I'm attempting to use the lastmod function in conjunction with cheerio (or gulp-cheerio) to pull a value from a meta tag from each file in the stream to populate this lastmod entry.
var gulp = require('gulp');
var cheerio = require('cheerio');
var sitemap = require('gulp-sitemap');
gulp.task('sitemap', function() {
return gulp.src(['/prod/**/*.html',])
.pipe(sitemap({
siteUrl: 'https://www.somewhere.com',
lastmod: function(file) {
var $ = cheerio.load(body);
var lastmodValue = $('head > meta[name="dc.date"]').attr('content');
return lastmodValue.toString().trim();
}
}))
.pipe(gulp.dest('/prod'));
});
This works fine if I hardcode a var before the return and comment out the cheerio statements. Gulp-sitemap lastmod instructions here: https://github.com/pgilad/gulp-sitemap#lastmod
Found my issue - Here's the code that works:
.pipe(sitemap({
siteUrl: 'https://www.somewhere.com',
lastmod: function(file) {
var $ = cheerio.load(file.contents);
var lastmodValue = $('head > meta[name="dc.date"]').attr('content');
return lastmodValue.toString().trim();
}
}))
Related
I'm seeing a similar issue as this post (gulp-concat twice the content).
However, I'm dumping the concatenated file into a different directory, so it's not pulling in the resulting concatenated file into task, yet I'm seeing the contents of each file doubling up for some reason.
My gulp file is as follows:
/* jshint node: true */
module.exports = function (gulp, options, plugins) {
var merge = require('merge-stream');
var uglify = require('gulp-uglify');
var pump = require('pump');
var gp_concat = require('gulp-concat');
var gp_rename = require('gulp-rename');
var gp_ignore = require('gulp-ignore');
var ngAnnotate = require('gulp-ng-annotate');
var paths = require('../paths');
var utils = require('../utils');
var base = [
paths.APP,
paths.ETC,
paths.DESIGN
];
gulp.task('scripts:clean', function () {
var srcOptions = {
read: false
};
var tasks = base.map(function (folder) {
return gulp.src(folder + '/**/' + paths.GENERATED_SUBPATH + '/js/**/*.js', srcOptions)
.pipe(plugins.clean({force: true}));
});
return merge(tasks);
});
gulp.task('compress', function () {
var filesToInclude = ['**/app/components/**/*.js'
];
var excludeCondition = '**/*.spec*.js'
var fileToDest = paths.GLOBAL + '/'+paths.GENERATED_SUBPATH + '/js';
return gulp.src(filesToInclude)
.pipe(gp_ignore.exclude(excludeCondition))
.pipe(ngAnnotate({add: true}))
.pipe(gp_concat('all.concat.js'))
.pipe(gulp.dest('dist'))
.pipe(gp_rename('all.min.js'))
.pipe(uglify())
.pipe(gulp.dest(fileToDest));
});
gulp.task('scripts:build', ['scripts:clean', 'compress']);
};
Can someone help me understand why the
var filesToInclude = ['**/app/components/**/*.js];
would bring in each file twice? I've checked the files and no, the files are not duplicated anywhere in there.
It seems that the issue was with the definition of the filesToInclude, with it starting out with a wildcard. Since the fileToDest puts the file in a separate target directory, but the structure is the same, the process picks up the files twice.
