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 $(':submit'), watch for the :
Related
i have a form and i wanna use jQuery with it. but when i submit it it just doesnt use the code at all.
<form id="form">
<label>Name</label><br />
<input type="text" name="name"><br /><br />
<label>Message</label><br />
<textarea name="message"></textarea><br /><br />
<button type="submit">Submit form</button>
</form>
<script src="https://code.jquery.com/jquery-3.6.3.js" integrity="sha256-nQLuAZGRRcILA+6dMBOvcRh5Pe310sBpanc6+QBmyVM=" crossorigin="anonymous">
//have to use jQuery to get the form data
jQuery(document).ready(function($){
$('#form').submit(function(e){
e.preventDefault();
alert('hello');
var data = $(this).serialize();
$.ajax({
url: 'http://localhost:8888/wp-json/v1/contact_form/submit',
type: 'POST',
data: data,
success: function(response){
console.log(response);
}
});
});
});
</script>
i used a alert to see if it works but it just does nothing.
when i submit i get to a blank page. so the e.preventDefaults also doesnt work.
i tried using copilot but it didnt do anything usefull
You must declare your scripts in different tags.
<script src="https://code.jquery.com/jquery-3.6.3.js" integrity="sha256-nQLuAZGRRcILA+6dMBOvcRh5Pe310sBpanc6+QBmyVM=" crossorigin="anonymous"></script>
<script>
//have to use jQuery to get the form data
jQuery(document).ready(function($){
$('#form').submit(function(e){
e.preventDefault();
alert('hello');
var data = $(this).serialize();
$.ajax({
url: 'http://localhost:8888/wp-json/v1/contact_form/submit',
type: 'POST',
data: data,
success: function(response){
console.log(response);
}
});
});
});
</script>
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 $_POST["All"] does not exist. It is all. (php)
Second, you send a GET request not a POST one. (jQuery)
Third, format your date into json. A good way to do this is to create a variable right after compare_id.push, it's more readable, as so :
var json_data = {"my_array" : [1,2, "bonjour", 4]};
Your problem is mostly related to "how to debug". I think you should print what's happening along the way to figure out what's happening.
I have below code. when I click reply link, it will show a comment box.
I want 2 points.
or give me best way to do my work.
When 1 comment box open other must be hide.
when I click on send button correct value should send vie serialize values.
These are the codes
PHP code
<?PHP
for($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!
the problem is that i am unable to insert data in my database without reloading
i have tested it without the note_sql.js and my note_sql.php works fine but there seems to be a problem in my js file if some body could point me in the right direction it would be wonderful
Index.php
<form id="noteform" action="note_sql.php" method="POST">
<input type="text" name="note"></input>
<input id="sub" type="submit" value="Save Note" />
</form>
<span id="result_note"><span>
<script type ="text/javascript" src="jquery/note_sql.js"></script>
note_sql.js
$("#sub").click(function(){
var data = $("#noteform: input").serializeArray();
$.post($("#noteform").attr("action"),data,function(info){$("result_note").html(info); });
clearInput();
});
$("#noteform").submit(function(){
return false;
});
function clearInput(){
$("#noteform :input").each(function(){
$(this).val('');
});
}
Use Jquery Ajax the working code is given below :
please add ID to Note Form Element :
<pre>
<script>
$(document).ready(function () {
$("#sub").click(function () {
var name = $("#note").val();
var message = $("#message").val();
$.ajax({
type: "post",
url: "note_sql.php",
data: "name=" + name,
success: function (data) {
alert(data);
}
});
});
});
</script>
</pre>
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>