setInterval does not call a functions after the specified time

Question

Inside one JS function I call another,.. I want to call it every 30 seconds

function showPopup() {
    $.get("/Feedback.aspx", function (data) {
        if (post_HasError(data))
            return;
        initPopup("popup-common", "feedback", data);
    });
    setInterval(AddFormToSession(3), 30000);
}
function AddFormToSession(form) {
    alert(1);
  var url1 = form == 3 ? "Feedback.aspx/AddFormToSession" : "Request.aspx/AddFormToSession";
    $.ajax
    ({
         type: "POST",
         async: true,
         url: url1,
         data: "{'funcParam':'" + $('#aspnetForm').serialize() + "'}",
         contentType: "application/json; charset=utf-8",
         dataType: "json",
         success: function(msg)
         {
             console.log(msg.d);
         }
     });
  }

is called only the first time

`setInterval(AddFormToSession(3), ..)` - is immediately invoking the callback function and using that as the argument; and yes, this *is* a duplicate — user2864740, Jun 23 '14 at 04:32
http://stackoverflow.com/questions/2037203/why-is-my-function-call-that-should-be-scheduled-by-settimeout-executed-immediat , http://stackoverflow.com/questions/15378302/settimeout-does-not-delay , http://stackoverflow.com/questions/4120781/settimeout-ignores-timeout-fires-immediately?lq=1 — user2864740, Jun 23 '14 at 04:33

Jashwant · Accepted Answer · 2014-06-23T06:34:28.433

4

You do not call a function in first argument of setInterval. Calling function returns the value from that function, not the reference to that function. In your case, it is returning undefined.

You just pass a reference of function. You may use anonymous function and then call from there.

setInterval(function () {
 AddFormToSession(3);
}, 30000);

From comments:

The function object is passed, not a "reference"
A function call as the first argument is fine if it returns a function.

edited Jun 23 '14 at 06:34

answered Jun 23 '14 at 04:35

Jashwant

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Nit: the function object *is* passed, not a "reference" – user2864740 Jun 23 '14 at 04:45
Well, the *value* of the first argument is set to a reference to the function object. A function call as the first argument is fine if it returns a function: `setTimeout(function(){return function(){...}}(), 0)`. :-) – RobG Jun 23 '14 at 04:56

score 0 · Answer 2 · answered Jun 23 '14 at 04:32

0

setInterval accepts a function name (i.e. AddFormToSession). What you've provided is a return value from a function call.

answered Jun 23 '14 at 04:32

Traveling Tech Guy

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setInterval does not call a functions after the specified time

2 Answers2