I'm trying to alert a value using jquery when it is generated from a javascript code, the value is generated on the input box the problem is jquery cannot detect the changes, mands i tested are "change, input" but when i manually input a value jquery triggers
sample code:
value is dynamically generated on the javascript and pushed / inserted to the inputbox, the value is not manually generated
javascript:
$('#displayDID').on("input" ,function() {
var work = $(this).val();
alert(work);
});
the value of the id="displayDID" changes but jquery cannot detect it after execution.
sample fiddle /
I'm trying to alert a value using jquery when it is generated from a javascript code, the value is generated on the input box the problem is jquery cannot detect the changes, mands i tested are "change, input" but when i manually input a value jquery triggers
sample code:
value is dynamically generated on the javascript and pushed / inserted to the inputbox, the value is not manually generated
javascript:
$('#displayDID').on("input" ,function() {
var work = $(this).val();
alert(work);
});
the value of the id="displayDID" changes but jquery cannot detect it after execution.
sample fiddle http://jsfiddle/SU7bU/1/
javascript
jquery
input
alert
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edited Sep 19, 2014 at 19:32
tshepang
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asked Feb 14, 2014 at 10:33
user3261755user32617558111 gold badge22 silver badges66 bronze badges5
it works fine: jsfiddle/paRPL/1
– caramba
CommentedFeb 14, 2014 at 10:37
What about to change "input" to 'keyup' if you want to alert every changes, or to 'change' if you want to alert the changes after focusout
– Pavlo
CommentedFeb 14, 2014 at 10:37
it works fine on manual input, but doesnt when the value is pushed from javascript
– user3261755
CommentedFeb 14, 2014 at 10:38
Try this one: $('#displayDID').val('some value').change(); - this should fire you onchange event.
– Pavlo
CommentedFeb 14, 2014 at 10:42
I think this is what you are looking for: jsfiddle/LGAWY/141 more information on this link, see Davids answer: stackoverflow./questions/1443292/…
– caramba
CommentedFeb 14, 2014 at 12:57
Add a ment
|
3 Answers
3
Reset to default
3
add trigger to it
$('#gen').on('click',function() {
$('#field').val('val').trigger("change");
});
$(document).on("change",'#field' ,function() {
var work = $(this).val();
alert(work);
});
http://jsfiddle/ytexj/
It is because you have added the script before input is ready.Due to which, event is not getting set on that element.write the code on document ready:
$(document).ready(function(){
$('#displayDID').on("input" ,function() {
var work = $(this).val();
alert(work);
});
})
or use event delegation:
$(document).on("input",'#displayDID' ,function() {
var work = $(this).val();
alert(work);
});
Ok I will propose you one answer and let's see if it solve your problem.
If you use keyup, and you insert 3350 value, you will display 4 alert, and I think you want to display only 1 alert.
$(document).ready(function() {
var interval;
$("#variazioneAnticipo").on("input", function() {
var variazioneAnticipo = $("#variazioneAnticipo").val();
clearInterval(interval);
interval = setTimeout(function(){ showValue(variazioneAnticipo) }, 1000);
});
});
function showValue(value) {
alert("ANTICIPO VARIATO: " + value);
}
I explain it, when you input anything into the input, will trigger a setTimeout in 1 sec, if you press another key under this time, we clear the timeOut and we triiger again, so you will only have 1 alert 1 sec after the last input insert.
