how to send and receive objects or values using JQuery-AJAX, JSON and php?

Question

How can I send and receive values?

What is it that I'm doing wrong?

jQuery: on click of save button

$("#budget_submit").click(function(){       
        var budget = $("#budget_cardno").val();
        $.ajax({
            type: 'POST',
            dataType: "json",
            url: 'budget.php',
            data: budget,
            success: function(data){
                if(data.success == true)
                alert(data.message);
        }
        });
    });

html:

<form method="post" id="budgetform">
   <fieldset>
      <ul>
    <li>
      <label>CARD NUMBER</label>
      <input name="budget_cardno" id="budget_cardno" />
    </li>
    <li>
      <label>MONTHLY BUDGET</label>
      <input name="budget_monthly" id="budget_monthly" />
    </li>
    <li>
      <input class="submit" id="budget_submit" name="budget_submit" type="submit" value="Save"/>
        </li>
      </ul>
   </fieldset>
</form>

Php:

<?php 
$dbhost = '#';
$dbuser = '#';
$dbpass = '#';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
    die('Could not connect: ' . mysql_error());
}

$cardNo = $_POST['budget'];     #This statement does not work.
if($cardNo == "")
{
    mysql_close($conn);
    $data = array('success' => true, 'message' => 'ENTER DETAILS');
    echo json_encode($data);
}

else
{
}
?>

how can I send the value budget to the PHP file and store it in $cardNo and return the message?

I don't want to check for the condition at the client side because I have other conditions to check which needs data from database.

you can check the $_POST received by the php script by using echo print_r($_POST); it will tell you the key and the value of the data received (if any). I don'T see any trouble at first glance. You might want to consider using cosole tools like firebug or something similar to see what is sent and retuned. — Louis Loudog Trottier
– Louis Loudog Trottier, Commented Nov 18, 2012 at 2:59

Louis Loudog Trottier · Accepted Answer · 2012-11-18 03:03:25Z

1

actually this seems weird to me.

if($cardNo == "")
{
    mysql_close($conn);
    $data = array('success' => true, 'message' => 'ENTER DETAILS');
    echo json_encode($data);
}

This will trigger only if the value is EMPTY. So if you send a value, it won't trigger since the statement is false. Change your == to != so it state: if $cardNo IS NOT EMPTY like so:

if($cardNo != "")
{
    mysql_close($conn);
    $data = array('success' => true, 'message' => 'ENTER DETAILS');
    echo json_encode($data);
}

answered Nov 18, 2012 at 3:03

Louis Loudog Trottier

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2 Comments

Louis Loudog Trottier Over a year ago

further more, i sugest you start using mysqli instead of mysql as it is (or soon will be) deprecated.

Newbie Over a year ago

I'm just checking whether the code works for empty value. The actual condition is different from the one I posted.

jbx · Accepted Answer · 2012-11-18 04:03:13Z

1

I think the problem is that you're treating the incoming data as a normal form, when in fact you're sending JSON. PHP is therefore not finding any $_POST parameter with name 'budget'.

You need to get the raw post data using file_get_contents('php://input'); and then use json_decode() to convert it into an object. From it you could then get the budget parameter.

So something like this should work:

$formdata = file_get_contents('php://input');
$formdata_obj = json_decode($formdata);

$cardNo = $formdata_obj->budget;

If you're actually just sending one value (as you seem to be doing) the first line alone might suffice, because the whole data would be the value itself rather than a JSON encoded object:

$cardNo = file_get_contents('php://input');

edited Nov 18, 2012 at 4:03

answered Nov 18, 2012 at 3:06

jbx

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8 Comments

Louis Loudog Trottier Over a year ago

Just saw the same thing over here.

Newbie Over a year ago

@jbx: how to use file_get_contents('php://input');

Newbie Over a year ago

@jbx: $requestBody = file_get_contents('php://input'); $requestBody = json_decode($requestBody); $cardNo = $requestBody -> budget;

jbx Over a year ago

Yeah, but it has to be json_decode() not jsonDecode(), and don't use the same variable for 2 different things ($requestBody) it just confuses you. Why don't you copy my example?

jbx Over a year ago

If it still doesn't work put some 'echo' lines here and there to see what you are receiving in the file_get_contents (to make sure that you are really receiving the JSON), and maybe use var_export() to view the contents of the object generated from json_decode(). We've put you on the right track, we can't debug your application!

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