Closed DeadcatDev closed 8 years ago
Move cache after less so the contents will change when b.less is changed
Not working - project have 'tree structure' with one entry root (point), so it must be rebuild from bottom to top every time :(( Probably need refactor and add some gulp-concat to it to use cache. Or maybe you have any idea, now its looks like:
index.less:
@import a.less
@import variables.less
@import c.less
c.less be like:
@import users.less
@import companies.less
@import [...]
and so companies.less be like:
@import invoices.less
@import morethings.less
and so and so.
When i want to rebuild invoices.less from cache there is no variables.less in it so i cant rebuild files like
gulp-task(src[ALL-LESS-FILES->toCSS], fnc(){
allLess.files to manyCss.files and then
concat (them)
minify etc.
});
but must do simply
gulp.task(src[index.less],fnc(){
less index -> index.css
minify
});
Thinikn' about doing more flat file structure like
invoices.less:
@import variables.less
@depedencies
for build it separately (from top to bottom) and than -> cache -> concat -> minify -> write.
What you thing about it? Build time is to high for me (like 2sec.) so need cache badly :) Sorry for my engrish ;)
Probably not the right module for your problem, look at some of the other plugins that handle incremental compilation in gulp
Hi,
i have 'main' index.less file which @import some additional files like
@import a.less @import b.less @import c. less and so...
So i have a gulp watcher
on all less files, but my build function is like:
The problem is that gulp-cached 'caching' only the 'index.less' file for it's name and contents, so if i update b.less - the cache will return the same file so nothing will change at all.
Any suggestion to run cache for every file?