I'm trying to render each file in my gulp source files with it's own json file, but I can't figure out how to access the current filename in the pipe function.
var gulp = require('gulp');
var handlebars = require('handlebars');
var gulpHandlebars = require('gulp-compile-handlebars');
gulp.task('compile-with-sample-data', function () {
var options = {}
return gulp.src('./src/**/*.html')
.pipe(gulpHandlebars({ data: require('./data/' + filename +'.json') }, options))
.pipe(gulp.dest('./build/'));
});
Where I can get it to work with the same file each time by just using require('./data/orders-complete.json'):
var gulp = require('gulp');
var handlebars = require('handlebars');
var gulpHandlebars = require('gulp-compile-handlebars');
gulp.task('compile-with-sample-data', function () {
var options = {}
return gulp.src('./src/**/*.html')
.pipe(gulpHandlebars({ data: require('./data/orders-complete.json') }, options))
.pipe(gulp.dest('./build/'));
});
It's not clear how I would do this.
Use gulp-tap, it enables you to get the file name and even change it for downstream pipes.
Related
I am trying to understand the following code from the browserify-css repo:
var gulp = require('gulp');
var gutil = require('gulp-util');
var path = require('path');
var browserify = require('browserify');
var sourceStream = require('vinyl-source-stream');
var fse = require('fs-extra');
var bundleStream = browserify()
.add('src/index.js')
.transform(require('browserify-css'), {
rootDir: 'src',
processRelativeUrl: function(relativeUrl) {
var stripQueryStringAndHashFromPath = function(url) {
return url.split('?')[0].split('#')[0];
};
var rootDir = path.resolve(process.cwd(), 'src');
var relativePath = stripQueryStringAndHashFromPath(relativeUrl);
var queryStringAndHash = relativeUrl.substring(relativePath.length);
//
// Copying files from '../node_modules/bootstrap/' to 'dist/vendor/bootstrap/'
//
var prefix = '../node_modules/';
if (_.startsWith(relativePath, prefix)) {
var vendorPath = 'vendor/' + relativePath.substring(prefix.length);
var source = path.join(rootDir, relativePath);
var target = path.join(rootDir, vendorPath);
gutil.log('Copying file from ' + JSON.stringify(source) + ' to ' + JSON.stringify(target));
fse.copySync(source, target);
// Returns a new path string with original query string and hash fragments
return vendorPath + queryStringAndHash;
}
return relativeUrl;
}
})
.bundle();
bundleStream
.pipe(sourceStream(bundleFile))
.pipe(gulp.dest(browserifyConfig.dest));
I don't understand the part
_.startsWith(relativePath, prefix)
Where is the underscore coming from? It's supposed to be javascript executed by a task runner. I've found that in the NodeJS REPL the underscore character outputs the result of the last executed expression, but that functionality can't be used inside scripts. It's also not an underscore.js instance because it is not being declared anywhere. startsWith is a String method.
So what am I missing?
That code is using the lodash library. You can see in this section of the readme that they're importing lodash with var _ = require('lodash');
I have several typescript files, some of them export a const named APIS.
I'm trying to access those exports (I want to concatenated all of them to a single file), but it doesn't seem to work. I'm obviously doing something wrong, but I'm not sure what.
For example, I have a folder named services, with 2 files: service1.ts, service2.ts.
service1.ts:
...
export const APIS = [ { "field1" : "blabla" } ];
service2.ts: does not contain the APIS var.
This is my gulpfile.js:
var gulp = require('gulp');
var concat = require('gulp-concat');
var map = require('gulp-map');
gulp.task('default', function() {
return gulp.src('.../services/*.ts')
.pipe(map(function(file) {
return file.APIS;
}))
.pipe(concat('all.js'))
.pipe(gulp.dest('./test/'));
});
When I run this task, I get nothing. When I added console.log(file.APIS); to the map function, I get undefined for all the values (although it is defined in service1.ts!).
This is following to: Extracting typescript exports to json file using gulp
EDIT: OK, so I tried saving the exports in a .js file instead of a .ts file, and now I can access those vars using require:
gulp.task('default', function() {
return gulp.src('./**/*.service.export.js')
.pipe(map(function(file) {
var fileObj = require(file.path);
...
}))
Now if I try console.log(fileObj.APIS); I get the correct values. What I'm still confused about is how I can pass these value on, and create a single file out of all these vars. Is it possible to push them into an array?
This will not work as you think it would work. Gulp itself knows nothing about typescript files, that file is a vinyl-file and has no knowledge about the typescript code within its content.
