Is using mixin with no parameter is more efficient than using extend
I have gone through many articles online and totally got confused.
Could someone clarify on below
Is using mixin with no parameter is more efficient than using extend
html5 css3
add a comment |
I have gone through many articles online and totally got confused.
Could someone clarify on below
Is using mixin with no parameter is more efficient than using extend
html5 css3
Are you using SASS?
– Mav
Nov 24 '18 at 3:45
yes.We are using SAAS.
– user3190467
Nov 24 '18 at 4:15
add a comment |
I have gone through many articles online and totally got confused.
Could someone clarify on below
Is using mixin with no parameter is more efficient than using extend
html5 css3
I have gone through many articles online and totally got confused.
Could someone clarify on below
Is using mixin with no parameter is more efficient than using extend
html5 css3
html5 css3
asked Nov 24 '18 at 3:23
user3190467user3190467
43
43
Are you using SASS?
– Mav
Nov 24 '18 at 3:45
yes.We are using SAAS.
– user3190467
Nov 24 '18 at 4:15
add a comment |
Are you using SASS?
– Mav
Nov 24 '18 at 3:45
yes.We are using SAAS.
– user3190467
Nov 24 '18 at 4:15
Are you using SASS?
– Mav
Nov 24 '18 at 3:45
Are you using SASS?
– Mav
Nov 24 '18 at 3:45
yes.We are using SAAS.
– user3190467
Nov 24 '18 at 4:15
yes.We are using SAAS.
– user3190467
Nov 24 '18 at 4:15
add a comment |
1 Answer
1
active
oldest
votes
Extending a class will add the styles just once, and add all the selectors that the style needs to be applied to alongside where it's defined. A mixin on the other hand, will copy the styles individually to each selector where you've included it.
Here's what I mean.
Mixin
@mixin mixin(){
color: blue;
}
.selector-1{
@include mixin();
}
.selector-2{
@include mixin();
}
Compiles to
.selector-1 {
color: blue;
}
.selector-2 {
color: blue;
}
Extending a class
.extend{
color: blue;
}
.selector-1{
@extend .extend;
}
.selector-2{
@extend .extend;
}
Compiles to
.extend, .selector-1, .selector-2 {
color: blue;
}
So in terms of efficiency, extending a class will result in a smaller compiled file size, which is why Sass offers the extend
feature in the first place.
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Extending a class will add the styles just once, and add all the selectors that the style needs to be applied to alongside where it's defined. A mixin on the other hand, will copy the styles individually to each selector where you've included it.
Here's what I mean.
Mixin
@mixin mixin(){
color: blue;
}
.selector-1{
@include mixin();
}
.selector-2{
@include mixin();
}
Compiles to
.selector-1 {
color: blue;
}
.selector-2 {
color: blue;
}
Extending a class
.extend{
color: blue;
}
.selector-1{
@extend .extend;
}
.selector-2{
@extend .extend;
}
Compiles to
.extend, .selector-1, .selector-2 {
color: blue;
}
So in terms of efficiency, extending a class will result in a smaller compiled file size, which is why Sass offers the extend
feature in the first place.
add a comment |
Extending a class will add the styles just once, and add all the selectors that the style needs to be applied to alongside where it's defined. A mixin on the other hand, will copy the styles individually to each selector where you've included it.
Here's what I mean.
Mixin
@mixin mixin(){
color: blue;
}
.selector-1{
@include mixin();
}
.selector-2{
@include mixin();
}
Compiles to
.selector-1 {
color: blue;
}
.selector-2 {
color: blue;
}
Extending a class
.extend{
color: blue;
}
.selector-1{
@extend .extend;
}
.selector-2{
@extend .extend;
}
Compiles to
.extend, .selector-1, .selector-2 {
color: blue;
}
So in terms of efficiency, extending a class will result in a smaller compiled file size, which is why Sass offers the extend
feature in the first place.
add a comment |
Extending a class will add the styles just once, and add all the selectors that the style needs to be applied to alongside where it's defined. A mixin on the other hand, will copy the styles individually to each selector where you've included it.
Here's what I mean.
Mixin
@mixin mixin(){
color: blue;
}
.selector-1{
@include mixin();
}
.selector-2{
@include mixin();
}
Compiles to
.selector-1 {
color: blue;
}
.selector-2 {
color: blue;
}
Extending a class
.extend{
color: blue;
}
.selector-1{
@extend .extend;
}
.selector-2{
@extend .extend;
}
Compiles to
.extend, .selector-1, .selector-2 {
color: blue;
}
So in terms of efficiency, extending a class will result in a smaller compiled file size, which is why Sass offers the extend
feature in the first place.
Extending a class will add the styles just once, and add all the selectors that the style needs to be applied to alongside where it's defined. A mixin on the other hand, will copy the styles individually to each selector where you've included it.
Here's what I mean.
Mixin
@mixin mixin(){
color: blue;
}
.selector-1{
@include mixin();
}
.selector-2{
@include mixin();
}
Compiles to
.selector-1 {
color: blue;
}
.selector-2 {
color: blue;
}
Extending a class
.extend{
color: blue;
}
.selector-1{
@extend .extend;
}
.selector-2{
@extend .extend;
}
Compiles to
.extend, .selector-1, .selector-2 {
color: blue;
}
So in terms of efficiency, extending a class will result in a smaller compiled file size, which is why Sass offers the extend
feature in the first place.
answered Nov 24 '18 at 5:05
MavMav
554520
554520
add a comment |
add a comment |
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Are you using SASS?
– Mav
Nov 24 '18 at 3:45
yes.We are using SAAS.
– user3190467
Nov 24 '18 at 4:15