I've created one Ajax function like this:
function searchPlanAjax(url){
jQuery('#loader').css('display','block');
var plan = jQuery('.rad').val();
var site = url+"fbilling_details/subscription";
jQuery.ajax({
type: "POST",
url: site,
data: "plan="+plan,
//data: "business_name="+val+"&bs="+bs,
success: function(msg){
jQuery('#loader').css('display','none');
}
});
}
Which I'm calling on radio button's onclick:
<?php echo $form->text(BILLING_DETAIL.'][id_plan', array('type'=>'radio', 'value'=>$val[PLAN]['id'], 'id'=>'id_plan_'.$val[PLAN]['id'], 'class'=>'rad','onclick'=>'searchPlanAjax('."'".SITE_NAME.ROOT_FOLDER_NAME."'".')')); ?>
The Ajax call will be processed in controller of cakePHP like this:
if(!empty($_REQUEST['plan'])){
$search_plan = $this->{PLAN}->find("all",array("conditions"=>array("id=".$_REQUEST['plan'])));
$this->set('search_plan',$search_plan);
}
but I couldn't get value in $search_plan variable. thanks if anybody can help.
Look at conditions in your find query
if(!empty($_REQUEST['plan'])){
$search_plan = $this->{PLAN}->find("all",array("conditions"=>array("id" => $_REQUEST['plan'])));
$this->set('search_plan',$search_plan);
}
Related
I would like this function:
After I select an option I want to update right away the database and then show it right away also inside a certain div. How can I achieve this?
Here's my HTML:
<select id="category1" name="cat1">
<option>Hardware</option>
<option>Software</option>
<option>Network</option>
</select>
AJAX code:
$('#category1').on('change',function(){
var data = $("cat1").val();
var ticket = '<?php echo $ticket_details[0]["cTicketNo"];?>';
if(data){
jQuery.ajax({
type: "POST",
dataType: 'json',
url: "<?php echo base_url();?>User/TicketView/update_cate",
data: {data: data,ticket: ticket},
success:function(response){
alert('successful');
},
error: function(response){
alert('tangina!');
}
});
}
});
Controller:
function update_cate(){
$this->TicketView_m->update_cat1();
redirect($_SERVER['HTTP_REFERER']);
}
Model:
function update_cat1(){
$ticket_id = $_POST['ticket'];
var_dump($cat = $_POST['data']);
$this->db->set('vCategory',$cat);
$this->db->where('cTicketNo',$ticket_id);
$this->db->update('tconcerns');
}
THanks for all the help!
Drop that redirect($_SERVER['HTTP_REFERER']);, You don't need that.
First of all, AJAX will allow you to send data to server without redirecting to another page.
Based on your AJAX dataType. It is JSON. So for you to get the result, You have to encode it to JSON like json_encode(['result' => 'things you want to show']).
Then you can access it in your response like this:
success:function(response){
alert('successful');
var res = response.result; //Access the JSON
$("#certainDiv").html( res ); //Print to the div you want
}
And it is also better to get the post request in your controller then pass it to your model:
function update_cate(){
$ticket_id = $this->input->post('ticket');
$cat1 = $this->input->post('cat1');
$this->TicketView_m->update_cat1($ticket_id, $cat1);
echo json_encode(['result' => 'things you want to show']);
}
I am fairly new to Laravel and ajax in general, what I am trying to implement is to pass the value of an input field through an ajax get request.
My request looks like this:
function getInfo() {
$.ajax({
url: "info",
dataType: "json"
}).success(function(data){
$('#result').append(JSON.stringify(data));
}).fail(function(){alert('Unable to fetch data!');;
});
}
$('#infoSubmit').click(getInfo);
I have setup a route for my function in laravel that works like this
public/info/Variable <--
When I add a variable after info/
I get the data for that variable (e.g profile name)
I need to pass this variable from an inputfield to ajax request to something like this:
url: "info/+$inputfieldVariable"
Change:
url: "info",
TO:
url: "info/" + $('input-field-selector').val(),
Not sure about the correctness of your JS code: Shouldn't you be using done instead of success?