I have a folder of HTML files that contain a comment at the top with metadata. I would like to run one gulp-replace operation if the metadata matches one regex, and another gulp-replace operation if it doesn't match, then continue on with the rest of the tasks pipeline. If tried various iterations using gulp-if but it always results in "TypeError: undefined is not a function" errors
import gulp from 'gulp';
import plugins from 'gulp-load-plugins';
const $ = plugins();
function preprocess() {
var template_data = new RegExp('<!-- template_language:(\\w+)? -->\n', 'i');
var handlebars = new RegExp('<!-- template_language:handlebars -->', 'i');
var primaryColor = new RegExp('#dc002d', 'gi');
var mailchimpColorTag = '*|PRIMARY_COLOR|*';
var handlebarsColorTag = '{{PRIMARY_COLOR}}';
var replaceCondition = function (file) {
return file.contents.toString().match(handlebars);
}
return gulp.src('dist/**/*.html')
.pipe($.if(
replaceCondition,
$.replace(primaryColor, handlebarsColorTag),
$.replace(primaryColor, mailchimpColorTag)
))
.pipe($.replace, template_data, '')
.pipe(gulp.dest('dist'));
}
What's the most efficient way to go about this?
gulp-filter was the answer. Whereas gulp-if can be used to decide whether a particular operation should be applied to the whole stream, gulp-filter can be used to decide which files in a stream an operation should be applied to.
import gulp from 'gulp';
import plugins from 'gulp-load-plugins';
const $ = plugins();
function preprocess() {
var template_language = new RegExp('<!-- template_language:(\\w+)? -->\n', 'i');
var handlebars = 'handlebars';
var primaryColor = new RegExp('#dc002d', 'gi');
var handlebarsColorTag = '{{PRIMARY_COLOR}}';
var handlebarsCondition = function (file) {
var match = file.contents.toString().match(template_language);
return (match && match[1] == handlebars);
}
var handlebarsFilter = $.filter(handlebarsCondition, {restore: true});
var mailchimpColorTag = '*|PRIMARY_COLOR|*';
var mailchimpCondition = function (file) {
return !handlebarsCondition(file);
}
var mailchimpFilter = $.filter(mailchimpCondition, {restore: true});
return gulp.src('dist/**/*.html')
.pipe(handlebarsFilter)
.pipe($.replace(primaryColor, handlebarsColorTag))
.pipe($.debug({title: 'Applying ' + handlebarsColorTag}))
.pipe(handlebarsFilter.restore)
.pipe(mailchimpFilter)
.pipe($.replace(primaryColor, mailchimpColorTag))
.pipe($.debug({title: 'Applying ' + mailchimpColorTag}))
.pipe(mailchimpFilter.restore)
.pipe($.replace(template_language, ''))
.pipe(gulp.dest('dist'));
}
I'm wanting to only load one JS file which uses jQuery code, but am confused about how to best do it. The thing I'm worried about is doing something sloppy like the below to solve the issue of loading all the scripts under $(document).ready(function(){});
gulp.task('compile-js', function() {
gulp.src(['./js/initialization.js', './stuff.js'./js/end.js'])
.pipe(concat('script.js'))
.pipe(gulp.dest('./public/javascripts/'));
});
where initialization.js and end.js are for the wrapping of the document.ready function (I know lol, hence asking)
Is there a better way of doing it?
Write a gulp file, lets call it 'jquery-noconflict.js'
var through = require('through2');
var gutil = require('gulp-util');
var fs = require('fs');
module.exports = function(){
var stream = through.obj(function(file, enc, cb) {
if (file.isStream()) {
this.emit('error', new PluginError(PLUGIN_NAME, 'Streams are not supported!'));
return cb();
}
if (file.isBuffer()) {
var contents = file.contents.toString();
file.contents = Buffer.concat([new Buffer('jQuery(document).ready(function(){'), file.contents, new Buffer('})')]);
}
cb(null, file);
}, function(){
})
return stream;
};
You might need to 'npm install through2'
now in your gulpfile.js
var gulp = require('gulp');
var concat = require('gulp-concat');
var jquery = require('./jquery-noconflict');
gulp.task('compile-js', function(){
gulp.src('./stuff.js')
.pipe(concat('script.js'))
.pipe(jquery())
.pipe(gulp.dest('./public/javascripts/'))
})
I am trying to use Object.assign from ES6, but it is always undefined. Here is the file where I am using it:
var Dispatcher = require('./dispatcher.js');
export default Object.assign(Dispatcher.prototype, {
handleViewAction(action) {
this.dispatch({
source: 'VIEW_ACTION',
action: action
});
}
});
And here is the gulp task I am using to transpile the code:
var browserify = require('browserify');
var reactify = require('reactify');
var source = require('vinyl-source-stream')
var to5 = require('6to5ify');
module.exports = function(gulp, config) {
gulp.task('browserify', function() {
browserify(config.app.entry, {debug: true})
.add(require.resolve('6to5/polyfill'))
.transform(to5)
.transform(reactify)
.bundle()
.pipe(source(config.app.bundleName))
.pipe(gulp.dest(config.app.bundle));
});
};
My Object is valid, but assign is undefined. What am I doing wrong?
