I have below code. when I click reply link, it will show a comment box.
I want 2 points.
or give me best way to do my work.
When 1 comment box open other must be hide.
when I click on send button correct value should send vie serialize values.
These are the codes
PHP code
<?PHP
for($i = 1; $i < 10; $i++)
{
?>
<div>comments text etc etc...</div>
Reply
<div class="reply-comment-form" style="display:none;">
<form class="comment_form" method="post">
<textarea name="comment_box" rows="6" class="span10"></textarea> <br /><br />
<input type="button" onClick="send_comment()" class="btn btn-primary" value="send" />
</form>
</div>
<?PHP
}
?>
Jquery code
<script>
$(function(){
$('.reply-comment').on('click', function(e){
e.preventDefault();
$(this).next('.reply-comment-form').show();
});
});
function send_comment()
{
$.ajax({
type: "POST",
data : $('.comment_form').serialize(),
cache: false,
url: 'test.php',
success: function(data){
}
});
}
</script>
test.php file no need. I am checking values through firebug.
please help me to clarify this problem.
or give me best way to do my work.
I am stuck since 2 days.
Thank you for your valuable time.
For the first one
$('.reply-comment').on('click', function(e){
e.preventDefault();
// Hide others
$(".reply-comment-form").hide();
// Show the one which is required
$(this).next('.reply-comment-form').show();
});
And for second, do a .on("submit"... on the form and it will serialize the right input fields only.
UPDATE:
<form class="comment_form" method="post">
<textarea name="comment_box" rows="6" class="span10"></textarea> <br /><br />
// Change type to submit and remove onclick
<input type="submit" class="btn btn-primary" value="send" />
</form>
jQuery:
$(".comment_form").on("submit", function(e){
e.preventDefault(); // Here
var _data = $(this).serialize();
$.ajax({
type: "POST",
data : _data,
cache: false,
url: 'test.php',
success: function(data){
console.log(data);
}
});
});
I found a solution. #void helped me for this.
$(".test").on("click", function(){
var _data = $(this).parent().serialize();
$.ajax({
type: "POST",
data : _data,
cache: false,
url: 'test.php',
success: function(data){
}
});
});
Thanks!
Related
I am trying to submit data to the database using AJAX. I have one array and I have to pass the value of the array to PHP using AJAX to display all the related records.
<form id="search-form" method="POST">
<input value="4869" name="compare_id[]" type="hidden">
<input value="4884" name="compare_id[]" type="hidden">
<input value="5010" name="compare_id[]" type="hidden">
<input type="button" id="search-button" name="search-button" value="search">
</form>
<div id="response"></div>
AJAX
<script>
$(document).ready(function(){
$('#search-button').click(function(){
$.ajax( {
type: 'POST',
url: 'response.php',
data: $('#search-form').serialize(),
dataType: 'json',
success: function(response) {
$('#response').html(response);
//alert(response);
}
});
});
});
</script>
PHP
$sql='SELECT Name, Email FROM request WHERE Id IN (' .( is_array( $_POST['compare_id'] ) ? implode( ',', $_POST['compare_id']) : $_POST['compare_id'] ).')';
$records = array();
$query=$conn->query($sql);
if ($query->num_rows > 0) {
while($row=$query->fetch_assoc()){
$records[]=$row;
}
}
echo json_encode($records);exit();
HTML
<form id="search-form" method="POST">
<input value="4869" name="compare_id[]" type="hidden">
<input value="4884" name="compare_id[]" type="hidden">
<input value="5010" name="compare_id[]" type="hidden">
<input type="button" id="search-button" name="search-button" value="search">
</form>
<div id="response"></div>
JS
<script>
$(document).ready(function(){
$('#search-button').click(function(){
$.ajax( {
type: 'POST',
url: 'response.php',
data: $('#search-form').serialize(),
dataType: 'json',
success: function(response) {
$('#response').html(response);
}
});
});
});
</script>
PHP
var_dump($_POST['compare_id']);
// it is already an array of ids. You can do whatever you want with it.
change your script as below. Your output is in array so you cant add it in div directly
<script>
$(document).ready(function(){
$('#search-button').click(function(){
$.ajax( {
type: 'POST',
url: 'action.php',
data: $('#search-form').serialize(),
dataType: 'json',
success: function(response) {
$('#response').html();
for(data in response) //loop over your data
{
$('#response').append(response[data].Email); //add email
}
//alert(response);
}
});
});
});
</script>
There are errors in your code. A good way to debug this is to print_r your POST value in your php script.