Edit
Based on your example, you can do something like this:
var gulp = require('gulp');
var concat = require('gulp-concat');
var map = require('gulp-map');
var fs = require('fs');
gulp.task('test', function ()
{
var allConstants = [];
var stream = gulp.src('./**/*.service.export.js')
.pipe(map(function(file)
{
var obj = require(file.path);
if (obj.APIS != null)
allConstants = allConstants.concat(obj.APIS);
return file;
}));
stream.on("end", function (cb)
{
// Do your own formatting here
var content = allConstants.map(function (constants)
{
return Object.keys(constants).reduce(function (aggregatedString, key)
{
return aggregatedString + key + " : " + constants[key];
}, "");
}).join(", ");
fs.writeFile('filename.txt', content, cb);
});
return stream;
});
Suggestion
If you want to collect multiple variables into a single file i.e. a common variables file I suggest gulp-replace.
Steps
Create a file, require it and use tags within that file to place your variables.
Advice
If you are already using services don't create an array. Instead create an object (JSON) where every property is a constant. i.e.
var constants = {
const_1: 0,
const_2: 1,
const_3: 2,
}
I have this gulpfile:
var gulp = require('gulp'),
concat = require('gulp-concat'),
uglify = require('gulp-uglify');
gulp.task('minifyJS', function() {
return gulp.src(['src/*.js'])
.pipe(uglify())
.pipe(gulp.dest('min'));
});
gulp.task('watch', function() {
gulp.watch(['src/*.js'], ['minifyJS']);
});
I want to know what file trigger the watcher and his absolute path.
For example: if my project is placed in /myCode and I change the file src/main.js, I want to see /myCode/src/main.js inside minifyJS task. Is there a way to do it?
Thank you for your time.
You can do it by using gulp-ng-annotate and gulp-changed:
var gulp = require('gulp');
var changed = require('gulp-changed');
var rename = require('gulp-rename');
var ngAnnotate = require('gulp-ng-annotate'); // just as an example
var SRC = 'src/*.js';
var DEST = 'src/';
//Function to get the path from the file name
function createPath(file) {
var stringArray = file.split('/');
var path = '';
var name = stringArray[1].split('.');
stringArray = name[0].split(/(?=[A-Z])/);
if (stringArray.length>1) {stringArray.pop()};
return {folder: stringArray[0], name: name[0]}
}
gulp.task('default', function () {
return gulp.src(SRC)
.pipe(changed(DEST))
// ngAnnotate will only get the files that
// changed since the last time it was run
.pipe(ngAnnotate())
.pipe(rename(function (path) {
var createdPath = createPath(path);
path.dirname = createdPath.folder;
path.basename: createdPath.name,
path.prefix: "",
path.suffix: "",
path.extname: ".min.js"
}))
.pipe(gulp.dest(DEST));
});
Result:
Use gulp-changed npm package.
$ npm install --save-dev gulp-changed
Try the below in gulp file, (I haven't tried)
var gulp = require('gulp'),
concat = require('gulp-concat'),
uglify = require('gulp-uglify'),
changed = require('gulp-changed');
gulp.task('minifyJS', function() {
return gulp.src(['src/*.js'])
.pipe(changed('min'))
.pipe(uglify())
.pipe(gulp.dest('min'));
});
gulp.task('watch', function() {
gulp.watch(['src/*.js'], ['minifyJS']);
});
see the documentation of this package https://www.npmjs.com/package/gulp-changed
Based on your comment to Julien's answer this should be fairly close to what you want, or at least get you going in the right direction:
var gulp = require('gulp'),
concat = require('gulp-concat'),
uglify = require('gulp-uglify'),
cache = require('gulp-cached'),
rename = require('gulp-rename'),
path = require('path');
function fileName(file) {
return file.dirname + path.sep + file.basename + file.extname;
}
gulp.task('minifyJS', function() {
return gulp.src(['src/*.js'])
.pipe(cache('minifyJS'))
.pipe(rename(function(file) {
var nameOfChangedFile = fileName(file);
if (nameOfChangedFile == './main.js') {
file.basename = 'main.min'
}
if (nameOfChangedFile == './userView.js') {
file.basename = 'user/userView.min'
}
console.log(nameOfChangedFile + ' -> ' + fileName(file));
}))
.pipe(uglify())
.pipe(gulp.dest('min'));
});
gulp.task('watch', function() {
gulp.watch(['src/*.js'], ['minifyJS']);
});
This uses gulp-cached to keep an in-memory cache of all the files in your src/ folder that have passed through the stream. Only files that have changed since the last invocation of minifyJS are passed down to the gulp-rename plugin.
The gulp-rename plugin itself is then used to alter the destination path of the changed files.
Note: the cache is empty on first run, since no files have passed through the gulp-cached plugin yet. This means that the first time you change a file all files in src/ will be written to the destination folder. On subsequent changes only the changed files will be written.