Babeljs (formerly 6to5) does not transform or polyfill Object.assign.
You will need to use another polyfill library. 2 popular libraries are core-js or object-assign.
With core-js you would do it like this:
require('core-js');
Object.assign({}, {a:1});
or
var core = require('core-js/library');
var result = core.Object.assign({}, {a:1});
With object-assign:
var _assign = require('object-assign');
var result = _assign({}, {a:1});
But there are of course a lot of other libraries which do the same.
I'm trying to write a minify function that can be used to minifiy html, css, and js depending on file type. I would like to use the existing gulp plugins for these 3 minification processes to do the actual minification. The problem I'm having is I don't know how to call a plugin on a single vinyl file. Here is what I have so far:
var cssmin = require('gulp-cssmin');
var htmlmin = require('gulp-minify-html');
var uglify = require('gulp-uglify');
var minifiers = {
js: uglify,
css: cssmin,
html: htmlmin
};
function minify(options) {
var options = options || {};
return tap(function(file){
var fileType = file.path.split('.').pop();
options = options[fileType] || options
var minifier = minifiers[fileType];
if(!minifier)
console.error("No minifier for " + fileType + " - " + file.path);
// WHAT DO I DO HERE? This doesn't work but I want to do something similar
file.pipe(minifier(options));
});
}
I would like to be able to call the minify function like this:
gulp.src(['test.html', 'test.css', 'test.js'])
.pipe(minify());
Use gulp-filter.
var gulpFilter = require('gulp-filter');
var jsFilter = gulpFilter('**/*.js');
var cssFilter = gulpFilter('**/*.css');
var htmlFilter = gulpFilter('**/*.html');
gulp.task('default', function () {
gulp.src('assets/**')
.pipe(jsFilter)
.pipe(uglify())
.pipe(jsFilter.restore())
.pipe(cssFilter)
.pipe(cssmin())
.pipe(cssFilter.restore())
.pipe(htmlFilter)
.pipe(htmlmin())
.pipe(htmlFilter.restore())
.pipe(gulp.dest('out/'));
});
Will work for single files too but globs are more futureproof :)
SOLUTION:
I ended up using gulp-filter to solve the issue, but it was fairly tricky to get it working in a reusable way. Here is my final code:
var cssmin = require('gulp-cssmin');
var htmlmin = require('gulp-htmlmin');
var uglify = require('gulp-uglify');
var lazypipe = require('lazypipe');
function getFilter(type) {
// create a filter for the specified file type
return filter('**/*.' + type);
}
var minify = function() {
var jsFilter = getFilter('js'),
cssFilter = getFilter('css'),
htmlFilter = getFilter('html');
var min = lazypipe()
.pipe(function(){return jsFilter;})
.pipe(uglify)
.pipe(jsFilter.restore)
.pipe(function(){return cssFilter;})
.pipe(cssmin)
.pipe(cssFilter.restore)
.pipe(function(){return htmlFilter;})
.pipe(htmlmin)
.pipe(htmlFilter.restore);
return min();
};
To run gulp plugin on a single file you need to do the following:
var stream = minifier(options);
stream.once('data', function(newFile) {
file.contents = newFile.contents;
})
stream.write(file);