Hidden Input Value not updating with Jquery












-2














When a user clicks a checkbox, I need the jquery update_flavors function to run and update the hidden input the value attribute with all the flavors.



Currently, the update_flavors function does not seem to get called.



Updated JSFiddle: https://jsfiddle.net/ecqp9nwm/7/



HTML



    <div class="row"> 
{% assign flavors = "apple, orange, cherry" | split: ","%}

{% for flavor in flavors %}
<div class="col-md-6">
<div class="form-check">
<label class="form-check-label" for="checkbox-{{flavor | handleize }}"><input name='contact[{{flavor| handleize }}]' type="checkbox" class="checkbox-flavor form-check-input" id="checkbox-{{flavor | handleize }}"> {{flavor}}</label>
</div>
</div>
{% endfor %}
</div>


<input type="hidden" id="checkbox-flavors" name="contact[flavor]" value="">


Jquery



<script>
function update_flavors() {
var allVals = ;
$('.checkbox-flavor :checked').each(function() {
allVals.push($(this).val());
});
$('#checkbox-flavors').val(allVals);
}


$(document).ready(function(){
$(".checkbox-flavor").click(function(){
update_flavors;
});
});

</script>


Solution:



I had a space in my selector! big oversight!
$('.checkbox-flavor :checked') should have been $('.checkbox-flavor:checked')
and I needed to add a () to update_flavors();



working and improved: https://jsfiddle.net/ecqp9nwm/12



function update_flavors() {
var allVals = $('.checkbox-flavor:checked').map(function() {
return $(this).val()
}).get();

$('#checkbox-flavors').val(allVals.join(', '));
}

$(document).ready(function() {
$(".checkbox-flavor").change(update_flavors);
});









share|improve this question




















  • 1




    stackoverflow.com/questions/9454645/… Don't repeat ids. Use classes instead.
    – Taplar
    Nov 20 at 22:56










  • Where do you have an element with the ID checkbox-flavor to be matched with $("#checkbox-flavor")?
    – j08691
    Nov 20 at 22:56










  • Oh, good point. flavor would be apple, orange or cherry (i guess?), so it's not even a dup id anyway. But still, you should use a class to avoid that issue.
    – Taplar
    Nov 20 at 22:58












  • @j08691 typo, corrected, but still does not run. See js fiddle
    – user2012677
    Nov 20 at 23:00






  • 1




    update_flavors(); You're missing your (), or get rid of the outer anonymous function
    – Taplar
    Nov 20 at 23:27


















-2














When a user clicks a checkbox, I need the jquery update_flavors function to run and update the hidden input the value attribute with all the flavors.



Currently, the update_flavors function does not seem to get called.



Updated JSFiddle: https://jsfiddle.net/ecqp9nwm/7/



HTML



    <div class="row"> 
{% assign flavors = "apple, orange, cherry" | split: ","%}

{% for flavor in flavors %}
<div class="col-md-6">
<div class="form-check">
<label class="form-check-label" for="checkbox-{{flavor | handleize }}"><input name='contact[{{flavor| handleize }}]' type="checkbox" class="checkbox-flavor form-check-input" id="checkbox-{{flavor | handleize }}"> {{flavor}}</label>
</div>
</div>
{% endfor %}
</div>


<input type="hidden" id="checkbox-flavors" name="contact[flavor]" value="">


Jquery



<script>
function update_flavors() {
var allVals = ;
$('.checkbox-flavor :checked').each(function() {
allVals.push($(this).val());
});
$('#checkbox-flavors').val(allVals);
}


$(document).ready(function(){
$(".checkbox-flavor").click(function(){
update_flavors;
});
});

</script>


Solution:



I had a space in my selector! big oversight!
$('.checkbox-flavor :checked') should have been $('.checkbox-flavor:checked')
and I needed to add a () to update_flavors();



working and improved: https://jsfiddle.net/ecqp9nwm/12



function update_flavors() {
var allVals = $('.checkbox-flavor:checked').map(function() {
return $(this).val()
}).get();