I recently started to use gulp to keep my dev project organized and I've run into a little something that I cant figure out. So this is my task :
gulp.task('jsassemble', function () {
return gulp
.src('vendor/proj/**/**/src/assets/js/*.js')
.pipe(concat('all.js'))
.pipe(gulp.dest('public/js'));
});
As you can see, it'll fetch every js file in vendor/proj/anyFolder/anySubFolder/src/assets/js, put them together, rename the newly created js 'all.js' and then put it in public/js. The problem is that I would like to have gulp to keep the folder hierarchy, for example :
Source = vendor/proj/anyFolder1/anySubFolder1/src/assets/js/*.js
Destination = public/js/anyFolder1/anySubFolder1/src/assets/js/all.js
Source = vendor/proj/anyFolder1/anySubFolder2/src/assets/js/*.js
Destination = public/js/anyFolder1/anySubFolder2/src/assets/js/all.js
Instead of simply having everything on those folder into a 1 and only public/js/all.js
Is there anyway to do it ? I've tried to google it first but I wasn't able to properly formulate my question in a few words and was given not-wanted results :/
You could create the function which keep your folder hierarchy. In this page (http://www.jamescrowley.co.uk/2014/02/17/using-gulp-packaging-files-by-folder/) you find the solution.
var fs = require('fs');
var path = require('path');
var es = require('event-stream');
var gulp = require('gulp');
var concat = require('gulp-concat');
var rename = require('gulp-rename');
var uglify = require('gulp-uglify');
var scriptsPath = './src/scripts/';
function getFolders(dir){
return fs.readdirSync(dir)
.filter(function(file){
return fs.statSync(path.join(dir, file)).isDirectory();
});
}
gulp.task('scripts', function() {
var folders = getFolders(scriptsPath);
var tasks = folders.map(function(folder) {
return gulp.src(path.join(scriptsPath, folder, '/*.js'))
.pipe(concat(folder + '.js'))
.pipe(gulp.dest(scriptsPath))
.pipe(uglify())
.pipe(rename(folder + '.min.js'))
.pipe(gulp.dest(scriptsPath));
});
return es.concat.apply(null, tasks);
});
Thanks to #caballerog who puts my on the right path, here's the explained code :
//get every folder from a 'pathTo/Something'
function getFolders(dir){
return fs.readdirSync(dir)
return fs.statSync(path.join(dir, file)).isDirectory();
}
var projectsRoot = 'vendor/proj/';
var pathToJsFiles = '/src/assets/js/';
var pathToPublic = 'public/js/';
gulp.task('scripts', function() {
var sites = [];
var pathToProjects = [];
// Fetching every folders in vendor/proj
projects = getFolders(projectsRoot);
// Fetching every subfolder in vendor/proj/something
for(index in projects){
sites.push(getFolders(projectsRoot + '/' + projects[index]));
}
// Pushing every projects/site that exists into an array
for(var i=0;i<projects.length;i++){
for(var j=0; j<sites.length; j++)
if(sites[i][j] != null)
pathToProjects.push(projects[i] + '/' + sites[i][j]);
}
// Fetching every JS on vendor/proj/pathToAProject/pathToJsFiles
// concatenate them together
// and sending them to pathToPublic/pathToAProject/all.js
var tasks = pathToProjects.map(function(pathToAProject) {
return gulp.src( projectsRoot + pathToAProject + pathToJsFiles + '/*.js')
.pipe(concat('all.js'))
.pipe(gulp.dest(pathToPublic + pathToAProject));
});
return es.concat.apply(null, tasks);
});
TL:DR = get every JS file in pathToPublic/someFolder/someFolder/PathToJS.
I'm using Gulp in a VS2015 project to run jscs on JavaScript files with the fix option set. The intention is to modify the same file that is read (viz., source and destination are the same).
var gulp = require('gulp');
var jscs = require('gulp-jscs');
var chmod = require('gulp-chmod');
var exec = require('gulp-exec');
var ourJsFiles = // an array of files and globbed paths
gulp.task('jscs', function (callback) {
ourJsFiles.forEach(function (fn) {
gulp.src(fn, { base: './' })
.pipe(jscs({
"preset": "google",
"maximumLineLength": 160,
"validateIndentation": 3,
"fix": true
}))
.pipe(gulp.dest('./'));
});
callback();
});
But I do not want to process any files that are read-only. Is there already a way to detect this in Gulp on Windows?
There is a plugin which allows you to work with subset of files: gulp-filter.
One of options is to pass filter function which will receive vinyl file object, so for e.g. you could use stat.mode property of that object which holds permissions and do something like:
var filter = require('gulp-filter');
...
var writableFiles = filter(function (file) {
//https://github.com/nodejs/node-v0.x-archive/issues/3045
var numericPermission = '0'+(e.stat.mode & parseInt('777', 8)).toString(8);
return numericPermission[1]==='6'
});
...
gulp.src(....)
.pipe(writableFiles)