I have markdown files that look like this
---
name: Some Name
date: '2013-09-09'
isCool: true
---
really cool text
I want to have a gulp task that only lets markdown through that has a particular property, for example isCool = true.
So I would imagine something like this
gulp.src('source/content/*/*.md')
.pipe(mdPrune({
isCool: true
}))
.pipe(gulp.dest('build/content/cool'));
then only the markdown that had an isCool attribute in the header would end up the build/content/cool folder.
gulp-filter would work.
const filter = require('gulp-filter');
gulp.task('default', function () {
// return true if want the file in the stream
const myFilter = filter(function (file) {
let contents = file.contents.toString();
return contents.match('isCool: true');
});
return gulp.src(['./src/*.md'])
.pipe(myFilter)
.pipe(gulp.dest('md'));
});
That will allow the file through if isCool: true is anywhere in the file. If that is a problem, just work on the regex to restrict it to the line after the date entry for example.
[The filter could also be defined outside of any task if it might be reused elsewhere or you just prefer it that way.
Related
I'm using GulpJS and gulp-token-replace to replace variables within an HTML file, such as title and meta tags. It's working great, but I'm stuck with only one configuration file (config-title-meta.json). I want to have an individual configuration file for each folder, so that I might be able to edit "main" variables for all html pages, but customize others, without having to edit each individual page. Am I using the correct tool for this?
gulpfile.js:
var replace = require('gulp-token-replace');
gulp.task('compile-html', function() {
var config = require('./config-title-meta.json');
return gulp.src([
'./**/*.html',
'!header.html', // ignore
'!footer.html' // ignore
])
.pipe(fileinclude({
prefix: '##',
basepath: '#file'
}))
.pipe(replace({global:config}))
.pipe(gulp.dest(Paths.scripts.dest));
});
I've read Get the current file name in gulp.src(), and it seems like it's approaching what I am attempting to do, but I need help.
Consider the following function in a gulpfile.js:
function inline() {
return gulp.src('dist/**/*.html')
.pipe($.if(PRODUCTION, inliner('dist/css/app.css')))
.pipe(gulp.dest('dist'));
}
And inliner(), to be thorough (also in the gulpfile):
function inliner(css) {
var css = fs.readFileSync(css).toString();
var mqCss = siphon(css);
var pipe = lazypipe()
.pipe($.inlineCss, {
applyStyleTags: false,
removeStyleTags: false,
removeLinkTags: false
})
.pipe($.replace, '<!-- <style> -->', `<style>${mqCss}</style>`);
return pipe();
}
These functions take an external CSS file and inline them into the respective HTML for email.
I really want to know how to do something like this:
function inline() {
return gulp.src('dist/**/*.html')
.pipe($.if(PRODUCTION, inliner('dist/css/' + file.name + '.css')))
.pipe(gulp.dest('dist'));
}
And you might ask yourself, "why?" Well, I don't have just one CSS file. If everything from app.css was to be inlined, there would be a lot more styles applied than were actually necessary.
So I want to inline:
email1.css ---- to -------> email1.html
email2.css ---- to -------> email2.html
email3.css ---- to -------> email3.html
And so on. Essentially, I want to get the name of the HTML file being processed at that moment in the Gulp Stream, save it as a variable, and then pass it into the inliner('dist/css/' + file.name + '.css') bit. I've exhausted every bit of Gulp Knowledge I have and have come up completely and utterly blank.
Basically what you need to do is send each .html file in your stream down its own little sub stream with its own inliner(). The gulp-foreach plugin let's you do just that.
Then it's just a matter of determining the simple name of your file from its absolute path. The node.js built-in path.parse() got you covered there.
Putting it all together:
var path = require('path');
function inline() {
return gulp.src('dist/**/*.html')
.pipe($.if(PRODUCTION, $.foreach(function(stream, file) {
var name = path.parse(file.path).name;
return stream.pipe(inliner('dist/css/' + name + '.css'));
})))
.pipe(gulp.dest('dist'));
}
I know there is already a tonne of automated tools to create a style guide / pattern library but in the interest of learning I'd like to see if I can roll my own.
Compiling the SASS is straight forward. Same with the js. I can also see how to wrap blocks of HTML from multiple files with a class and compiled into a single file. Ideal for displaying all the 'partials' together on one page.
gulp.task('inject:wrap', function(){
return gulp.src('./_patterns/*/*/*.html')
/// get the partial html filename here and insert below ###
.pipe(inject.wrap('<div id="###" class="pattern">', '</div>'))
.pipe(concat('patterns.html'))
.pipe(gulp.dest('build'));
});
gulp.task('process', ['inject:wrap']);
What I struggling with is how I can get the filename of the block - let's say _button.html - and pass this to the wrapper as the element id "###" above. Which I can then use to build the style guides navigation / anchor links.
