How to pass parameters in jquery

Question

I am supposed to submit 2 values called str and name to another page.But only 1value is getting returned in the next page. When I include name and on next page if I say print_r($_POST) ,then even the first value is not getting printed. I have written a function as follows,which works because there is only one parameter.

function sendValue(str)
{
$.post(
"newsletter/subscribe.php", //Ajax file
{
     sendValue: str 
},
function(data){
        $('#display').html(data.returnValue);
           },
    "json"
);
}

But if I pass 2 values in that function and in $.post , I do sendValue:str,name then I am not getting even 1 value.

Answer 1

You can post 2 values like this:

function sendValue(str, name) {
  $.post("newsletter/subscribe.php",
         { 'string': str, 'name' : name },
         function(data){
           $('#display').html(data.returnValue);
         }, 
        "json");
}

The format for the data argument of $.post() is like this:

{ 'varName' : variable, 'var2Name', variable2, 'var3Name' : variable3 }

Answer 2

You cannot assign comma separated values to the same key within an object, as it breaks Javascript syntax. You will need two pairs , eg:

function sendValue(str)
{
    $.post(
    "newsletter/subscribe.php", //Ajax file
    {
        sendValue: str, anotherValue: anotherStr 
    },
    function(data){
        $('#display').html(data.returnValue);
    },
    "json"
    );
}

Then in your PHP script, you can access anotherValue as $_POST['anotherValue'] , and it will correctly show up in print_r($_POST) .

If you prefer, you can send a single string with key value pairs as follows:

$.post( "newsletter/subscribe.php", "theValue=something&anotherValue=somethingelse",
        function(data){
            $('#display').html(data.returnValue);
        },"json");