Develop Reference

show values entered in form as a table before getting submitted

First, I do not have much talent in ajax so please consider that while answering. Here I want to get all the values which I inserted into the form in a table before getting submitted but unfortunately am not getting the results I want. So far I have done this. Please have a look.
<script>
var form_data={ agent_name: $('#agent_name').val(),
type: $('input[name="type"]').val(),
number: $('#number').val(),
quantity: $('#quantity').val()
}
$.ajax({
type: 'POST',
url: '<?php echo base_url();?>admin_control/ajax_data',
data: form_data,
dataType:"json", //to parse string into JSON object,
success: function(data){
if(data){
var len = data.length;
alert(len);
var txt = "";
if(len > 0){
for(var i=0;i<len;i++){
if(data[i].agent_name && data[i].type){
txt += $('#table').append('<tr class="item-row"><td><input type="hidden" id="add_type" name="add_type[]" value="super">Super</td><td><input type="hidden" id="add_amount" name="add_amount[]" value="10">745</td><td><input type="hidden" class="add_quantity" id="add_quantity" name="add_quantity[]" value="10">10</td><td name="add_amount" id="add_amount">100</td><td><input type="checkbox" class="add_checkbox" name="layout" id="add_checkbox" value="1" checked></td></tr>');
}
}
if(txt != ""){
$("#table").append(txt).removeClass("hidden");
}
}
}
},
error: function(jqXHR, textStatus, errorThrown){
alert('error: ' + textStatus + ': ' + errorThrown);
}
});
return false;
</script>
Here I want to pass the values of form_data in to the table I had written and how can we pass that and did it compulsory to use the url in ajax while using this am getting an error like error: parsererror: SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data.
Here is my controller
public function ajax_data()
{
$array = array("agent_name" => "admin","number"=>"785","type"=>"super","quantity"=>"10");
$data['json'] = $array;
echo json_encode($data);
}

The Ajax post will call the controller method for post values. The ajax code usually correct but you have not defined url.
url: '',
Firstly define a method (ex: HomeController) then set ajax url parameter like
url: '#Url.Action("Save","Home")' or url: 'Save/Home'
[HttpPost]
public JsonResult Save(string agent_name, string type , int number,int quantity)
{
// return json
}
Note :
The data types you send from the form with the method types must be
the same.

Well, just to give you some hints to work on: The Ajax call could look like this:
$.ajax({
type: 'POST',
url: 'echodata.php',
data: form_data,
dataType:"json",
success: function(data){
$('#table').append('<tr><td>'+data.agent_name+'</td>' // + more columns ...
+'</tr>');}
);
The php script should really be something that processes the incoming information (passed as the $_POST array) into some other data that will ultimately be echoed as JSON string.
This is just a trivial version of echodata.php for testing:
<?php
echo json_encode($_POST);
?>
EDIT:
While you have fixed your URL attribute and have added a "controller" on the server your Ajax success function still expects something it will not get. Your PHP script delivers a JSON string that will be parsed into as a JavaScript object and not an array. The object will not have the length property your success function tries to use. Instead you should be looking for the json property.
So, there is no point for the for loop, instead you can use data.json.agent_name, data.json.number and data.json.type (etc.) properties directly. Your current .append() also only adds a static string to your table. This should of course be changed to use the freshly received values.

AJAX not coming up a success even though its updating the database

My php is updating the table but not refreshing in javascript have tried several different ways of doing this and nothing is working.
PHP
$sql = "UPDATE INTOXDM.ASTP_FORM SET SUPERVISOR_EID = '".$newSuper."' WHERE FORMID = '".$formId."'";
$row = $xdm->fetch($sql);
$return["color"] = $row['APPRENTICE_SIGNATURE'];
$return["json"] = json_encode($return);
echo json_encode($return);
?>
Javascipt
var data = {
"formId": formID,
"newSuper": newSuper
};
data = $.param(data);
$.ajax({
type: "POST",
dataType: "json",
url: "src/GetInfo.php",
data: data,
success: function() {
location.reload();
}
});

