uncalled javascript function start during other function execution

Question

I have a fiunction that set value to form field from Object and during execution another uncalled function start and I do not why....

function loadRecord(i) {// Function for display records which are retrived from database.
    var item = dataset.item(i);
    $("#data_id").val((item['id']).toString());
    $("#source").val((item['source']).toString());
    $("#s_date").val((item['s_date']).toString());
    $("#lat").val((item['lat']).toString());
    $("#lon").val((item['lon']).toString());
    $("#acc").val((item['acc']).toString());
    $("#image").attr("src", item['photo']);
    $("#photo").val((item['photo']).toString());
    $("#exif").val((item['exif']).toString());
>>>>$("#wateres").val((item['wateres']).toString()).change(); //the other function sart here!
    $("#top").val((item['top']).toString()).change();
    $("#pump").val((item['pump']).toString()).change();
    ..... the function continue normally

The form fields are identified by $("input_id") the only difference in the row that cause the unwanted behaviour is that is the first select of the form. But if I comment this row everything work well and the next row (is a select to)

The other function is probably called by a click event handler, which could be bound by HTML `onchange` attribute, or a jQuery `$("#top").change(...)` call, or many other variants of how events can be listened to. Without more information we cannot know.... — trincot, Sep 11 '16 at 20:28

score 1 · Answer 1 · answered Sep 11 '16 at 20:29

1

Maybe you define some event on that item?

Like

$("#wateres").on("change", function(e){});

Because when you apply change() it Bind an event handler to the "change" JavaScript event, or trigger that event on an element.

From docs https://api.jquery.com/change/

So check your code for that listener.

Hope this helps.

answered Sep 11 '16 at 20:29

Mykola Borysyuk

3,373
1
18
24

Yes I've that on my form ` – Francesco Giunta Sep 11 '16 at 20:36
So you have your answer... When you run .change() it triggers onchange="wrForm(this.value)" and it runs that function. Glad that it help you. – Mykola Borysyuk Sep 11 '16 at 20:38
Maybe this will help. http://stackoverflow.com/questions/13343566/set-select-option-selected-by-value .. Just use yourID val your value. – Mykola Borysyuk Sep 11 '16 at 20:43
Or this http://stackoverflow.com/questions/1311287/jquery-setting-the-selected-attribute-on-a-select-list – Mykola Borysyuk Sep 11 '16 at 20:45

uncalled javascript function start during other function execution

1 Answers1