Here's a sample code I've got, uses jade template language (which takes care of injections, partials, evaluation etc. by itself); There are two tasks, one generates static HTML pages, other pre-compiles templates to be used as runtime template functions wrapped in AMD
// preprocess & render jade static templates
gulp.task('views:preprocess', function () {
return gulp.src([ 'source/views/*.jade', '!source/views/layout.jade' ])
.pipe(plumber()) // plumber, because why not?
.pipe(data(function (file) {
// prepare data to be passed to the template
// here we can use the file name to map specific data to each file
return _.assign(settingsData, { timestamp: timestamp });
}))
// render template with data
.pipe(jade())
.pipe(gulp.dest('destination'));
});
// precompile jade runtime templates
gulp.task('views:precompile', function () {
// grab folder names
var folders = fs.readdirSync('source/templates').filter(function (file) {
return fs.statSync(path.join('source/templates', file)).isDirectory();
});
// create a separate task for each folder
var tasks = folders.map(function (folder) {
return gulp.src(path.join('source/templates', folder, '*.jade'))
.pipe(plumber())
// pre-compile the template as functions, for runtime
.pipe(jade({
client: true
}))
// wrap it in AMD, so we can use stuff like require.js to fetch them later
.pipe(wrap({
moduleRoot: 'source/templates',
modulePrefix: 'templates',
deps: [ 'jade' ],
params: [ 'jade' ]
}))
// concat all the templates in each folder to a single .js file
.pipe(concat(folder + '.js'))
.pipe(uglify())
.pipe(header(banner, { package: packageData }))
.pipe(gulp.dest('destination/scripts/templates'));
});
return merge(tasks);
});
Modules I've used are merge-stream, path, gulp, fs, gulp-data, gulp-jade, gulp-plumber etc.
Didn't quite understand what you're trying to achieve, but I hope this gives you some clues.
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
}
In my gulp.js file I'm streaming all HTML files from the examples folder into the build folder.
To create the gulp task is not difficult:
var gulp = require('gulp');
gulp.task('examples', function() {
return gulp.src('./examples/*.html')
.pipe(gulp.dest('./build'));
});
But I can't figure out how retrieve the file names found (and processed) in the task, or I can't find the right plugin.
I'm not sure how you want to use the file names, but one of these should help:
If you just want to see the names, you can use something like gulp-debug, which lists the details of the vinyl file. Insert this anywhere you want a list, like so:
var gulp = require('gulp'),
debug = require('gulp-debug');
gulp.task('examples', function() {
return gulp.src('./examples/*.html')
.pipe(debug())
.pipe(gulp.dest('./build'));
});
Another option is gulp-filelog, which I haven't used, but sounds similar (it might be a bit cleaner).
Another options is gulp-filesize, which outputs both the file and it's size.
If you want more control, you can use something like gulp-tap, which lets you provide your own function and look at the files in the pipe.
I found this plugin to be doing what I was expecting: gulp-using
Simple usage example: Search all files in project with .jsx extension
gulp.task('reactify', function(){
gulp.src(['../**/*.jsx'])
.pipe(using({}));
....
});
Output:
[gulp] Using gulpfile /app/build/gulpfile.js
[gulp] Starting 'reactify'...
[gulp] Finished 'reactify' after 2.92 ms
[gulp] Using file /app/staging/web/content/view/logon.jsx
[gulp] Using file /app/staging/web/content/view/components/rauth.jsx
Here is another simple way.
var es, log, logFile;
es = require('event-stream');
log = require('gulp-util').log;
logFile = function(es) {
return es.map(function(file, cb) {
log(file.path);
return cb(null, file);
});
};
gulp.task("do", function() {
return gulp.src('./examples/*.html')
.pipe(logFile(es))
.pipe(gulp.dest('./build'));
});
You can use the gulp-filenames module to get the array of paths.
You can even group them by namespaces:
var filenames = require("gulp-filenames");
gulp.src("./src/*.coffee")
.pipe(filenames("coffeescript"))
.pipe(gulp.dest("./dist"));
gulp.src("./src/*.js")
.pipe(filenames("javascript"))
.pipe(gulp.dest("./dist"));
filenames.get("coffeescript") // ["a.coffee","b.coffee"]
// Do Something With it
For my case gulp-ignore was perfect.
As option you may pass a function there:
function condition(file) {
// do whatever with file.path
// return boolean true if needed to exclude file
}
And the task would look like this:
var gulpIgnore = require('gulp-ignore');
gulp.task('task', function() {
gulp.src('./**/*.js')
.pipe(gulpIgnore.exclude(condition))
.pipe(gulp.dest('./dist/'));
});
If you want to use #OverZealous' answer (https://stackoverflow.com/a/21806974/1019307) in Typescript, you need to import instead of require:
import * as debug from 'gulp-debug';
...
return gulp.src('./examples/*.html')
.pipe(debug({title: 'example src:'}))
.pipe(gulp.dest('./build'));
(I also added a title).