating a to-do list and I want to be able to click on a div with a "to-do" and directly delete it from the database. This is my main page.
<div class="hoofder">
<h2>Voegtoe:</h2>
</div>
<?php include('connect.php'); ?>
<form method="post" action="add.php">
<div class="input">
<label>Taak</label>
<input type="text" name="taak">
</div>
<div class="input">
<button type="submit" name="add_todo class="btn">Register</button>
<?php include('getthings.php'); ?>
<script src="jquery-3.2.1.min.js"></script>
<script>
$(document).ready ( function () {
alert("1");
$(document).on("click", ".lijst", function () {
alert("2.");
var del_id = $(this).attr('id').substring(1);
alert(del_id);
var $ele = $(this).parent();
$.ajax({
type:'POST',
url:'delete.php',
data:{del_id:del_id},
success: function(data){
if(data=="YES"){
$ele.fadeOut().remove();
}else{
alert("can't delete the row");
}
}
});
});
});
alert("end");
</script>
</form>
</body>
</html>
delete.php:
<?php
include 'connect.php';
include 'add.php';
$delete_this = $_POST['del_id'];
$qry = "DELETE FROM tedoen WHERE id ='$delete_this'";
$result = mysqli_query($conn, $qry);
?>
When I load the page the "alert(del_id);" comes back as undefined.
This is a problem as I need it not to be undefined ofcourse. Do you have any idea why?

Ajax form submit using mongodb

i am trying to use ajax submit form but for some reason it doesn't work for me, any suggestions how to fix it.
I'm getting the alert message when I submit but it takes me to another page, what am i doing wrong with ajax request ?
<!DOCTYPE html>
<html>
<head>
<title>Get Data From a MySQL Database Using jQuery and PHP</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
$(document).ready(function(){
// AJAX forms
$("#search_form").submit(function(e){
e.preventDefault();
//var data = $(this).serialize();
var method = $(this).attr("method");
var action = $(this).attr("action");
var username = $('#username').val();
$.ajax({
url: 'process.php',
type: 'POST',
data: { name: username },
cache: false,
success: function(data){
$('#results').html(data);
}
})
})
});
</script>
</head>
<body>
<span>Search by name: </span>
<form method="POST" action="process.php" id="search_form">
<input type="text" id="username" name="name">
<input type="submit" id="submit" value="Search">
</form>
<div id="results"></div>
</body>
</html>
process.php
<?php
error_reporting(E_ALL);
ini_set('display_errors', '1');
// Check if $_POST is set
if ( empty ( $_POST['name'] ) ) {
echo "Something wrong!";
exit;
}
$name = $_POST['name'];
$m = new MongoClient();
//echo "Connection to database successfully";
// select a database
$address = array(
'name'=>$name,
'city' => 'test',
'state' => 'test2',
'zipcode' => 'test3'
);
$db = $m->local;
//echo "Database mydb selected";
$collection = $db->user;
//echo "Collection selected succsessfully";
$collection->insert($address);
$user = $collection->findOne(array('name' => $name));
?>
<ul>
<li><?php echo $user['name']; ?>: <?php echo $user['city']; ?></li>
<script>
alert('test 1234');
</script>
</ul>
I had to change:
$("#search_form").submit(function(e){
to:
$(document).on('submit', '#search_form', function(){
Now it works fine.
<!DOCTYPE html>
<html>
<head>
<title>Get Data From a MySQL Database Using jQuery and PHP</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
$(document).ready(function() {
// AJAX forms
$(document).on('submit', '#search_form', function() {
//e.preventDefault();
//var data = $(this).serialize();
var method = $(this).attr("method");
var action = $(this).attr("action");
var username = $('#username').val();
$.ajax({
type: 'POST',
url: 'process.php',
data: {
name: username
},
cache: false,
success: function(data) {
$('#results').html(data);
}
})
return false;
});
});
</script>
</head>
<body>
<span>Search by name: </span>
<form method="POST" action="process.php" id="search_form">
<input type="text" id="username" name="name">
<input type="submit" id="submit" value="Search">
</form>
<div id="results"></div>
</body>
</html>