I'd start by modifing the code like this:
var data = {
"formId": formID,
"newSuper": newSuper
};
// No need for serialization here,
// the 'data' parameter of jQuery.ajax accepts JS object
// data = $.param(data);
$.ajax({
type: "POST",
dataType: "json",
url: "src/GetInfo.php",
data: data,
// As suggested by Rocket Hazmat, try to add an error callback here
error: function(jQueryXHR, textStatus, errorMessage) {
console.log("Something went wrong " + errorMessage);
},
success: function(jsonResponse) {
// Try to reference the location object from document/window
// wd = document or window as seen here http://stackoverflow.com/questions/2624111/preferred-method-to-reload-page-with-javascript
// Also watch out, usually browsers require a user confirmation before reloading if the page contains POST data
// One of these should be fine
wd.location.assign(wd.location.href) : go to the URL
wd.location.replace(wd.location.href) : go to the URL and replace previous page in history
wd.location.reload(<true/false/blank>) : reload page from server/cache/cache
}
});
Also, this might be a shot in the dark but the parameter dataType gave me problems sometime in the past, so if you are sure about the json returned by your php script, you could use the eval function to jsonify the response
$.ajax({
...
// Remove data type
// dataType: "json",
...
success: function(plainTextResponse) {
// Eval response, NOT SAFE! But working
var jsonResponse = eval('('+ plainTextResponse +')');
...
}
});

Your ajax is expecting json data and your php is sending malformed json string. Send a correct json string and your script will work fine.
Your php json_encode should be like this:
$data = json_encode($return);
echo $data;

Making Key Value pair for the form elements in JavaScript

I have a module where the forms created are dynamic. So the number of inputs can defer always. Also, the array key can also defer.
My current method of posting form is this:
name = form_options[option_1] value = 1
On submitting the form using POST, I get the form as array in $_POST, which looks like this.
form_options(
option_1 => 1
)
But, now I am trying to implement the same thing using AJAX. So, I would need a common module to get all form values.
I found a way to do it.
var objectResult = $('#options_form').serializeArray();
console.log(objectResult);
This gives me a result like this:
0: Object
name: "form_options[option_1]"
value: "1"
How can parse this result to get an array like $_POST array, which I can send as data in AJAX.
P.S: All the form elements have name field as form_options[key]

You should use this for get post data in PHP file.
// You can use like this
var objectResult = $('#options_form').serializeArray();
$.ajax({
type: "POST", // Enter Request type GET/POST
url: 'action.php', // Enter your ajax file URL here,
dataType: 'json', // If you are using dataType JSON then in php file use die( json_encode($resultArray) );
data: objectResult, // Put your object here
beforeSend: function(){
alert('before');
},
error: function(data) {
console.log(data);
},
success: function(response){
console.log(response);
}
});
// In php file get values like this way
$_POST['form_options']

try like this,
In JQuery:
var objectResult = $('#options_form').serializeArray();
$.ajax({
type: 'POST',
url: 'yoururl'',
data: objectResult ,
success:function(data){
alert(data);
}
});
In your php:
<?php echo var_dump($_POST);?>

You can use jquery serialize with ajax directly, there is no need to use serializeArray:
$.ajax({
type:"post",
url:"formHandleSkript.php",
data: $("#options_form").serialize(),
success: function(response){
$(".result").html(response);
}
});
For more information see http://api.jquery.com/serialize/

Ajax Error Potentially Due to Incorrect Data Object Being Passed

Hi I am new to ajax and I am attempting to pass a Json to a Database, but I am not that far yet. Currently I am attempting to be verified that the data I am passing is being done successfully. However, I always drop into the ajax error method. I will upload my code and the way the data looks and then the error.
Thank you for your help!
<script>
function updateTable()
{
alert("Do i try to update table?");
document.getElementById("testLand").innerHTML = "Post Json";
//echo new table values for ID = x
}
function popupClick (){
var popupObj = {};
popupObj["Verified_By"] = $('#popupVBy').val();
popupObj["Date_Verified"] = $('#popupDV').val();
popupObj["Comments"] = $('#popupC').val();
popupObj["Notes"] = $('#popupN').val();
var popupString = JSON.stringify(popupObj);
alert(popupString);
$.ajax({
type: "POST",
dataType: "json",
url: "popupAjax.php",
data: popupObj,
cache: false,
success: function(data)
{
alert("Success");
updateTable();
},
error: function(data)
{
alert("there was an error in the ajax");
alert(JSON.stringify(data));
}
});
}
</script>
JSON Being Passed shown in var popupString:
Error:
popupAjax.php file (warning it's testy)
<?php
echo "Testing tests are testy";
?>

You are specifying the dataType as json. But this is the returned data type, not the type of the data you are sending.
You are returning html / text so you can just remove the dataType line:
type: "POST",
url: "popupAjax.php",
If you do want to return json, you need to build your datastructure on the server-side and send it at the end. In your test-case it would just be:
echo json_encode("Testing tests are testy");
But you could send a nested object or array as well.
As an additional note, you can use .serialize() on your form (if you use a form...) so that jQuery automatically builds an object that you can send in the ajax method. Then you don't have to do that manually.