First $_POST["All"] does not exist. It is all. (php)
Second, you send a GET request not a POST one. (jQuery)
Third, format your date into json. A good way to do this is to create a variable right after compare_id.push, it's more readable, as so :
var json_data = {"my_array" : [1,2, "bonjour", 4]};
Your problem is mostly related to "how to debug". I think you should print what's happening along the way to figure out what's happening.
I'm trying to send an input value to a php script and have the returned value posted to a div, using ajax, but I can't seem to get this right. Any help/suggestions would be appreciated. Thanks!!
This is what I have by now, but console says: "Failed to load resource: the server responded with a status of 404 (Not Found)".
test1.php:
<script>
$.ajax({
type: 'POST',
url: 'test2.php',
data: {url: $('#id1').val()},
success: function (data)
{
$(document).ready(function(){$("#content").load("test2.php");});
}
});
</script>
<form name="input">
<input type="text" id="id1">
<input type="submit">
</form>
<div id="content"></div>
test2.php:
<?php
$string=$_POST['id1'];
require_once('connect.php');
$inf = "SELECT * FROM `comments` WHERE date='$string'";
$info = mysql_query($inf);
while($info2 = mysql_fetch_object($info)) {echo $info2->username.$info2->date;}
?>
<script>
$(document).ready(function() {
$('#submit').click(function(e) {
e.preventDefault();
$.ajax({
type: 'POST',
url: 'test2.php',
data: {id1: $('#id1').val()},
success: function(data)
{
$("#content").html(data);
}
});
});
});
</script>
<form name="input">
<input type="text" id="id1">
<input type="submit" id="submit">
</form>
<div id="content"></div>
When you submit the ajax request, you're already submitting your content to test2.php, so you don't need to load it again. In the success function, you can append the result to the div from the callback.
$(document).on('click','#submit',function(e) {
e.preventDefault();
$.post('test2.php',{url: $('#id1').val()},function(data){
$("#content").html(data);
}
});
});
404 (Not Found) Error is for page not found. Please make sure that file test2.php is exist in same folder. Check url.
Also you can copy the URL from console and paste it in the browser URL to check the url correct or incorrect.
jQuery
<script>
$(document).ready(function() {
$('#submit').click(function(e) {
e.preventDefault();
$.ajax({
type: 'POST',
url: 'test2.php',
data: {id1: $('#id1').val()},
success: function(data)
{
$("#content").html(data);
}
});
});
});
</script>
HTML
<form name="input">
<input type="text" id="id1">
<input type="submit" id="submit">
</form>
You could try this:
<script>
$('#submitBtn').on('click',function(){
$.ajax({
type: 'POST',
url: 'test2.php',
data: {url: $('#id1').val()},
success: function (data)
{
$("#content").html(data);
}
});
return false;
});
</script>
<form name="input">
<input type="text" id="id1">
<input id="submitBtn" type="submit">
</form>
<div id="content"></div>
How to send some message to other php file? I should to see "Loading" or result from my imput. I try to find some answer even this code is from this place, but still dosn't work.
I have:
<form>
something<input name="sthis" type="text" />
<input type="submit" value="submit" id="submit" />
</form>
<script type="text/javascript">
$(function(){
$('submit').click(function(){
$('#container').append('loading');
var sthis = $('#sthis').val();
$.ajax({
url: 'form1.php' ,
type: 'POST',
data: 'sthis: ' + sthis,
success: function(result){
$('#container').append('<p>' + result + '</p>')
}
});
return false;
});
});
});
</script>
Form1.php
<?php
$str = $_POST['sthis'];
echo $str;
}
?>
Any ideas?
Try this, I think should be like
data: {"sthis": sthis},
<form id="form1">
<input name="sthis" type="text" />
<input type="button" value="submit" id="submit" />
</form>
<script type="text/javascript">
$('#submit').click(function(){
$('#container').append('loading');
var data = $('#form1').serialize();
$.ajax({
url: 'form1.php' ,
type: 'POST',
data: data,
success: function(result){
$('#container').append('<p>' + result + '</p>')
}
});
return false;
});
</script>
This will send all data from your form to php file
So there are some things you have to change in your script:
$(function(){
$(':submit').click(function(event){
event.stopPropagation();
$('#container').append('loading');
var sthis = $('#sthis').val();
$.ajax({
url: 'form1.php' ,
type: 'POST',
data: {'sthis': sthis},
success: function(result){
$('#container').append('<p>' + result + '</p>');
}
});
return false;
});
});
First there is an extra }); at the end you can remove.
Your input field must have an id with the name sthis, not just a name, so you can access it with $("#sthis"), like this:
<input name="sthis" type="text" id="sthis" />
3rd, change your the data line so it look like this:
data: {'sthis': sthis},
Also, to capture the event of the button you have to use $(':submit'), watch for the :
Start learning Ajax (and jQuery), I encounter the issue with post form data, which I could not resolve myself.
