I have this code that changes the styling of the checkbox "tick" mark on selecting the checkbox,
.x-input-checkbox:checked + .x-field-mask:after {
//Your style here
}
P.S. The above code is only for the check mark not the whole field.
but I dont know how to change the background color of the checkbox field on selecting it using purely CSS (the whole field)
You have to manipulate the .x-field-mask class. .x-field-mask:after is only used to draw the tick.
.x-field-mask {
background-color: green;
}
should do the trick.
Related
I'm using Vuetify's v-btn button component with a variety of colors set via the color prop. Once a user clicks the button, I set disabled to true so they can't click it again, but the button loses its color and gets greyed out.
Is there any way to disable the button without changing its color to grey?
Instead of disabled prop you could use your custom class with pointer-events: none, e.g.
.disable-events {
pointer-events: none
}
<v-btn :class="{'disable-events': customCondition}">
Then add additional styling to that class if needed.
I do it by removing v-btn--disabled and playing with vuetify's css classes.
Still grey but with colored text solution
The button will still be grey, but text will be colored, like that you have a visual effect showing that the button is disabled but still have a colored part.
I, personally, also had some custom opacity to disabled buttons.
HTML
<v-btn id="btnA" :disabled="true" color="success">Success</v-btn>
CSS
button.v-btn[disabled] {
opacity: 0.6;
}
JS
created(){
// Trick to remove class after initialising form
this.$nextTick(() => {
document.getElementById('btnA').classList.remove('v-btn--disabled')
})
}
CodePen
Same display solution
If you really want, the same display you will have to remove [color]--text and add [color] class (and sometimes add white--text class for readability).
JS
created(){
// Trick to remove class after initialising form
this.$nextTick(() => {
document.getElementById('btnA').classList.remove('v-btn--disabled')
document.getElementById('btnA').classList.remove('success--text')
document.getElementById('btnA').classList.add('success')
})
}
CodePen
As Vuetify allready use important! in .v-btn--disabled it's not possible to just override this class. But with the use of a higher level selector like id (example: #custom-disabled which selects id="custom-disabled") you can. This doesen't keep the original colors but you are at least able to override the class to your liking.
<v-btn :disabled="true" id="custom-disabled">
Button
</v-btn>
<style>
#custom-disabled.v-btn--disabled {
background-color: red !important;
}
</style>
For light and dark theme:
<style>
#custom-disabled.v-btn--disabled.theme--light {
background-color: red !important;
}
#custom-disabled.v-btn--disabled.theme--dark {
background-color: brown !important;
}
</style>
Okay so you can do it by disabling the pointer events as mentioned in other comments but if someone is using a keyboard they can still tab to the control and if you are writing automated tests the button can still be clicked.
You can manually override the style and change the disabled button colour in the css however this will potentially be a problem if you are manually setting the color through the color="" property on v-btn based off a theme (because your application supports branding for different clients for example) because Vuetify doesn't just override the color, it stops adding the color altogether.
So my solution was to simply set the button color via a style attribute and set the important flag (to override the disabled important flag) note that you will need to change the text color as well.
<v-btn
:style="{
color: `${getTxtColor()} !important`,
backgroundColor: `${getBtnColor()} !important`
}"
:disabled="status"
#click="doSomething"
>
Click Here
</v-btn>
This approach should play nice with testing, themeing, and will not allow users to tab to the button accidentally.
I've got to create our own buttons using Bootstrap's btn class. I need to override default colors for the button text in particular. I know about .button-variant but I cannot use it (the corresponding LESS file is not included in project build and I can't make such changes). Here is my LESS:
.some-company-control(#text-color, #hover-text-color) {
color: #text-color;
&:hover,
&:active {
color: #hover-text-color;
}
}
The problem is after a button is clicked it gets the default Bootstrap button text color. When I add &:focus it overrides Bootstrap's defaults but after it is clicked and not hovered it still remains as if it is clicked. I would like to disable the styling when a button is still focused but not hovered anymore.
Thanks everyone for any suggestions!
Try to add this rule:
&:focus:not(:hover) {
color: #text-color;
}
I found this example on the web on how to implement custom properties accessible from QSS for custom QWidgets: https://qt-project.org/wiki/Qt_Style_Sheets_and_Custom_Painting_Example
Does anyone know how can I implement the widget so that I can have different colors for hover or pressed states?