$('#checkbox-flavors').val(allVals.join(', '));
}

$(document).ready(function() {
$(".checkbox-flavor").change(update_flavors);
});









share|improve this question




















  • 1




    stackoverflow.com/questions/9454645/… Don't repeat ids. Use classes instead.
    – Taplar
    Nov 20 at 22:56










  • Where do you have an element with the ID checkbox-flavor to be matched with $("#checkbox-flavor")?
    – j08691
    Nov 20 at 22:56










  • Oh, good point. flavor would be apple, orange or cherry (i guess?), so it's not even a dup id anyway. But still, you should use a class to avoid that issue.
    – Taplar
    Nov 20 at 22:58












  • @j08691 typo, corrected, but still does not run. See js fiddle
    – user2012677
    Nov 20 at 23:00






  • 1




    update_flavors(); You're missing your (), or get rid of the outer anonymous function
    – Taplar
    Nov 20 at 23:27
















-2












-2








-2







When a user clicks a checkbox, I need the jquery update_flavors function to run and update the hidden input the value attribute with all the flavors.



Currently, the update_flavors function does not seem to get called.



Updated JSFiddle: https://jsfiddle.net/ecqp9nwm/7/



HTML



    <div class="row"> 
{% assign flavors = "apple, orange, cherry" | split: ","%}

{% for flavor in flavors %}
<div class="col-md-6">
<div class="form-check">
<label class="form-check-label" for="checkbox-{{flavor | handleize }}"><input name='contact[{{flavor| handleize }}]' type="checkbox" class="checkbox-flavor form-check-input" id="checkbox-{{flavor | handleize }}"> {{flavor}}</label>
</div>
</div>
{% endfor %}
</div>


<input type="hidden" id="checkbox-flavors" name="contact[flavor]" value="">


Jquery



<script>
function update_flavors() {
var allVals = ;
$('.checkbox-flavor :checked').each(function() {
allVals.push($(this).val());
});
$('#checkbox-flavors').val(allVals);
}


$(document).ready(function(){
$(".checkbox-flavor").click(function(){
update_flavors;
});
});

</script>


Solution:



I had a space in my selector! big oversight!
$('.checkbox-flavor :checked') should have been $('.checkbox-flavor:checked')
and I needed to add a () to update_flavors();



working and improved: https://jsfiddle.net/ecqp9nwm/12



function update_flavors() {
var allVals = $('.checkbox-flavor:checked').map(function() {
return $(this).val()
}).get();

$('#checkbox-flavors').val(allVals.join(', '));
}

$(document).ready(function() {
$(".checkbox-flavor").change(update_flavors);
});









share|improve this question















When a user clicks a checkbox, I need the jquery update_flavors function to run and update the hidden input the value attribute with all the flavors.



Currently, the update_flavors function does not seem to get called.



Updated JSFiddle: https://jsfiddle.net/ecqp9nwm/7/



HTML



    <div class="row"> 
{% assign flavors = "apple, orange, cherry" | split: ","%}

{% for flavor in flavors %}
<div class="col-md-6">
<div class="form-check">
<label class="form-check-label" for="checkbox-{{flavor | handleize }}"><input name='contact[{{flavor| handleize }}]' type="checkbox" class="checkbox-flavor form-check-input" id="checkbox-{{flavor | handleize }}"> {{flavor}}</label>
</div>
</div>
{% endfor %}
</div>


<input type="hidden" id="checkbox-flavors" name="contact[flavor]" value="">


Jquery



<script>
function update_flavors() {
var allVals = ;
$('.checkbox-flavor :checked').each(function() {
allVals.push($(this).val());
});
$('#checkbox-flavors').val(allVals);
}


$(document).ready(function(){
$(".checkbox-flavor").click(function(){
update_flavors;
});
});

