Add/remove class based on checkboxes
up vote
1
down vote
favorite
I have a small form that is kind of working. There are 2 options, A and B.
If A and B are NO then disable the button. If A OR B are YES then remove the disable class from the button. That's what I'm aiming for, this is what I have:
$('.check-opt').change(function () {
if ($('input[name="step_2"],[name="step_4"]:checked').val() == "yes") {
$('#order_btn_id').removeClass('disabled');
} else {
$('#order_btn_id').addClass('disabled');
}
});
It's kind of working in the sense that it removes the disabled class when choosing A OR B, but when you select both A and B as NO then the disabled class isn't added back in. Where have I gone wrong?
Many thanks!
jquery
edited Nov 22 at 14:23
Rory McCrossan
240k29203244
asked Nov 22 at 11:26
Edward Hill
297
add a comment |
up vote
1
down vote
favorite
I have a small form that is kind of working. There are 2 options, A and B.
If A and B are NO then disable the button. If A OR B are YES then remove the disable class from the button. That's what I'm aiming for, this is what I have:
$('.check-opt').change(function () {
if ($('input[name="step_2"],[name="step_4"]:checked').val() == "yes") {
$('#order_btn_id').removeClass('disabled');
} else {
$('#order_btn_id').addClass('disabled');
}
});
It's kind of working in the sense that it removes the disabled class when choosing A OR B, but when you select both A and B as NO then the disabled class isn't added back in. Where have I gone wrong?
Many thanks!
jquery
edited Nov 22 at 14:23
Rory McCrossan
240k29203244
asked Nov 22 at 11:26
Edward Hill
297
You can usetoggleClassinstead of add/remove class.
– Shubham Baranwal
Nov 22 at 11:28
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a small form that is kind of working. There are 2 options, A and B.
If A and B are NO then disable the button. If A OR B are YES then remove the disable class from the button. That's what I'm aiming for, this is what I have:
$('.check-opt').change(function () {
if ($('input[name="step_2"],[name="step_4"]:checked').val() == "yes") {
$('#order_btn_id').removeClass('disabled');
} else {
$('#order_btn_id').addClass('disabled');
}
});
It's kind of working in the sense that it removes the disabled class when choosing A OR B, but when you select both A and B as NO then the disabled class isn't added back in. Where have I gone wrong?
Many thanks!
jquery
edited Nov 22 at 14:23
Rory McCrossan
240k29203244
asked Nov 22 at 11:26
Edward Hill
297
I have a small form that is kind of working. There are 2 options, A and B.
If A and B are NO then disable the button. If A OR B are YES then remove the disable class from the button. That's what I'm aiming for, this is what I have:
$('.check-opt').change(function () {
if ($('input[name="step_2"],[name="step_4"]:checked').val() == "yes") {
$('#order_btn_id').removeClass('disabled');
} else {
$('#order_btn_id').addClass('disabled');
}
});
It's kind of working in the sense that it removes the disabled class when choosing A OR B, but when you select both A and B as NO then the disabled class isn't added back in. Where have I gone wrong?
Many thanks!
jquery
jquery
edited Nov 22 at 14:23
Rory McCrossan
240k29203244
asked Nov 22 at 11:26
Edward Hill
297
edited Nov 22 at 14:23
Rory McCrossan
240k29203244
asked Nov 22 at 11:26
Edward Hill
297
edited Nov 22 at 14:23
Rory McCrossan
240k29203244
edited Nov 22 at 14:23
Rory McCrossan
240k29203244
edited Nov 22 at 14:23
Rory McCrossan
240k29203244
240k29203244
asked Nov 22 at 11:26
Edward Hill
297
asked Nov 22 at 11:26
Edward Hill
297
asked Nov 22 at 11:26
Edward Hill
297
297
You can usetoggleClassinstead of add/remove class.
– Shubham Baranwal
Nov 22 at 11:28
add a comment |
You can usetoggleClassinstead of add/remove class.
– Shubham Baranwal
Nov 22 at 11:28
You can use
toggleClass instead of add/remove class.– Shubham Baranwal
Nov 22 at 11:28
You can use
toggleClass instead of add/remove class.– Shubham Baranwal
Nov 22 at 11:28
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
The issue with your logic is that you're calling val() on a collection of elements, so it will only read the value of the first one.
