<div id="success_hold">
<?php
if($row['mandate_end']!=date('Y-m-d') && $row['job_position']=='unhold')
{
echo '<span class="badge badge-success">Open</span>';
}
else
{
echo '<span class="badge badge-warning">Hold</span>';
}
?>
</div>
<script>
$(document).ready(function(){
var flag = 1;
$(".hold").click(function(){
job_id = this.id;
if(flag == 1)
{
$.ajax({
type:"POST",
data:{"job_id":job_id},
url:"<?php echo base_url(); ?>pausehold",
success:function(data){
$("#success_hold").load("#success_hold");
}
});
flag = 0;
}
else
{
$.ajax({
type:"POST",
data:{"job_id":job_id},
url:"<?php echo base_url(); ?>resumehold",
success:function(data){
$("#success_hold").load("#success_hold");
}
});
flag = 1;
}
});
});
</script>
I have div where id="success_hold". Now, what happens when I click on class="hold" I am updating my data. Now, I want to refresh my <div id="success_hold"> without reload whole page where I am using .load() function But the problem is that when I use .load() the function shows the whole page in success_hold. So, How can I fix this issue? I want to refresh only success_hold.
Thank You
The problem is because load() is used to make another AJAX request and place the a portion of the HTML retrieved from that request in the target element.
As you're already making an AJAX request to get the data, which is presumably HTML, you simply need to append() data to the container.
Also note that the only difference between the two sides of your condition is the URL the request is sent to, so you can easily DRY this code up:
$(document).ready(function() {
var flag = true;
$(".hold").click(function() {
var url = '<?php echo base_url(); ?>' + (flag ? 'pausehold' : 'resumehold');
flag = !flag;
$.ajax({
type: "POST",
data: { job_id: this.id },
url: url,
success: function(data) {
$("#success_hold").append(data);
}
});
});
});
If data contains the entire page, then you should change that PHP file to return only the relevant HTML as it will help to speed up the request. If, for whatever reason, you can't do that then you can extract the required element from the response like this:
$("#success_hold").append($(data).find('#success_hold'));
success:function(data){
$("#success_hold").load("#success_hold");
}
This is your success method in ajax. You want to pur the content of data into your div #success_hold ?
If you want that, just do this :
success:function(data){
$("#success_hold").html(data);
}
Related
I want to send data from javascript to another php page where I want to display it. I found that I need to use Ajax to pass the data to php so I tried myself.
My file where is the javascript:
$('#button').on('click', function () {
$.jstree.reference('#albero').select_all();
var selectedElmsIds = [];
var selectedElmsIds = $('#albero').jstree("get_selected", true);
var i = 0;
$.each(selectedElmsIds, function() {
var nomenodo = $('#albero').jstree('get_selected', true)[i].text;
//var idnodo = selectedElmsIds.push(this.id);
var livellonodo = $('#albero').jstree('get_selected', true)[i].parents.length;
//console.log("ID nodo: " + selectedElmsIds.push(this.id) + " Nome nodo: " + $('#albero').jstree('get_selected', true)[i].text);
//console.log("Livello: " + $('#albero').jstree('get_selected', true)[i].parents.length);
i++;
$.ajax({
type: "POST",
data: { 'namenodo': nomenodo,
'levelnodo': livellonodo
},
success: function(data)
{
$("#content").html(data);
}
});
});
});
I want to send the data to another php page which consists of:
<?php echo $_POST["namenodo"]; ?>
But when I try to go to the page there's no data displayed.
This is a very basic mistake I think every beginner (including me) does while posting a data using ajax to another php page.
Your ajax code is actually posting the data to lamiadownline.php (if you are using the variables correctly) but you can't get that data by simply using echo.
Ajax post method post data to your php page (lamiadownline.php) but when you want to echo the same data on the receiver page (lamiadownline.php), you are actually reloading the lamiadownline.php page again which makes the $_POST["namenodo"] value null.
Hope this will help.
First of all you won't be able to see what you have post by browsing to that page.
Secondly, is this
<?php echo $_POST["namenodo"]; ?>
in the current page?
Otherwise, specify the url
$.ajax({
url: "lamiadownline.php",
type: "POST",
data: { 'namenodo': nomenodo,
'levelnodo': livellonodo},
success: function(data) {
$("#content").html(data);
}
});
//try this
$.ajax({
type: "POST",
url:"Your_php_page.php"
data: { namenodo: nomenodo levelnodo: livellonodo},
success: function(data)
{
$("#content").html(data);
}
});
my website has an option of country like for different country the website layout is different. it is running on the basis of sessions if session is not set the user will be redirected to index to select a country then will be redirected from the page where he originally came from. here's the code
my session_check_client.php file that is included in every file except index
<?php
session_start();
if(!isset($_SESSION['country']))
{
$actual_link = "http://$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]";
header("location:index.php?return_uri=$actual_link");
}
?>
now what happens is when i go back to home page i wanna check whether this requested has some return parameter or just user has visited he website for the first time. there are two button for two countries of which i am showing the code.