JSON Response {"d":"128.00"} but displaying "128"

I have been working on a shopping cart that the user can add/remove order items as they please and am returning an updated sub-total via a webservice using jQuery $.ajax
Here is how I am calling the webservice and setting the sub-total with the response.
//perform the ajax call
$.ajax({
url: p,
data: '{' + s + '}',
success: function(sTotal) {
//order was updated: set span to new sub-total
$("#cartRow" + orderID).find(".subTotal").text(sTotal);
},
failure: function() {
//if the orer was not saved
//console.log('Error: Order not deleted');
}
});
The response I am getting seems perfectly fine:
{"d":"128.00"}
When I display the total on the page it displays as 128 rather than 128.00
I am fully sure it is something very simple and silly but I am so deep into it now I need someone with a fresh brain to help me out!!
Cheers :)
EDIT
I am also using $.ajaxSetup to set the correct contentType:
$.ajaxSetup({
type: "POST",
contentType: "application/json; charset=utf-8",
data: "{}",
dataFilter: function(data) {
var msg;
if (typeof (JSON) !== 'undefined' &&
typeof (JSON.parse) === 'function')
msg = JSON.parse(data);
else
msg = eval('(' + data + ')');
if (msg.hasOwnProperty('d'))
return msg.d;
else
return msg;
}
});

That is because the value is treated as a number, while you want it treated as a string.
When you are using '.. .text(sTotal)', you are actually calling the .toString() method on the Number object wrapping the primitive sTotal. And since this is a whole number, it displays it without decimals.
You need to use a format the number as a string prior to calling .text(foo) for the number to be formatted like that.
This will give you two decimals
var a=1/3;
a = a.toString();
switch(a.lastIndexOf(".")){
case -1:
a+=".00";
break;
case a.length-2:
a+="0";
break;
default:
a=a.substring(0, a.indexOf(".") + 3);
}
alert(a);

I don't see anywhere in this code where you access the d property of the response.
Perhaps you mean to do this?
$("#cartRow" + orderID).find(".subTotal").text(sTotal.d);
// --------------------------------------------------^^
EDIT
Ok, I see the problem. You're returning JSON but not defining a dataType in the $.ajax() call. This means that jQuery sees your application/json mimetype and interprets the response as JSON. 128.00 in JSON is a Number, not a String. However, "128.00" would be a String.
In order to keep this working, You need to format the response before printing it (as others have suggested), or adjust your endpoint to return a valid JSON string.
Here's my test to prove the solution
<div id="test">
Subtotal <span class="subTotal"></span>
</div>
<script type="text/javascript" src="http://www.google.com/jsapi"></script>
<script type="text/javascript" charset="utf-8">
google.load("jquery", "1.4.2");
</script>
<script type="text/javascript" charset="utf-8">
$.ajax({
url: 'test.php',
data: {},
success: function(sTotal) {
//order was updated: set span to new sub-total
$("#test").find(".subTotal").text(sTotal);
}
});
</script>
and test.php
<?php
header( 'Content-type: application/json' );
echo '128.00';
Output
Subtotal 128
But when I change test.php to be this
<?php
header( 'Content-type: application/json' );
echo '"128.00"';
The expected output is generated
Subtotal 128.00
Or, you could alternatively tell jQuery to treat the response as text by specifying a dataType parameter, for example
$.ajax({
url: 'test.php',
data: {},
dataType: 'text', // <---- here
success: function(sTotal) {
//order was updated: set span to new sub-total
$("#test").find(".subTotal").text(sTotal);
}
});
EDIT 2
Ok, after messing with this some more, I see what's going on. The dataFilter handler you defined converts the response into JSON itself, and in this case, returns the string 128.00. However, jQuery still applies the intellgent-guessed dataType (which is JSON) to this value before sending it to the success handler.
There are a multitude of ways to fix this, all of which depend on what other AJAX calls your application relies on this setup for. The quick-fix I applied in my test was to do this
$.ajaxSetup({
type: "POST",
contentType: "application/json; charset=utf-8",
data: "{}",
// define the text data type so that we return data.d, jQuery doesn't parse it as JSON again
dataType: 'text',
dataFilter: function(data) {
data = $.parseJSON( data ); // Use jQuery's parsing
if (data.hasOwnProperty('d'))
{
return data.d;
}else{
return data;
}
}
});
But that may not work across the board for you

Please try this:
$("#cartRow" + orderID).find(".subTotal").text(sTotal.toFixed(2));
HTH