我有一个项目,其中有大约30个css主题。这意味着我有下一个css文件结构:
src/
themes/
default/
a.scss
b.scss
rockStar/
a.scss
b.scss
oneMoreTheme/
a.scss
b.scss
dist/
themes/
default/
styles.css
rockStar/
styles.css
oneMoreTheme/
styles.css这里只是吞咽文件的一个例子:
var gulp = require('gulp'),
glob = require('glob'),
path = require('path'),
_ = require('underscore'),
$ = require('gulp-load-plugins')(),
options = {};
options.themes = [
'default',
'rockStar',
'oneMoreTheme'
];
gulp.task('styles', function () {
_.each(options.themes, function(themeName, themeKey) {
gulp.src('src/themes/' + themeName + '/**/*.scss')
.pipe($.concat('styles.scss'))
.pipe($.sass())
.pipe(gulp.dest('dist/themes/' + themeName + '/'));
});
});
gulp.task('watch', function () {
gulp.watch('src/**/*.*', ['styles']);
});在我的gulp文件中,我有一个任务“样式”,它从每个主题编译scss文件,并将编译后的文件放到dist文件夹中。我有任务“监视”,当任何scss文件形成任何源主题变化时,该任务运行“样式”任务。它工作,但它需要很多时间,因为有很多的主题!我的任务“监视”如何检测哪些主题文件更改,并只运行任务“样式”此已更改的主题?
发布于 2015-05-13 18:14:30
这确实是一个艰难的问题,但这里有一个解决办法。请参阅代码中的评论以获得解释。
第1版
var gulp = require('gulp');
var merge = require('merge2');
var $ = require('gulp-load-plugins')();
var path = require('path');
var options = {};
options.themes = [
'default',
'rockStar',
'oneMoreTheme'
];
// we extract the task itself to a new function
// this allows us to reuse it
var styleTask = function(themeName) {
return gulp.src('src/themes/' + themeName + '/**/*.scss')
.pipe($.concat('styles.scss'))
.pipe($.sass())
.pipe(gulp.dest('dist/themes/' + themeName + '/'));
}
// we adapt the style task to use that function
// please note that I switched _.each for merge
// this allows you to work with streams!
gulp.task('styles', function() {
var tasks = themes.map(styleTask);
return merge(tasks);
});
// here we set up a new watcher. Instead of running 'styles'
// we filter the theme directory from the file that has changed
// and run the styleTask function
gulp.task('default', function() {
var watcher = gulp.watch('src/themes/**/*.scss', function(e) {
var theme = path
.relative(__dirname, e.path)
.substr('src/themes/'.length)
.split('/')[0];
console.log('rebuilding ' + theme);
return styleTask('theme');
});
});第2版
// same as above except the default task. we save the theme
// we want to build in a file
var singleTheme;
// and call the styleTask function should it be set
gulp.task('single-style', function(done) {
if(singleTheme) {
return styleTask(singleTheme);
} else {
done();
}
});
// here we have a watcher that calls single-style, but before calling
// it gets the right themename.
gulp.task('default', function() {
var watcher = gulp.watch('src/themes/**/*.scss', 'single-style');
watcher.on('change', function(e) {
singleTheme = path
.relative(__dirname, e.path)
.substr('src/themes/'.length)
.split('/')[0];
console.log('rebuilding ' + theme);
})
})希望这能帮上忙。
如果您想要运行更多的任务并在任务结束时进行状态调用,请更新,请转到version 2。无法添加所有要运行的任务。
gulp.task('default', function() {
var watcher = gulp.watch('src/themes/**/*.scss', ['single-style', 'another-task']);
watcher.on('change', function(e) {
singleTheme = path
.relative(__dirname, e.path)
.substr('src/themes/'.length)
.split('/')[0];
console.log('rebuilding ' + theme);
})
})而不是gulp.run,您可以使用gulp.start。
https://stackoverflow.com/questions/30218442
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