Current stylesheet looks like this:
SWidget
{
qproperty-lineColor: yellow;
qproperty-rectColor: red;
}
I want to be able to have something like this:
SWidget:hover
{
qproperty-lineColor: blue;
qproperty-rectColor: green;
}
SWidget:pressed
{
qproperty-lineColor: orange;
qproperty-rectColor: violet;
}
Note: I know it is possible to implement mouse events and change the colors using qproperties specific to the mouse events, for example:
SWidget
{
qproperty-lineColor: yellow;
qproperty-rectColor: red;
qproperty-lineColor-hover: orange;
qproperty-rectColor-hover: violet;
}
but I would like to be able to make it work using the original qss/css way.
Regards!
When user hover or pressed widget, you should change some widget property. And make different QSS selector for that properties.
After change property you should make unpolish\polish of application styles.
qApp->style()->unpolish( this );
qApp->style()->polish( this );
Where "this" current window pointer. That "magic" code will help to have affect of appropriate QSS selector.
In my web app (c#/MVC3), I have a huge set of checkboxes in a table. Rather than a table of checkboxes, I'd like for it to look like a wall of toggle buttons. To the user I want it to look like a wall of buttons and when they click one it is 'checked' and the button changes color.
I wasn't sure if there was CSS that could make a checkbox do this (look like a button and change colors on check rather than show a check mark), or if I would have to use some combination of buttons and javascript/jquery and hidden checkboxes or what.
The jQuery UI Button widget can handle that:
http://jqueryui.com/button/#checkbox
Yes, it is definitely possible to do what you want with pure CSS.
I think you should check out the jsFiddle mentioned on this question.
Radio buttons are generated by the operating system and cannot be easily styled.
If you wany something different you need to generate it using CSS/images and JavaScript.
First of all, I'd actually avoid doing this for usability concerns but if you still want to then read on.
This is actually quite tricky to achieve but it is possible. My solution avoids the need to assign individual IDs to your check-boxes.
Essentially, you will need an image sprite for the "on" and "off" states which you will position with the CSS background-position property, using a toggle class. Then, the following jQuery will allow you to not only swap the image state, but also confirm the respective checkbox as checked or unchecked for use of the form. Do note, that the "actual" checkbox is hidden from view but the functionality remains.
<form>
<input type="checkbox" class="custom" />
</form>
<style type="text/css">
.checkbox {
clear:left;
float:left;
background:url('your_image');
background-position:top;
width:20px;
height:20px;
display:block;
}
.toggled {
background-position:bottom !important;
}
</style>
$(document).ready(function () {
var checkboxes = $('form .custom'),
custom = $('<span></span>').addClass('checkbox');
checkboxes.before(custom);
checkboxes.css('visibility', 'hidden');
$('.checkbox').click(function () {
$(this).toggleClass('toggled');
var isChecked = $(this).next(':checkbox');
var value = isChecked.prop('checked') ? 'true' : 'false';
if (value == 'false') {
isChecked.prop('checked', true);
} else {
isChecked.prop('checked', false);
}
});
});
You will, of course, have to edit the CSS to suit your exact needs. I hope this helps as this task was deceptively non-trivial.
I would like to style my selected button.
I would like to display a light-blue border around the image of my selected button to show which page the user is on. (or just use the same hover image as the selected button image when the button is pushed.)
I didn't have success with the css link selectors :visited, :focus, or :selected.
Does this require a javascript solution?
thanks for any pointers!
i usually just a extra class name called selected
<div class="button selected">Button 1</div>
<div class="button">Button 2</div>
.selected {
border: 1px solid #0000ff;
}
It depends on how you display your page (using ajax or refresh on every click). If you are using javascript to load the page content than you just put an extra classname using javascript when the button is clicked.
you should use :active pseudo class in css to achieve what you want.
jQuery Solution with your CSS
You would probably want to check first if it is selected, that way this solution works with things like Twitter Bootstrap, where you can make any element act like a button:
$(function () {
$('div.button').click(function(){
if ($(this).hasClass('selected') {
$(this).removeClass('selected');
//Insert logic if you want a type of optional click/off click code
}
else
{
$(this).addClass('selected');
//Insert event handling logic
}
})
});
You will, in fact, need to use javascript. I did this in a project a while back, by iterating through the links in the navbar, and setting a class called "selected" on the one the user is currently visiting.
If you use jQuery, you can accomplish it like this:
$(function() {
$('#navbar li').each(function() {
if ($(this).children('a').attr('href') == window.location.pathname)
{
$(this).addClass('active');
}
});
})
The CSS Pseudo-selector :active won't still be active after a pagereload.