</script>


Solution:



I had a space in my selector! big oversight!
$('.checkbox-flavor :checked') should have been $('.checkbox-flavor:checked')
and I needed to add a () to update_flavors();



working and improved: https://jsfiddle.net/ecqp9nwm/12



function update_flavors() {
var allVals = $('.checkbox-flavor:checked').map(function() {
return $(this).val()
}).get();

$('#checkbox-flavors').val(allVals.join(', '));
}

$(document).ready(function() {
$(".checkbox-flavor").change(update_flavors);
});






jquery






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 21 at 2:28

























asked Nov 20 at 22:43









user2012677

1,39141436




1,39141436








  • 1




    stackoverflow.com/questions/9454645/… Don't repeat ids. Use classes instead.
    – Taplar
    Nov 20 at 22:56










  • Where do you have an element with the ID checkbox-flavor to be matched with $("#checkbox-flavor")?
    – j08691
    Nov 20 at 22:56










  • Oh, good point. flavor would be apple, orange or cherry (i guess?), so it's not even a dup id anyway. But still, you should use a class to avoid that issue.
    – Taplar
    Nov 20 at 22:58












  • @j08691 typo, corrected, but still does not run. See js fiddle
    – user2012677
    Nov 20 at 23:00






  • 1




    update_flavors(); You're missing your (), or get rid of the outer anonymous function
    – Taplar
    Nov 20 at 23:27
















  • 1




    stackoverflow.com/questions/9454645/… Don't repeat ids. Use classes instead.
    – Taplar
    Nov 20 at 22:56










  • Where do you have an element with the ID checkbox-flavor to be matched with $("#checkbox-flavor")?
    – j08691
    Nov 20 at 22:56










  • Oh, good point. flavor would be apple, orange or cherry (i guess?), so it's not even a dup id anyway. But still, you should use a class to avoid that issue.
    – Taplar
    Nov 20 at 22:58












  • @j08691 typo, corrected, but still does not run. See js fiddle
    – user2012677
    Nov 20 at 23:00






  • 1




    update_flavors(); You're missing your (), or get rid of the outer anonymous function
    – Taplar
    Nov 20 at 23:27










1




1




stackoverflow.com/questions/9454645/… Don't repeat ids. Use classes instead.
– Taplar
Nov 20 at 22:56




stackoverflow.com/questions/9454645/… Don't repeat ids. Use classes instead.
– Taplar
Nov 20 at 22:56












Where do you have an element with the ID checkbox-flavor to be matched with $("#checkbox-flavor")?
– j08691
Nov 20 at 22:56




Where do you have an element with the ID checkbox-flavor to be matched with $("#checkbox-flavor")?
– j08691
Nov 20 at 22:56












Oh, good point. flavor would be apple, orange or cherry (i guess?), so it's not even a dup id anyway. But still, you should use a class to avoid that issue.
– Taplar
Nov 20 at 22:58






Oh, good point. flavor would be apple, orange or cherry (i guess?), so it's not even a dup id anyway. But still, you should use a class to avoid that issue.
– Taplar
Nov 20 at 22:58














@j08691 typo, corrected, but still does not run. See js fiddle
– user2012677
Nov 20 at 23:00




@j08691 typo, corrected, but still does not run. See js fiddle
– user2012677
Nov 20 at 23:00




1




1




update_flavors(); You're missing your (), or get rid of the outer anonymous function
– Taplar
Nov 20 at 23:27






update_flavors(); You're missing your (), or get rid of the outer anonymous function
– Taplar
Nov 20 at 23:27














1 Answer
1






active

oldest

votes


















0














Currently your checkboxes have no values to be added. They just have labels.



