I'm using Gulp and I have a folder that has several .js files in in, of which I only need one of them; I read up on negating files within Gulp and from what I understood you should remove them first and then afterwards you can add them back in case you didn't want to negate them all.
I have this code for example:
var js_scripts = [
'js/dev/lib/**/*.js',
'js/dev/plugins/**/*.js',
'!js/dev/plugins/fancybox/*.js',
'js/dev/plugins/fancybox/jquery.fancybox-1.3.4.pack.js',
'!js/dev/plugins/inner/*.js',
'!js/dev/plugins/jquery.bxslider/**/*.js',
'js/dev/plugins/jquery.bxslider/jquery.bxslider.min.js',
// We have to set the bootstrap lines separately as some need to go before others
'js/dev/bootstrap/collapse.js',
'js/dev/bootstrap/dropdown.js',
'js/dev/bootstrap/tab.js',
'js/dev/bootstrap/transition.js',
'js/dev/scripts.js'
];
gulp.task('scripts', function() {
return gulp.src(js_scripts)
.pipe(sourcemaps.init())
.pipe(concat('scripts.js'))
.pipe(sourcemaps.write('../maps'))
.pipe(gulp.dest('./js'));
});
gulp.task('uglify', ['scripts'], function() {
return gulp.src(js_scripts)
.pipe(gulpif('!**/*.min.js', uglify({mangle: false})))
.pipe(concat('scripts.min.js'))
.pipe(gulp.dest('./js'));
});
However I have just noticed that both js/dev/plugins/fancybox/jquery.fancybox-1.3.4.pack.js and js/dev/plugins/jquery.bxslider/jquery.bxslider.min.js have not been included in the output files even though I added them back in after initial removal.
According to the docs it states that:
Note that globs are evaluated in order, which means this is possible:
// exclude every JS file that starts with a b except bad.js
gulp.src(['*.js', '!b*.js', 'bad.js'])
So the fact that I first include the whole plugins directory and then remove everything inside js/dev/plugins/jquery.bxslider/ and then re-add the file js/dev/plugins/jquery.bxslider/jquery.bxslider.min.js for example then should should include that file?
I noticed if I removed this line: !js/dev/plugins/jquery.bxslider/**/*.js then it gets included.
What am I doing wrong here?
Related
I use Gulp 4 with gulp-sass. What I try to do is the following.
The code I provide here is just a fragment of the whole code, to show the important parts.
The watch should like it does, watch all .scss files.
In style(), only the current file is going to be processed.
If I save custom/components/header/header.scss, then only custom/components/header/header.scss should be processed, not all files.
The saved file should then have a filename like assets/css/dist/header/header.css
Both src and dest is unknown in this case because I don't have a single file to grab on to.
I see now that I also need to remove custom/components from the dest, but the important thing is that I can get the current file to start working with that.
gulp.watch('custom/components/**/*.scss', style);
function style() {
return gulp
.src()
.pipe(sass())
.on('error', sass.logError)
.pipe(gulp.dest('assets/css/dist'));
}
I just figure it out. It's possible to do it like this.
let watcher = gulp.watch(css.watch);
watcher.on('change', function(path) {
console.log(path);
style();
}
The setup is as simple as this:
gulp.task('rev-js', function() {
return gulp.src('/js/main.js, {base: '.'})
.pipe(newer('_build'))
.pipe(rev())
.pipe(gulp.dest('_build'))
.pipe(rev.manifest())
.pipe(gulp.dest('_build/rev/js'));
});
gulp-newer obviously doesn't work here since the destination file gets a different name. Any workaround to make gulp-newer (or gulp-changed) work in this case?
In the gulp-newer options documentation I read that it supports passing in a configuration object instead of the destination. In that configuration object you can specify a mapping function from old to new files. So instead of
newer('_build')
you can write
newer({dest: '_build', map: mappingFn})
The mapping function takes the relative name of the file and expects it to return a translated name - see the index.js file. You can define a function that uses the previously generated rev-manifest.json manifest to look up the correct filename. Id put something along these lines in your build script (not tested):
gulp.task('rev-js', function() {
// get the existing manifest
// todo: add logic to skip this if file doesn't exist
var currentManifest = JSON.parse(fs.readFileSync('rev-manifest.json', 'utf8'));
// mapping function for gulp-newer
function mapToRevisions(relativeName) {
return currentManifest[relativeName]
}
return gulp.src('/js/main.js, {base: '.'})
.pipe(newer({dest: '_build', map: mapToRevisions}))
.pipe(rev())
.pipe(gulp.dest('_build'))
.pipe(rev.manifest())
.pipe(gulp.dest('_build/rev/js'));
});
May I suggest gulp-newy in which you can manipulate the path and filename in your own function. Then, just use the function as the callback to the newy(). This gives you complete control of the files you would like to compare.
This will allow 1:1 or many to 1 compares.
newy(function(projectDir, srcFile, absSrcFile) {
// do whatever you want to here.
// construct your absolute path, change filename suffix, etc.
// then return /foo/bar/filename.suffix as the file to compare against
}
I'm using gulp-compass to compile scss files. However, I'm concatenating output into a single file.
This all works fine, but I'm noticing that compass itself is writing the individual files to the output directory.
I'm left with the individual files, as well as the concatenated result.
