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();
}
Related
I want to make gulp work with files without extension also. I tried to do it the following ways:
function someTask() {
return gulp.src('./src/**/*.{html,,xml,txt}')
.pipe(gulp.dest('./build'));
}
function someTask() {
return gulp.src('./src/**/*.{html,!*,xml,txt}')
.pipe(gulp.dest('./build'));
}
function someTask() {
return gulp.src('./src/**/*.{html,[^*],xml,txt}')
.pipe(gulp.dest('./build'));
}
but I haven't got any results. Any suggestions?
I am still trying to find a better answer, but one route is to get all the files and then eliminate the extentions you don't want. Like
var debug = require('gulp-debug'); // nice, will list all the files in the stream
return gulp.src(['./src/**/*', '!./src/**/*.{js,png}'])
.pipe(debug())
.pipe(gulp.dest('./build'));
So here {js,png} list the possible file extensions in your src directories you don't want.
Hopefully there is a better answer.
[EDIT]: Maybe there is, try this:
return gulp.src(['./src/**/*.{html,xml,txt}', './src/**/!(*\.*)'])
This part ./src/**/!(*\.*) appears to match only files with no . in them, i.e., no extention.
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?
I would like to process a series of folders through the build pipeline. For each folder I start another gulp pipeline that will create some source files. Finally I'd like to pipe them all together through the zip() plugin to produce a final artifact.
The following pseudocode shows my approach. However, I cannot push the result of the gulp.src to the current pipeline.
return gulp.src(someFolder)
.pipe(through(function(folder, enc, cb){
// how to add the result of the following line to my
// own pipeline????
this.push(gulp.src(path.join(folder, "whatever"))
.pipe(somePlugin()));
}))
.pipe(zip(....))
.pipe(gulp.dest(....));
Any hints how to combine those pipelines together?
This is exactly what gulp-foreach is intended for. Using that the equivalent of the code you posted would look like this:
var foreach = require('gulp-foreach');
return gulp.src(someFolder)
.pipe(foreach(function(stream, folder) {
return gulp.src(path.join(folder.path, "whatever"))
.pipe(somePlugin());
}))
.pipe(zip(....))
.pipe(gulp.dest(....));
Note that the above will discard all files from the original gulp.src(someFolder). They will not be included in the ZIP. If you want to keep those around as well you can use merge-stream:
var foreach = require('gulp-foreach');
var merge = require('merge-stream');
return gulp.src(someFolder)
.pipe(foreach(function(stream, folder) {
return merge(stream,
gulp.src(path.join(folder.path, "whatever"))
.pipe(somePlugin()));
}))
.pipe(zip(....))
.pipe(gulp.dest(....));
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
}
I need to apply a build task for specific files. For finding them, I use the typical template. But I can't understood how to pass the arguments (file path) from gulp.src.
Desirable solution.
gulp.task('bundles', function() {
gulp.src('bundles/**/*.js').
pipe(gulp.start('build', file.path));
});
gulp.task('build', function (path) {
// use here
});
Question is a bit stale and I am not sure I totally understand what you're trying to achieve here, but I think what you're looking for is lazypipe
You might want to clarify your question if that's not what you're looking for
Example Usage:
var lazypipe = require('lazypipe'),
g = require('gulp-load-plugins')({lazy: true}),
jsTransformPipe = lazypipe()
.pipe(g.jshint) // <-- Notice the notation: g.jshint, not g.jshint()
.pipe(g.concat, 'bundle.js'), // <-- Notice how the param is passed to g.concat, as a second param to .pipe()
jsSourcePipe = lazypipe()
.pipe(gulp.src, './**/*.js');
gulp.task('bundle', function() {
jsSourcePipe()
.pipe(jsTransformPipe()) // <-- You execute the lazypipe by calling it as a function
.pipe(gulp.dest('../build/');
});
With lazypipe you basically create a pipe for future use; hope this help
(Can't comment because of rep, sorry)
I assume that your sample code isn't filled with everything, but why don't you merge those tasks and use your gulp.src() in your build task instead of calling another task.
Maybe it's useful for you but with what you're showing I can't find an explanation for why you do this instead of simply going with something like :
gulp.task('build', function (path) {
gulp.src('bundles/**/*.js)
//Your code for this task
});
Of course, it removes the bundles task, but it's not useful as is.
Don't hesitate to comment if I'm wrong and I'll try to help you as much as I can.
First off, gulp.task('build', function (path) won't ever work. The only valid argument for gulp tasks is a callback to signal asynchronous task completion. If you tried to do run the above, gulp would expect path to be a function and the task would never complete unless that function was called. In this example, the 'build' task should be a regular function called from the 'bundles' pipe, not a task.
The better question would be: How do I run a custom function inside a gulp pipe? Plugins like gulp-tap might get you close, but it's not difficult to create what is essentially an inline gulp plugin to call your function.
Gulp pipes receive a through2 object stream containing a vinyl file object, an encoding and a callback. Here's a basic skeleton for calling any arbitrary function against the files in a gulp pipe:
var gulp = require('gulp');
var through = require('through2');
gulp.task('stack', function() {
return gulp.src('./src/*.js')
.pipe(through.obj(function(file, enc, cb) {
// file.path is the full path to the file
myBuildFunction(file.path);
cb(null, file);
}))
.pipe(gulp.dest('./build/'));
})
This can be incredibly powerful. To modify the file's contents, just change the file.contents buffer. To rename or relocate the file, change file.path. Everything can be done in gulp's native pipes.