The easiest way to achieve what you need is to use toggleClass() along with a boolean to determine if the class should be added or removed. You can set the state of that boolean depending on whether or not both checkboxes are unchecked, like this:
$('.check-opt').change(function() {
var disabled = !$('input[name="step_2"]').prop('checked') && !$('input[name="step_4"]').prop('checked');
$('#order_btn_id').toggleClass('disabled', disabled);
});.disabled {
background-color: #CCC;
color: #666;
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<input type="checkbox" name="step_2" class="check-opt" />
<input type="checkbox" name="step_4" class="check-opt" />
<button id="order_btn_id" class="disabled">Order</button>This could potentially be simplified to check if any .check-opt element is checked, but this depends on how many groups of checkboxes you have in the DOM:
var disabled = $('.check-opt:checked').length == 0;
Also note that if you want to completely disable the button it would be worth setting prop('disabled', true) as well as adding the class to it.
answered Nov 22 at 11:30
Rory McCrossan
240k29203244
Wondeful, that sorted it! Thankyou so much, ill mark it as the answer in 8mins..
– Edward Hill
Nov 22 at 11:33
add a comment |
up vote
0
down vote
Use two selectors for your query:
var a = $('input[name="step_2"]:checked').val() == "yes";
var b = $('[name="step_4"]:checked').val() == "yes";
$('#order_btn_id').toggleClass('disabled', a || b)
answered Nov 22 at 11:30
Justinas
26.9k33457
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The issue with your logic is that you're calling val() on a collection of elements, so it will only read the value of the first one.
The easiest way to achieve what you need is to use toggleClass() along with a boolean to determine if the class should be added or removed. You can set the state of that boolean depending on whether or not both checkboxes are unchecked, like this:
$('.check-opt').change(function() {
var disabled = !$('input[name="step_2"]').prop('checked') && !$('input[name="step_4"]').prop('checked');
$('#order_btn_id').toggleClass('disabled', disabled);
});.disabled {
background-color: #CCC;
color: #666;
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<input type="checkbox" name="step_2" class="check-opt" />
<input type="checkbox" name="step_4" class="check-opt" />
<button id="order_btn_id" class="disabled">Order</button>This could potentially be simplified to check if any .check-opt element is checked, but this depends on how many groups of checkboxes you have in the DOM:
var disabled = $('.check-opt:checked').length == 0;
Also note that if you want to completely disable the button it would be worth setting prop('disabled', true) as well as adding the class to it.
answered Nov 22 at 11:30
Rory McCrossan
240k29203244
Wondeful, that sorted it! Thankyou so much, ill mark it as the answer in 8mins..
– Edward Hill
Nov 22 at 11:33
add a comment |
up vote
5
down vote
accepted
The issue with your logic is that you're calling val() on a collection of elements, so it will only read the value of the first one.
The easiest way to achieve what you need is to use toggleClass() along with a boolean to determine if the class should be added or removed. You can set the state of that boolean depending on whether or not both checkboxes are unchecked, like this:
$('.check-opt').change(function() {
var disabled = !$('input[name="step_2"]').prop('checked') && !$('input[name="step_4"]').prop('checked');
$('#order_btn_id').toggleClass('disabled', disabled);
});.disabled {
background-color: #CCC;
color: #666;
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<input type="checkbox" name="step_2" class="check-opt" />
<input type="checkbox" name="step_4" class="check-opt" />
<button id="order_btn_id" class="disabled">Order</button>This could potentially be simplified to check if any .check-opt element is checked, but this depends on how many groups of checkboxes you have in the DOM:
var disabled = $('.check-opt:checked').length == 0;
Also note that if you want to completely disable the button it would be worth setting prop('disabled', true) as well as adding the class to it.
answered Nov 22 at 11:30
Rory McCrossan
240k29203244
Wondeful, that sorted it! Thankyou so much, ill mark it as the answer in 8mins..
– Edward Hill
Nov 22 at 11:33
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The issue with your logic is that you're calling val() on a collection of elements, so it will only read the value of the first one.
The easiest way to achieve what you need is to use toggleClass() along with a boolean to determine if the class should be added or removed. You can set the state of that boolean depending on whether or not both checkboxes are unchecked, like this:
$('.check-opt').change(function() {
var disabled = !$('input[name="step_2"]').prop('checked') && !$('input[name="step_4"]').prop('checked');
$('#order_btn_id').toggleClass('disabled', disabled);
});.disabled {
background-color: #CCC;
color: #666;
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<input type="checkbox" name="step_2" class="check-opt" />
<input type="checkbox" name="step_4" class="check-opt" />
<button id="order_btn_id" class="disabled">Order</button>This could potentially be simplified to check if any .check-opt element is checked, but this depends on how many groups of checkboxes you have in the DOM:
var disabled = $('.check-opt:checked').length == 0;
Also note that if you want to completely disable the button it would be worth setting prop('disabled', true) as well as adding the class to it.
answered Nov 22 at 11:30
Rory McCrossan
240k29203244
The issue with your logic is that you're calling val() on a collection of elements, so it will only read the value of the first one.