AJAX comment system Validation problems

So i am haveing this page where it is displaying articles andunderneet each article it will have a textarea asking allowing the user to insert a comment.I did the AJAX and it works fine.Some of the validation works fine aswell(Meaning that if the textarea is left empty it will not submit the comment and display an error).The way i am doing this validation is with the ID.So i have multi forms with the same ID.For the commets to be submited it works fine but the validtion doesnt work when i go on a second form for exmaple it only works for the first form
AJAX code
$(document).ready(function(){
$(document).on('click','.submitComment',function(e) {
e.preventDefault();
//send ajax request
var form = $(this).closest('form');
var comment = $('#comment');
if (comment.val().length > 1)
{
$.ajax({
url: 'ajax_comment.php',
type: 'POST',
cache: false,
dataType: 'json',
data: $(form).serialize(), //form serialize data
beforeSend: function(){
//Changeing submit button value text and disableing it
$(this).val('Submiting ....').attr('disabled', 'disabled');
},
success: function(data)
{
var item = $(data.html).hide().fadeIn(800);
$('.comment-block_' + data.id).append(item);
// reset form and button
$(form).trigger('reset');
$(this).val('Submit').removeAttr('disabled');
},
error: function(e)
{
alert(e);
}
});
}
else
{
alert("Hello");
}
});
});
index.php
<?php
require_once("menu.php");
?>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js" type="text/javascript"></script>
<script src="comments.js" type="text/javascript" ></script>
<?php
$connection = connectToMySQL();
$selectPostQuery = "SELECT * FROM (SELECT * FROM `tblposts` ORDER BY id DESC LIMIT 3) t ORDER BY id DESC";
$result = mysqli_query($connection,$selectPostQuery)
or die("Error in the query: ". mysqli_error($connection));
while ($row = mysqli_fetch_assoc($result))
{
$postid = $row['ID'];
?>
<div class="wrapper">
<div class="titlecontainer">
<h1><?php echo $row['Title']?></h1>
</div>
<div class="textcontainer">
<?php echo $row['Content']?>
</div>
<?php
if (!empty($row['ImagePath'])) #This will check if there is an path in the textfield
{
?>
<div class="imagecontainer">
<img src="images/<?php echo "$row[ImagePath]"; ?>" alt="Article Image">
</div>
<?php
}
?>
<div class="timestampcontainer">
<b>Date posted :</b><?php echo $row['TimeStamp']?>
<b>Author :</b> Admin
</div>
<?php
#Selecting comments corresponding to the post
$selectCommentQuery = "SELECT * FROM `tblcomments` LEFT JOIN `tblusers` ON tblcomments.userID = tblusers.ID WHERE tblcomments.PostID ='$postid'";
$commentResult = mysqli_query($connection,$selectCommentQuery)
or die ("Error in the query: ". mysqli_error($connection));
#renderinf the comments
echo '<div class="comment-block_' . $postid .'">';
while ($commentRow = mysqli_fetch_assoc($commentResult))
{
?>
<div class="commentcontainer">
<div class="commentusername"><h1>Username :<?php echo $commentRow['Username']?></h1></div>
<div class="commentcontent"><?php echo $commentRow['Content']?></div>
<div class="commenttimestamp"><?php echo $commentRow['Timestamp']?></div>
</div>
<?php
}
?>
</div>
<?php
if (!empty($_SESSION['userID']) )
{
?>
<form method="POST" class="post-frm" action="index.php" >
<label>New Comment</label>
<textarea id="comment" name="comment" class="comment"></textarea>
<input type="hidden" name="postid" value="<?php echo $postid ?>">
<input type="submit" name ="submit" class="submitComment"/>
</form>
<?php
}
echo "</div>";
echo "<br /> <br /><br />";
}
require_once("footer.php") ?>
Again the problem being is the first form works fine but the second one and onwaord dont work properly
try this:
var comment = $('.comment',form);
instead of
var comment = $('#comment');
That way you're targeting the textarea belonging to the form you're validating
ps.
remove the id's from the elements or make them unique with php, all element id's should be unique

Getting value from contenteditable div

Up to this point, I've been using a textarea as the main input for a form. I've changed it to use a contenteditable div because I wanted to allow some formatting.
Previously, when I had the textarea, the form submitted fine with Ajax and PHP. Now that I've changed it to use a contenteditable div, it doesn't work anymore and I can't tell why.
HTML:
<form>
<div name="post_field" class="new-post post-field" placeholder="Make a comment..." contenteditable="true"></div>
<input name="user_id" type="hidden" <?php echo 'value="' . $user_info[0] . '"' ?>>
<input name="display_name" type="hidden" <?php echo 'value="' . $user_info[2] . '"' ?>>
<ul class="btn-toggle format-post">
<button onclick="bold()"><i class="fa-icon-bold"></i></button>
<button onclick="italic()"><i class="fa-icon-italic"></i></button>
</ul>
<div class="post-buttons btn-toggle">
<button class="btn-new pull-right" type="submit">Submit</button>
</div>
</form>
JQuery Ajax:
$(document).ready(function() {
$(document).on("submit", "form", function(event) {
event.preventDefault();
$.ajax({
url: 'php/post.php',
type: 'POST',
dataType: 'json',
data: $(this).serialize(),
success: function(data) {
alert(data.message);
}
});
});
});
PHP (post.php): Just your typical checks and echo a message back. This is just a snippet of the code.
<?php
$user_id = $_POST["user_id"];
$display_name = $_POST["display_name"];
$post_content = $_POST["post_field"];
$array = array('message' => $post_content);
echo json_encode($array);
?>
For some reason, it's not sending back the post content anymore ever since I added the contenteditable div.
Please help!
The contents of the div are not serialized. You would have to add them on your own.
var data = $(this).serialize();
data += "post_field=" + encodeURIComponent($("[name=post_field]").html());

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