JavaScript:
function getInfo() {
var myFieldsValue = {};
var $inputs = $("#myForm :input");
$inputs.each(function() {
myFieldsValue[this.name] = $(this).val();
});
$.ajax({
url: "info",
dataType: "json",
data: myFieldsValue,
type: "GET" // Even if its the default value... looks clearer
success: function(data){
$('#result').append(JSON.stringify(data));
},
error: function(){
alert('Unable to fetch data!');
}
});
return false;
}
$('#infoSubmit').click(getInfo);
Untested but should be something like that
I have modified the code
to POST prodID to ProductsList.php
// its a dynamically generated drop menu
while($rowmnu2=mysql_fetch_assoc($resulmnusub2))
{
echo '<li><a id="'.$rowmnu2['liid'].'" href="#" onclick="passto(this.id)">'.$rowmnu2['title'].'</a></li>
';
}
and here is my ajax function :
function passto(val){
//window.location.href="ProductsList.php?idd=" + val;
$.ajax({
url: 'ProductsList.php',
type: "POST",
data: ({prodID: val}),
success: function(data){
//or if the data is JSON
window.location.href="ProductsList.php";
}
});
}
the passed element to the function is an integer
in the ProductsList.php I have
<?php
if(!$_POST['prodID']) die("There is no such product!");
echo $_POST['prodID'];
?>
and I get There is no such product! while there should be an INT #
why is that ?
any one knows? all the bellow suggestions are not responding correctly
$(document).ready(function() {
$("a").click(function(event) {
myid = $(this).attr('id');
$.ajax({
type: "POST",
url: "ProductsList.php",
data: {prodID: myid},
dataType: "json",
complete:function(){
window.location("ProductsList.php");
}
});
});
});
if you want to POST id , you can change:
...onclick="passto(this)"...
to
...onclick="passto(this.id)"...
That behavior is normal because you are requesting ProductsList.php twice. the first time with an AJAX request using $.ajax. for that time the id is sent correctly. The problem is that you request ProductsList.php again just after AJAX complete using window.location.href="ProductsList.php"; without sending anything. So the result is as expected, a page printing There is no such product!
You can fix the problem by replacing window.location.href="ProductsList.php"; by this one :
$('body').html(data);
or any other instruction to use properly the returned data.
You can either use my edited code or just edit yours :
echo '<li ><a id="'.$rowmnu2['link'].'" href="#">'.$rowmnu2['title'].'</a></li>';
JS part :
$('a').click(function() {
var val = $( this ).attr('id');
$.ajax({
type: "POST",
url: "ProductsList.php",
data: {prodID:val},
complete:function(){
$('body').html(data);
}
});
});
Actually the following function works fine, but now I need to add other variable in order to return from the php file the right statement.
function sssssss1(page) {
loading_show();
$.ajax({
type: "GET",
url: "load_data.php",
data: "page=" + page,
success: function (msg) {
$("#search").ajaxComplete(function (event, request, settings) {
loading_hide();
$("#search").html(msg);
});
}
});
}
I need to add the following two variable to be read by my php file. I have tried different solution, but nothing seem working
var form2 = document.myform2;
var dataString1 = $(form2).serialize();
How to add those variable in my existing function? Any idea?
You can send object as data,
this line:
data: "page="+page,
could be
data: {mypage:"page="+page, form2:document.myform2, dataString1:$(form2).serialize()}
and your PHP can get it like:
$page = $_GET['mypage'];
$form2 = $_GET['form2'];
$dataString = $_GET['dataString1'];
Hope it Help.
My imaginery code:
$(document).ready(function() {
$("#sub").click(function() {
info['moto'] = $("#moto").val();
info['motox'] = $("#motox").val();
$.ajax({
type: "POST",
url: "index.php",
data: "arr="+info,
success: function(msg){
$('.answer').html(msg);
}
})
})
})
How could I make, that after receiving it in .php file I could use POST method like this: $_POST['moto'] and $_POST['motox'] or something like that? What should I change? Thanks.
Just:
data: info,
(And you need to initialize info as an object in the first place: var info = {})
Check out jQuery serialize(), it does all the work for you if you are working with form inputs.
I thinkyoushould go also the same way like jquery-ajax-data-array-from-php