This site is such a pain in the hind quarters to post any sort of solution to, but try something like this:



function updateFlavors(e) {
console.log("update");
var checks = $(".checkbox-flavor:checked");
var vals = $.map(checks, function(elem, i) {
var inp = $(elem);
return inp.val();
});

$('#checkbox-flavors').val(vals.join(", "));}

$(document).ready(function() {
$(".checkbox-flavor").each(function() {
var inp = $(this);
inp.val(inp.parent().find("label").text());
})
$(".check_holder").on("change", ".checkbox-flavor", updateFlavors);
});


By using the change handler, you update when an item is checked or unchecked.
Also, by assigning the handler to a parent element, you only have one handler running instead of one for every input element.






share|improve this answer





















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    0














    Currently your checkboxes have no values to be added. They just have labels.



    This site is such a pain in the hind quarters to post any sort of solution to, but try something like this:



    function updateFlavors(e) {
    console.log("update");
    var checks = $(".checkbox-flavor:checked");
    var vals = $.map(checks, function(elem, i) {
    var inp = $(elem);
    return inp.val();
    });

    $('#checkbox-flavors').val(vals.join(", "));}

    $(document).ready(function() {
    $(".checkbox-flavor").each(function() {
    var inp = $(this);
    inp.val(inp.parent().find("label").text());
    })
    $(".check_holder").on("change", ".checkbox-flavor", updateFlavors);
    });


    By using the change handler, you update when an item is checked or unchecked.
    Also, by assigning the handler to a parent element, you only have one handler running instead of one for every input element.






    share|improve this answer


























      0














      Currently your checkboxes have no values to be added. They just have labels.



      This site is such a pain in the hind quarters to post any sort of solution to, but try something like this:



      function updateFlavors(e) {
      console.log("update");
      var checks = $(".checkbox-flavor:checked");
      var vals = $.map(checks, function(elem, i) {
      var inp = $(elem);
      return inp.val();
      });

      $('#checkbox-flavors').val(vals.join(", "));}

      $(document).ready(function() {
      $(".checkbox-flavor").each(function() {
      var inp = $(this);
      inp.val(inp.parent().find("label").text());
      })
      $(".check_holder").on("change", ".checkbox-flavor", updateFlavors);
      });


      By using the change handler, you update when an item is checked or unchecked.
      Also, by assigning the handler to a parent element, you only have one handler running instead of one for every input element.






      share|improve this answer
























        0












        0








        0






        Currently your checkboxes have no values to be added. They just have labels.



        This site is such a pain in the hind quarters to post any sort of solution to, but try something like this:



        function updateFlavors(e) {
        console.log("update");
        var checks = $(".checkbox-flavor:checked");
        var vals = $.map(checks, function(elem, i) {
        var inp = $(elem);
        return inp.val();
        });

        $('#checkbox-flavors').val(vals.join(", "));}

        $(document).ready(function() {
        $(".checkbox-flavor").each(function() {
        var inp = $(this);
        inp.val(inp.parent().find("label").text());
        })
        $(".check_holder").on("change", ".checkbox-flavor", updateFlavors);
        });


        By using the change handler, you update when an item is checked or unchecked.
        Also, by assigning the handler to a parent element, you only have one handler running instead of one for every input element.






        share|improve this answer












        Currently your checkboxes have no values to be added. They just have labels.



        This site is such a pain in the hind quarters to post any sort of solution to, but try something like this:



        function updateFlavors(e) {
        console.log("update");
        var checks = $(".checkbox-flavor:checked");
        var vals = $.map(checks, function(elem, i) {
        var inp = $(elem);
        return inp.val();
        });

        $('#checkbox-flavors').val(vals.join(", "));}

        $(document).ready(function() {
        $(".checkbox-flavor").each(function() {
        var inp = $(this);
        inp.val(inp.parent().find("label").text());
        })
        $(".check_holder").on("change", ".checkbox-flavor", updateFlavors);
        });


        By using the change handler, you update when an item is checked or unchecked.
        Also, by assigning the handler to a parent element, you only have one handler running instead of one for every input element.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 21 at 3:25









        Devi8

        1




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