The easiest way to achieve what you need is to use toggleClass() along with a boolean to determine if the class should be added or removed. You can set the state of that boolean depending on whether or not both checkboxes are unchecked, like this:
$('.check-opt').change(function() {
var disabled = !$('input[name="step_2"]').prop('checked') && !$('input[name="step_4"]').prop('checked');
$('#order_btn_id').toggleClass('disabled', disabled);
});.disabled {
background-color: #CCC;
color: #666;
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<input type="checkbox" name="step_2" class="check-opt" />
<input type="checkbox" name="step_4" class="check-opt" />
<button id="order_btn_id" class="disabled">Order</button>This could potentially be simplified to check if any .check-opt element is checked, but this depends on how many groups of checkboxes you have in the DOM:
var disabled = $('.check-opt:checked').length == 0;
Also note that if you want to completely disable the button it would be worth setting prop('disabled', true) as well as adding the class to it.
$('.check-opt').change(function() {
var disabled = !$('input[name="step_2"]').prop('checked') && !$('input[name="step_4"]').prop('checked');
$('#order_btn_id').toggleClass('disabled', disabled);
});.disabled {
background-color: #CCC;
color: #666;
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<input type="checkbox" name="step_2" class="check-opt" />
<input type="checkbox" name="step_4" class="check-opt" />
<button id="order_btn_id" class="disabled">Order</button>$('.check-opt').change(function() {
var disabled = !$('input[name="step_2"]').prop('checked') && !$('input[name="step_4"]').prop('checked');
$('#order_btn_id').toggleClass('disabled', disabled);
});.disabled {
background-color: #CCC;
color: #666;
}<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<input type="checkbox" name="step_2" class="check-opt" />
<input type="checkbox" name="step_4" class="check-opt" />
<button id="order_btn_id" class="disabled">Order</button>answered Nov 22 at 11:30
Rory McCrossan
240k29203244
edited Nov 22 at 11:35
answered Nov 22 at 11:30
Rory McCrossan
240k29203244
answered Nov 22 at 11:30
Rory McCrossan
240k29203244
answered Nov 22 at 11:30
Rory McCrossan
240k29203244
240k29203244
Wondeful, that sorted it! Thankyou so much, ill mark it as the answer in 8mins..
– Edward Hill
Nov 22 at 11:33
add a comment |
Wondeful, that sorted it! Thankyou so much, ill mark it as the answer in 8mins..
– Edward Hill
Nov 22 at 11:33
Wondeful, that sorted it! Thankyou so much, ill mark it as the answer in 8mins..
– Edward Hill
Nov 22 at 11:33
Wondeful, that sorted it! Thankyou so much, ill mark it as the answer in 8mins..
– Edward Hill
Nov 22 at 11:33
add a comment |
up vote
0
down vote
Use two selectors for your query:
var a = $('input[name="step_2"]:checked').val() == "yes";
var b = $('[name="step_4"]:checked').val() == "yes";
$('#order_btn_id').toggleClass('disabled', a || b)
answered Nov 22 at 11:30
Justinas
26.9k33457
add a comment |
up vote
0
down vote
Use two selectors for your query:
var a = $('input[name="step_2"]:checked').val() == "yes";
var b = $('[name="step_4"]:checked').val() == "yes";
$('#order_btn_id').toggleClass('disabled', a || b)
answered Nov 22 at 11:30
Justinas
26.9k33457
add a comment |
up vote
0
down vote
up vote
0
down vote
Use two selectors for your query:
var a = $('input[name="step_2"]:checked').val() == "yes";
var b = $('[name="step_4"]:checked').val() == "yes";
$('#order_btn_id').toggleClass('disabled', a || b)
answered Nov 22 at 11:30
Justinas
26.9k33457
Use two selectors for your query:
var a = $('input[name="step_2"]:checked').val() == "yes";
var b = $('[name="step_4"]:checked').val() == "yes";
$('#order_btn_id').toggleClass('disabled', a || b)
answered Nov 22 at 11:30
Justinas
26.9k33457
answered Nov 22 at 11:30
Justinas
26.9k33457
answered Nov 22 at 11:30
Justinas
26.9k33457
answered Nov 22 at 11:30
Justinas
26.9k33457
26.9k33457
add a comment |
add a comment |
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You can use
toggleClassinstead of add/remove class.– Shubham Baranwal
Nov 22 at 11:28