First, here is the simple example where I collect data, post it, and displayed it on a new page:
<!-- index.html -->
<form id="myForm" action="test.php" method="post">
<input type="checkbox" id="ch1" name="Ch1" value="1">Check Box 1 <br />
<input type="checkbox" id="ch2" name="Ch2" value="2">Check Box 2 <br />
<input type="submit" id="myButton" />
</form>
<?php
// test.php
print_r($_POST);
?>
Then I tried with Ajax and jQuery. I slightly modified the form pasted above:
<!-- index.html -->
<form id="myForm">
<input type="checkbox" id="ch1" name="Ch1" value="1">Check Box 1 <br />
<input type="checkbox" id="ch2" name="Ch2" value="2">Check Box 2 <br />
<input type="button" id="myButton" value="Submit"/>
</form>
And here is the jQuery function:
// script.js
$(document).ready(function () {
$("#myButton").click(function () {
$.ajax({
cache: false,
type: 'POST',
url: 'test.php',
data: $("#myForm").serialize()
});
});
});
My question is, how to reimplement the jQuery function to echo the result in the test.php tab? I crawled the many resources, but i didn't find any solution.
Thanks in advance for any pointer.
Best, Andrej
You're going to need a success callback, which I'll assume will be some HTML...
$(document).ready(function(){
$("#myButton").click(function() {
$.ajax({
cache: false,
type: 'POST',
url: 'test.php',
data: $("#myForm").serialize(),
success: function(response) {
$("#someElement").html(response);
}
});
});
});
Try this.
$(document).ready(function(){
$("#myButton").click(function() {
$.ajax({
cache: false,
type: 'POST',
url: 'test.php',
data: $("#myForm").serialize(),
success: function(data){
alert(data);
}
});
});
});
I believe you're looking for the success in the .ajax options parameter.
$.ajax({
...
success: function(d){
alert(d);
}
});
im trying to achieve the following, in php i have a form like this:
<form method="post" id="form_1" action="php.php">
<input type="submit" value="add" name="sub"/>
<input type="submit" value="envoi" name="sub"/>
</form>
the form action file is:
<?php
if( $_POST["sub"]=="add"){ ?>
<script>
alert("")
</script>
<?php echo "ZZZZZZ"; ?>
<?php } ?>
so this means if i press sub with value add an alert prompt will come up, how can i do the same thing(differentiate both submit) but using a Ajax request:
the following code so does not work:
$(function(){
$('form#form_1').submit(function(){
var _data= $(this).serialize()
$.ajax({
type: 'POST',
url: "php.php?",
data:_data,
success: function(html){
$('div#1').html(html)
}
})
})
})
</script>
</head>
<body>
<div id="1" style="width: 100px;height: 100px;border: 1px solid red"></div>
<form method="post" id="form_1" action="javascript:;">
<input type="submit" value="add" name="sub"/>
<input type="submit" value="envoi" name="sub"/>
</form>
</body>
You could put the event handler on the buttons instead of on the form. Get the parameters from the form, and then add a parameter for the button, and post the form. Make sure the handler returns "false".
$(function() {
$('input[name=sub]').click(function(){
var _data= $('#form_1').serialize() + '&sub=' + $(this).val();
$.ajax({
type: 'POST',
url: "php.php?",
data:_data,
success: function(html){
$('div#1').html(html);
}
});
return false;
});
});
You have to explicitly add the "sub" parameter because jQuery doesn't include those when you call "serialize()".
In this case you need to manually add the submit button to the posted data, like this:
$(function(){
$('form#form_1 :submit').submit(function(){
var _data = $(this).closest('form').serializeArray(); //serialize form
_data.push({ name : this.name, value: this.value }); //add this name/value
_data = $.param(_data); //convert to string
$.ajax({
type: 'POST',
url: "php.php?",
data: _data,
success: function(html){
$('div#1').html(html);
}
});
return false; //prevent default submit
});
});
We're using .serializeArray() to get a serialized version of the form (which is what .serialize() uses internally), adding our name/value pair to that array before it gets serialized to a string via $.param().
The last addition is a return false to prevent the default submit behavior which would leave the page.
Lots of semicolon missing, see below
$(function(){
$('form#form_1').submit(function(){
var _data= $(this).serialize();
$.ajax({
type: 'POST',
url: "php.php?",
data:_data,
success: function(html){
$('div#1').html(html);
}
});
});
});
jQuery Form plugin provide some advance functionalities and it has automated some tasks which we have to do manually, please have a look at it. Also it provides better way of handling form elements, serialization and you can plug pre processing functions before submitting the form.