Passing parameters in a callback function using $.getJSON in Javascript

Question

Passing parameters in a callback function using $.getJSON in Javascript

I am currently using a $.getJSON call that is working perfectly fine, as demonstrated below.

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

However, I want to pass a variable along with it so that the PHP script can utilize its $_GET[''] value and customize the data.

I tried the following but couldn't get things to work. Any suggestions?

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

The PHP page looks somewhat like this after being stripped down. I attempted to retrieve the $_GET['value'], but it didn't function as expected.

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

     if ($result131)
     {

        //compose Json string in $temp

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

javascript json getjson

Answer 1

Answer №1

To improve your JSONUrl, I recommend excluding the callback=?

Answer 2

To improve your JSONUrl, I recommend excluding the callback=?

Answer 3

Answer №2

Instead of using query strings, consider passing your parameters into the data parameter when calling the function:

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/

You can then access these parameters using $_POST

Keep in mind that echo sends the expected JSON result back to your $.getJSON() method call, specifically to the success() function if it was successful. If you are aware of the JavaScript method name in the success() function and only need to pass it $temp, you can try the following approach:

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

In your PHP script, you can do:

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

Answer 4

Instead of using query strings, consider passing your parameters into the data parameter when calling the function:

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/

You can then access these parameters using $_POST

Keep in mind that echo sends the expected JSON result back to your $.getJSON() method call, specifically to the success() function if it was successful. If you are aware of the JavaScript method name in the success() function and only need to pass it $temp, you can try the following approach:

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

In your PHP script, you can do:

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

Answer 5

Answer №3

const jsonDataUrl = "http://www.anotherwebsite.com/datajson.php?callback=?";                   
$.getJSON(jsonDataUrl,{lasttimestamp: "",},function(data){....

Looks like it's functioning properly...

Answer 6

const jsonDataUrl = "http://www.anotherwebsite.com/datajson.php?callback=?";                   
$.getJSON(jsonDataUrl,{lasttimestamp: "",},function(data){....

Looks like it's functioning properly...

Passing parameters in a callback function using $.getJSON in Javascript

Answer №1

Answer №2

Answer №3

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