Is there any way to prevent that intermediate output?
gulp.task('compass:dev', function() {
return gulp.src(appPath + '/**/*.scss')
.pipe(plugins.compass({
css: distPath + '/css',
sass: appPath
}))
.pipe(plugins.concat('app.css'))
.pipe(gulp.dest(distPath + '/css'));
});
As mentioned before, gulp-compass has been blacklisted by the Gulp developers for violating against some "plugin rules" which have been established (for instance: you have to redefine input and output). Which means that you really, really shouldn't use it. However, gulp-ruby-sass as an option for allowing you to use compass imports. Consider this:
var sass = require('gulp-ruby-sass');
gulp.task('compass:dev', function() {
return sass(appPath + '/**/*.scss', { compass: true})
.pipe(plugins.concat('app.css'))
.pipe(gulp.dest(distPath + '/css'));
});
Depending on your setup, there still might be breaks.
I'm using gulp to copy all files from one dir to another using code like this:
gulp.src([ 'app/**/*' ]).pipe(gulp.dest('dist'));
Glob docs say * match all files, but in fact files which have names starting with dot, like .gitignore, are not copied.
How can it be worked around?
If you add the option dot: true, it should work. Eg:
gulp.task('something', function () {
return gulp.src([ 'app/**/*' ], {
dot: true
}).pipe(gulp.dest('dist'));
});
Reference
For instances where the glob pattern is the only available interface. This pattern will do the trick:
**/{,.,.*/**/,.*/**/.}*
This expands to become the following globs:
**/*
**/.*
**/.*/**/*
**/.*/**/.*
You can add app to the beginning for app/**/{,.,.*/**/,.*/**/.}*.
I'm probably trying to make gulp do something that's not idiomatic, but here goes.
I want my build task to only run if the source files are newer than the output file.
In gulp, it seems standard practice to create a build task that always runs, and then set up a watch task to only run that build task when certain files change. That's okay, but it means that you always build on the first run.
So, is it possible to do what I want? Here's what I've got so far (newer is gulp-newer):
gulp.task('build_lib', function() {
return gulp.src(["app/**/*.ts"])
.pipe(newer("out/outputLib.js")) //are any of these files newer than the output?
** NEED SOMETHING HERE **
how do I say, "If I got _any_ files from the step before, replace all of them with a single hardcoded file "app/scripts/LibSource.ts" "?
.pipe(typescript({
declaration: true,
sourcemap: true,
emitError: false,
safe: true,
target: "ES5",
out: "outputLib.js"
}))
.pipe(gulp.dest('out/'))
});
I tried using gulpif, but it doesn't seem to work if there are no files going into it to begin with.
.pipe(gulpif(are_there_any_files_at_all,
gulp.src(["app/scripts/LibSource.ts"])))
However, my condition function isn't even called because there are no files on which to call it. gulpif calls the truthy stream in this case, so LibSource gets added to my stream, which isn't what I want.
Maybe doing all of this in a single stream really isn't the right call, since the only reason I'm passing those files through the "gulp-newer" filter is to see if any of them is newer. I'm then discarding them and replacing them with another file. My question still stands though.
You can write your own through/transform stream to handle the condition like so:
// Additional core libs needed below
var path = require('path');
var fs = require('fs');
// Additional npm libs
var newer = require('gulp-newer');
var through = require('through');
var File = require('vinyl');
gulp.task('build_lib', function() {
return gulp.src(["app/**/*.ts"])
.pipe(newer("out/outputLib.js"))
.pipe(through(function(file) {
// If any files get through newer, just return the one entry
var libsrcpath = path.resolve('app', 'scripts', 'LibSource.ts');
// Pass libsrc through the stream
this.queue(new File({
base: path.dirname(libsrcpath),
path: libsrcpath,
contents: new Buffer(fs.readFileSync(libsrcpath))
}));
// Then end this stream by passing null to queue
// this will ignore any other additional files
this.queue(null);
}))
.pipe(typescript({
declaration: true,
sourcemap: true,
emitError: true,
safe: true,
target: "ES5",
out: "outputLib.js"
}))
.pipe(gulp.dest('out/'));
});
I know like, this question was posted over 4 years ago, however; I am sure this problem crosses the path of everyone, and although I think I understand the question that is being asked, I feel that there is an easier way to perform this task, off which, I posted a similar question recently on stackoverflow at New to GULP - Is it necessary to copy all files from src directory to dist directory for a project?
It uses gulp-changed, and for me, it worked like a charm, so for others who may look at this post for similar reasons, have a look at my post and see if it is what you are looking for.
Kind Regards
You don't need to build first. You can on your 'first run' only run the watch task from which you run all the other ones.
example:
// Create your 'watch' task
gulp.task( 'watch', function() {
gulp.watch( 'scripts/*.js', [ 'lint', 'test', 'scripts' ] );
gulp.watch( 'styles/sass/*.scss', [ 'sass_dev' ] );
} );
// On your first run you will only call the watch task
gulp.task( 'default', [ 'watch' ] );
This will avoid running any task on startup. I hope this will help you out.
May I suggest gulp-newy in which you can manipulate the path and filename in your own function. Then, just use the function as the callback to the newy(). This gives you complete control of the files you would like to compare.
This will allow 1:1 or many to 1 compares.
newy(function(projectDir, srcFile, absSrcFile) {
// do whatever you want to here.
// construct your absolute path, change filename suffix, etc.
// then return /foo/bar/filename.suffix as the file to compare against
}