passing /retrieving data from jquery to php function

Question

I have actually read all related answers to my question but I need a clear and simple example on how to properly implement my code below.

myHome.php

jquery

var url = "computeArea.php";
var data = $('thisForm').serialize();
$.post(url,data,function(response)); // how do i get the area being returned from  
                                       computeArea function? i need to save the 
                                       return value to a javascript variable

computeArea.php

function computeArea ($data){ // do i need to parse $data to make it an array?
    return $area;
}

im new to jquery and your help is very helpful. thank you!

Did you mean `$.post`? Also check question here http://stackoverflow.com/questions/14220321/how-to-return-the-response-from-an-ajax-call — elclanrs, Jul 12 '13 at 00:38
Are you returning that or echoing? Your computeArea function in computeArea.php — Rainulf, Jul 12 '13 at 00:40
i would like to return it and when it get back to myHome.php, i will use the returned value in my other functions. — iamhealed, Jul 12 '13 at 00:50
You are doing a POST request to computeArea.php. It must echo/print its response -- $area on this case. — Rainulf, Jul 12 '13 at 00:52

Mimo · Answer 1 · 2013-07-12T01:02:08.760

1

You can do:

$.post(url,data,function(response){
   alert(response)
});

ps: you are missing the . between $ and post.

In your php code you could do that:

 echo json_encode($area);

edited Jul 12 '13 at 01:02

answered Jul 12 '13 at 00:40

Mimo

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I added the `echo json_encode($val)` on the php bit, because with just return you won't get anything client side – Mimo Jul 12 '13 at 01:03
ok, i tried this one but my page will only reload. something is wrong with the $.post() because alert ("hello") wont popup. any idea? – iamhealed Jul 12 '13 at 02:43
This might be because you do the serialize on form submit event, which if not prevented issues the page reload. – Mimo Jul 12 '13 at 02:57
ok, the popup will now show but the response is empty. i even pass a literal string just to check if it will return the value but it did not. what could be missing here? – iamhealed Jul 12 '13 at 03:04

Rainulf · Answer 2 · 2013-07-12T01:26:24.333

0

You are misunderstanding the use of post requests. This will not call the computeArea function in computeArea.php and pass data as its parameter:

var data = $('thisForm').serialize();
$.post(url,data,function(response));

You can do this instead for computeArea.php:

$data = $_POST['watever_you_are_serializing'];
// Do computations, etc.
$area = 123; // Contains computed area
echo $area;  // Or json_encode($area);

If you need to call that function from computeArea.php, then you can create a new file for $.post request (eg. computeArea2.php) and include computeArea.php from there. It would be something like this:

include 'computeArea.php';
$data = $_POST['watever_you_are_serializing'];
echo computeArea($data);

edited Jul 12 '13 at 01:26

answered Jul 12 '13 at 01:07

Rainulf

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hello, i just made it this way to make it simpler..but i really need to post the value because the php function is a third party app. – iamhealed Jul 12 '13 at 01:16
If that's the case, then you can just call that function with the data you will get from $_POST as its parameter. And finally echo its return. – Rainulf Jul 12 '13 at 01:18

score 0 · Accepted Answer · answered Jul 12 '13 at 03:37

0

Do a simple teste.

jQuery:

$.post("url/to/file.php",{variable_name: "hello"/*(we'll give this value to variable_name*/)},
  function(response){
    if(response>0)
       alert('Something went wrong');
    else
       alert(response);
});

Now, on server side:

<?php
 if(isset($_POST['variable_name']) && $_POST['variable_name']!=="")
  echo $_POST['variable_name'];
 else
  echo 1;
?>

answered Jul 12 '13 at 03:37

calexandru

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im getting "Something went wrong" output..is there something that I might missed? – iamhealed Jul 12 '13 at 03:51

score -1 · Answer 4 · answered Jul 12 '13 at 00:39

-1

Something along the lines of:

var url  = "computeArea.php",
    data = $('thisForm').serialize(),
    new_variable;

$.post(url, data, function(response) {

    new_variable = response;

});

Though I presume there's a bit more to your PHP script, as otherwise $area isn't defined anywhere.

answered Jul 12 '13 at 00:39

christian.thomas

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by the way, regarding the $data , when my php function receives it, how can i convert it to an array? – iamhealed Jul 12 '13 at 01:03

passing /retrieving data from jquery to php function

4 Answers4