function canada(){
$.ajax({
type: 'post',
url: 'ajax_country.php?country=canada',
success: function (data) {
var $_GET = <?php echo json_encode($_GET);?>;
if($_GET){
//window.location.href=$_GET['return_uri'];
alert($_GET['return_uri']);
}
else {
window.location.href = "home.php";
}
}
});
}
function us(){
$.ajax({
type: 'post',
url: 'ajax_country.php?country=us',
success: function (data) {
var $_GET = <?php echo json_encode($_GET);?>;
if($_GET){
//window.location.href=$_GET['return_uri'];
alert($_GET['return_uri']);
}
else {
window.location.href = "home.php";
}
}
});
}
now the problem is when i am alerting the value of $_GET['return_uri'] it is giving me a false value
e.g my return_uri value is http://localhost/interfold/products2.php?category=Aprons&id=57725599688 it actually shows the whole value in return_uri in index page like http://localhost/interfold/products2.php?category=Aprons&id=57725599688 but when is get the url value using javascript it is onlye giving me the value http://localhost/interfold/products2.php?category=Aprons it is missing the $ and afterwards parts!!! any recommendations?
Since you are only using super global variables, you can directly print a JS variable above the function you are describing:
var actual_link = "<?php echo "http://$_SERVER[HTTP_HOST]$_SERVER[REQUEST_URI]"; ?>";
function postCountry( country ){
$.ajax({
type: 'post',
url: 'ajax_country.php?country=' + country,
success: function (data) {
if(actual_link){
//window.location.href=actual_link;
alert(actual_link);
}
else {
window.location.href = "home.php";
}
}
});
}
postCountry('us');
postCountry('canada');
I've got this variable $type and I want it to be month or year.
It should be changed by pressing a div.
I've tried creating an onclick event with an ajax call.
The ajax call and the variable are in the same script (index.php)
Inside the onclick function:
var curr_class = $(this).attr('class');
$.ajax({
type: "POST",
url: "index.php",
data: {
type: curr_class
},
dataType: 'text',
success: function(data) {
// Test what is returned from the server
alert(data);
}
});
But the alert returns the whole html page.
When I console.log the data (create a var data = { type:curr_class }) and console.log *that data* it returnstype = month` (which is correct)
while I just want it to return month or year
So on top of the page I can call
if(empty($_POST['type'])){
$type = 'month';
} else {
$type = $_POST['type'];
}
and change the PHP variable so I can use it in the rest of my script.
But how can I accomplish this?
With kind regards,
as you are sending request to the same page so as a result full page is return .You will have to send it to another page and from that page return the type variable
if(empty($_POST['type'])){
$type = 'month';
} else {
$type = $_POST['type'];
echo $type;
keep this code in separate file and make an ajax call to that page
//Try This It's Work
Get Value
Get Value
$(".btn-my").click(function(){
var curr_class = $(this).data('title');
$.ajax({
type: "POST",
url: "index.php",
data: {
type: curr_class
},
dataType: 'text',
success: function(data) {
// Test what is returned from the server
alert(data);
}
});
});
For a link like this
DELETE
i am using this ajax call
$(function() {
$(".delete_msgs").click(function() {
var msgid = $(this).attr("id");
var dataString4 = 'msgid=' + msgid;
$('a#'+msgid).html('<img src="img/loader.gif" class="loading" />');
$.ajax({
type: "POST",
url: "del_msg.php",
data: dataString4,
cache: false,
success: function(data){
if (data == 0) {
//alert('Sent!');
} else {
$('a#'+msgid).html(data);
$('.viewmessage'+msgid).hide();
}
}
});
return false;
});
});
The del_msg.php file contains that code:
if(isset($_POST)) {
$msgid = $_POST['msgid'];
$delmsg = mysql_query("DELETE FROM messages
WHERE id = $msgid") or die(mysql_error());
echo 'Message sent!';
} else {
echo 0;
}
My big question is, how can i protect my send throught AJAX function? Anyone can change the id of the #a element and, consequently, can delete any row from my database.
Is there a more secure way to call php with ajax, or at least, to hide my important variables sent via ajax?
Ajax code
$("#login_btn").click(function() {
var url = "core/login.php";
$.ajax({
type: "POST",
url: url,
data: $("#sr").serialize(),
success: function(data) {
var responseData = jQuery.parseJSON(data);
$('.oops').html('<div class="error">'+responseData.oops+'</div>');
alert(responseData.good);
if(responseData.good === 1) {
alert(good)
location.reload();
}
}
});
});
Php code if everything passes
else {
$_SESSION['id'] = $login;
$good = 1;
exit();
}
How come it's not freshing the page with location.reload? Should I be using .done .try?
Dont refresh, the whole idea of using ajax is that you dont need to.
You can set your sessions in the ajax php script and pull new content based on the change in session.
That being said, you are missing a ';' at
alert(good);