I use gulp-react to compile jsx to js. I need to save folder structure while compiling.
The code below works good for single folder of all files, but I need dumanic destination
var gulp = require('gulp');
var react = require('gulp-react');
gulp.task('default', function () {
return gulp.src('template.jsx')
.pipe(react())
.pipe(gulp.dest('dist')); // in this line need dumanic destination
});
Any Ideas?
Problem solved like this
gulp.src('./public/js/**/.')
Related
I am new to gulp so i don't know as much good gulp plugins. I wrote a code for minifying js, css and html using gulp and its plugins which is working fine. But now i am stuck in unminifying code. I don't know which plugins to use which can easily unminify code.
guplfile.js:
var gulp = require('gulp'),
uglify = require('gulp-uglify')
htmlmin = require('gulp-html-minifier')
csso = require('gulp-csso');
gulp.task('min_js', function () {
gulp.src('app/**/*.js')
.pipe(uglify())
.pipe(gulp.dest('min'))
});
gulp.task('min_html', function () {
gulp.src('app/**/*.html')
.pipe(htmlmin({ collapseWhitespace: true }))
.pipe(gulp.dest('min'))
});
gulp.task('min_css', function () {
gulp.src('app/**/*.css')
.pipe(csso())
.pipe(gulp.dest('min'))
});
gulp.task('minify_all', ['min_js', 'min_html', 'min_css']);
//pending
//gulp.task('unminify',[]);
Uglifying/Minifying is attended for production, you should not uglify your code while you are developing (except for testing purpose).
When you start gulp tasks, you have to make sure that you have in one part your "working code", that you will transform into a "destination code".
When you are doing this :
gulp.task('min_js', function () {
gulp.src('app/**/*.js')
.pipe(uglify())
.pipe(gulp.dest('min'))
});
The code on which you are working on is in the app folder, and your transformed code is in the min folder (it's the destination folder).
But, if the min directory is also used in development, just disable the uglify task in development (easier to debug a not-uglifyied file).
There is no need to un-minify your sources, there are still present in app folder.
I am a newbie of gulp, just started now. I am confused in giving src and destination to my task. What i am try to do is to create a new minify destination folder with same inside hierarchy like below.
Current Folder Structure (originals files):
app
-components
--post-creator
---post-creator.component.js
--post-rating
---post-rating.component.js
-routing_components
--app-routing
---app-routing.component.js
--user-routing
---user-routing.component.js
Minify Folder (where js file are minified):
minify_destination
-components
--post-creator
---post-creator.component.js
--post-rating
---post-rating.component.js
-routing_components
--app-routing
---app-routing.component.js
--user-routing
---user-routing.component.js
Gulpfile.js
var gulp = require('gulp'),
uglify = require('gulp-uglify');
gulp.task('default', function () {
gulp.src('app/components/*/*/*.js')
.pipe(uglify())
.pipe(gulp.dest(components_min))
});
Try:
gulp.src('app/**/*.js')
and everything else as you have it and let me know if that works for you. In this case your components_min should replace the app folder - everything before the glob - and maintain your folder structure.
I'm trying to watch my #imported scss files, but for some reason it stopped working. Can't remember changing anything:
var gulp = require('gulp');
var sass = require('gulp-sass');
//style paths
var sassFiles = 'assets/css/scss/styles.scss',
cssDest = 'assets/css/';
gulp.task('styles', function(){
gulp.src(sassFiles)
.pipe(sass().on('error', sass.logError))
.pipe(gulp.dest(cssDest));
});
gulp.task('watch',function() {
gulp.watch(sassFiles,['styles']);
});
My styles.scss:
#import
"fonts.scss",
"basics.scss",
"home.scss";
Whenever I change something in home.scss, I want it to watch, but it only watches styles.scss. Makes sense to me, but it used to work it the beginning..
From documentation
gulp.watch(glob[, opts], tasks)
A single glob or array of globs that indicate which files to watch
for changes.
You watch only that files which are in sassFiles. You can change this line in such way
var sassFiles = 'assets/css/scss/**.scss',
or
var sassFiles = ['assets/css/scss/styles.scss',
'assets/css/scss/fonts.scss','assets/css/scss/basics.scss',
'assets/css/scss/home.scss'],
And all scss-files in folder assets/css/scss/ will be watched.
If you would like to watch files by dependencies (#import) it is better to use webpack
I'm trying to use these two gulp plugins together:
gulp-html-minifier
gulp-inject-stringified-html
Or put differently, I'm trying to inject the contents of files containing html fragments into my javascript files after they're minified.
When I'm trying to run a straight up gulp build I get this:
Error: ENOENT: no such file or directory, open 'C:\path\to\.temp\template.html'
Here's a repro of my situation. My folder structure:
/src/app.js
/src/template.html
/gulpfile.js
/package.json
My gulpfile.js:
var gulp = require('gulp');
var injectHtml = require('gulp-inject-stringified-html');
var htmlmin = require('gulp-html-minifier');
gulp.task('minify', [], function() {
gulp.src('src/*.html')
.pipe(htmlmin())
.pipe(gulp.dest('.temp'));
});
gulp.task('default', ['minify'], function() {
gulp.src('src/*.js')
.pipe(injectHtml())
.pipe(gulp.dest('.build'));
});
The template.html file:
<div>My Template</div>
The app.js file:
var html = { gulp_inject: "../.temp/template.html" };
Now, if I run minify manually first, things will work as expected. From this I speculate I'm not using Gulp correctly. I reckon I'd need to pipe the result of htmlmin into the injectHtml method. But I fail to see how.
How can I get these two plugins to play together nicely?
You are missing a return in the minify task. It should look like that:
gulp.task('minify', [], function() {
return gulp.src('src/*.html')
.pipe(htmlmin())
.pipe(gulp.dest('.temp'));
});
Without return, the default task doesn't have any way to know that minify finished, so it may start before the minified html file was created.
I'm trying to get gulp to compile some sass, however I can't seem to get it to write out anything. I'm not seeing an error and using sass to compile myself is successful. Here is what my config looks like:
var sass = require('gulp-sass');
var baseOutputDir = '../publish/homepage';
var cssOutputDir = baseOutputDir + '/css';
gulp.task('sass', function () {
gulp.src(['./css-lib/Bootstrap/3.3.1/assets/stylesheets/_bootstrap.scss'])
.pipe(sass())
.on('error', function (err) { console.log(err.message); })
.pipe(gulp.dest(cssOutputDir));
})
The bootstrap is a clone of https://github.com/twbs/bootstrap-sass
I'm use to LESS and not Sass. A file with an underscore in front won't generate a css file.
I had the same problem. As #Steven pointed out _filename do not generate css files. Just rename/copy the _bootstrap.scss to something without leading underscore
gulp.src(['./css-lib/Bootstrap/3.3.1/assets/stylesheets/bootstrap.scss'])