Javascript $.getJSON callback and parameter passing

Question

I currently have a $.getJSON call which is working fine as shown below.

var jsonUrl = "http://www.somesite.co.uk/jsonusv.php?callback=?";                   
$.getJSON(jsonUrl,function(zippy){
...some code
}

However, I wish to pass a variable with it so that the PHP script can use its $_GET[''] value and tailor the data.

I tired the fooling but could not get things to work any ideas ?

var jsonUrl = "http://www.somesite.co.uk/jsonusv.php?callback=?&value=65";

The php page looks something like this this had been stripped down. I did try to detect the $_GET['value'] but it didn't work.

<?PHP
header("content-type: application/json");  
$theSqlquery = "SELECT * FROM table ORDER BY timestamp DESC LIMIT 20";   
$result131 = mysql_query($theSqlquery);

     if ($result131)
     {

        //make up Json string in $temp

    echo $_GET['callback'] . '(' . $temp . ');';
     }                 
?>

Have you tried using a different name for "value" ... your jQuery/JS code works just fine. However, maybe there's a problem with using "value" as the name of a key (in PHP)? — Joe Johnson
– Joe Johnson, Commented Nov 29, 2012 at 19:50
possiblty this can help you stackoverflow.com/questions/6555172/… — Ivaylo Nikolov
– Ivaylo Nikolov, Commented Aug 8, 2013 at 11:55

bluetoft · Accepted Answer · 2012-11-29 19:30:33Z

1

I would suggest removing the callback=? from your jsonUrl

answered Nov 29, 2012 at 19:30

bluetoft

5,4432 gold badges25 silver badges27 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

bluetoft Over a year ago

a url querystring starts with a ? If you're checking the 'callback' property and want to pass an empty value just leave ?callback=&value=65". I would imagine you're querystring is kind of jacked up with the question mark in the middle of it.

Nick Rolando · Accepted Answer · 2012-11-29 19:50:12Z

0

Try passing your parameters into the data parameter of the function call instead of query string:

var jsonUrl = "http://www.somesite.co.uk/jsonusv.php";
$.getJSON(jsonUrl, {
    callback: "your callback val",
    value: "65",
  },
function(zippy){
...some code
});

http://api.jquery.com/jQuery.getJSON/

Then you can access them with $_POST

Note, echo sends the supposed json result back to your $.getJSON() method call, e.g., success() if it was successful. If you know the js method name in the success() method, and only need to pass it $temp, try this

var jsonUrl = "http://www.somesite.co.uk/jsonusv.php";
$.getJSON(jsonUrl, {
    value: "65"
  },
function(zippy){
    callbackMethod(zippy[0]);
});

and in your php

$output = array();
$output[0] = $temp;
echo json_encode($output);

edited Nov 29, 2012 at 19:50

answered Nov 29, 2012 at 19:30

Nick Rolando

26.2k13 gold badges84 silver badges120 bronze badges

3 Comments

Behnam Esmaili Over a year ago

but he wanna access it with $_GET.

DevilCode Over a year ago

I am ok with POST but your method with no callback=? in the URL causes an Access-Control-Allow-Origin.

DevilCode Over a year ago

Also its still GET not post.

DevilCode · Accepted Answer · 2012-12-04 17:38:45Z

0

var jsonUrl = "http://www.somesite.co.uk/jsonusv.php?callback=?";                   
$.getJSON(jsonUrl,{lastdatetime: "",},function(zippy){....

Seems to work ...

answered Dec 4, 2012 at 17:38

DevilCode

1,1045 gold badges36 silver badges64 bronze badges

Collectives™ on Stack Overflow

Javascript $.getJSON callback and parameter passing

3 Answers 3

1 Comment

3 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related