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 . ');';
     }                 
?>

Answer 1

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

Seems to work …