I need to remove all the style definitions in a page using javascript, to obtain the same result as doing View > Page Style > No Style
in Firefox. Which i
Here is the ES6 goodness you can do with just one line.
1) To remove all inline styles (eg: style="widh:100px"
)
document.querySelectorAll('[style]')
.forEach(el => el.removeAttribute('style'));
2) To remove link external stylesheet (eg: <link rel="stylesheet"
)
document.querySelectorAll('link[rel="stylesheet"]')
.forEach(el => el.parentNode.removeChild(el));
3) To remove all inline style tags (eg: <style></style>
)
document.querySelectorAll('style')
.forEach(el => el.parentNode.removeChild(el));
Actually, document.querySelectorAll
returns NodeList
which has forEach method.
document.querySelectorAll('link[rel="stylesheet"], style')
.forEach(elem => elem.parentNode.removeChild(elem));
You can recursively iterate through all elements and remove the style
attribute:
function removeStyles(el) {
el.removeAttribute('style');
if(el.childNodes.length > 0) {
for(let child in el.childNodes) {
/* filter element nodes only */
if(el.childNodes[child].nodeType == 1)
removeStyles(el.childNodes[child]);
}
}
}
Or:
function removeStyles(el) {
el.removeAttribute('style')
el.childeNodes.forEach(x => {
if(x.nodeType == 1) removeStyles(x)
})
}
Usage:
removeStyles(document.body);
To remove linked stylesheets you can, in addition, use the following snippet:
const stylesheets = [...document.getElementsByTagName('link')];
for(let i in stylesheets) {
const sheet = stylesheets[i];
const type = sheet.getAttribute('type');
if(!!type && type.toLowerCase() == 'text/css')
sheet.parentNode.removeChild(sheet);
}
Or:
const sheets = [...document.getElementsByTagName('link')];
sheets.forEach(x => {
const type = x.getAttribute('type');
!!type && type.toLowerCase() === 'text/css'
&& x.parentNode.removeChild(x);
});
If you have jQuery, you can probably do something like
$('link[rel="stylesheet"], style').remove();
$('*').removeAttr('style');