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        Gulps gulp.watch 不會為新文件或已刪除文件觸發(fā)?

        Gulps gulp.watch not triggered for new or deleted files?(Gulps gulp.watch 不會為新文件或已刪除文件觸發(fā)?)

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                  本文介紹了Gulps gulp.watch 不會為新文件或已刪除文件觸發(fā)?的處理方法,對大家解決問題具有一定的參考價值,需要的朋友們下面隨著小編來一起學(xué)習(xí)吧!

                  問題描述

                  限時送ChatGPT賬號..

                  在全局匹配中編輯文件時,以下 Gulpjs 任務(wù)可以正常工作:

                  The following Gulpjs task works fine when editing files in the glob match:

                  // watch task.
                  gulp.task('watch', ['build'], function () {
                      gulp.watch(src + '/js/**/*.js', ['scripts']);
                      gulp.watch(src + '/img//**/*.{jpg,jpeg,png,gif}', ['copy:images']);
                      gulp.watch(src + '/less/*.less', ['styles']);
                      gulp.watch(src + '/templates/**/*.{swig,json}', ['html']);
                  });
                  
                  // build task.
                  gulp.task('build', ['clean'], function() {
                      return gulp.start('copy', 'scripts', 'less', 'htmlmin');
                  });
                  

                  但是,對于新文件或已刪除文件,它不起作用(未觸發(fā)).我有什么遺漏嗎?

                  However it doesn't work (it's not triggered) for new or deleted files. Is there something I'm missing?

                  編輯:即使使用 grunt-watch 插件,它似乎也不起作用:

                  EDIT: even using grunt-watch plugin it seems not working:

                  gulp.task('scripts', function() {
                      return streamqueue(
                          { objectMode: true },
                          gulp.src([
                              vendor + '/jquery/dist/jquery.min.js',
                              vendor + '/bootstrap/dist/js/bootstrap.min.js'
                          ]),
                          gulp.src([
                              src + '/js/**/*.js'
                          ]).pipe(plugins.uglify())
                      )
                      .pipe(plugins.concat(pkg.name + '.min.js'))
                      .pipe(gulp.dest(dest + '/js/'));
                  });
                  
                  gulp.task('watch', ['build'], function () {
                      plugins.watch({glob: src + '/js/**/*.js'}, function () {
                          gulp.start('scripts');
                      });
                  });
                  

                  編輯:已解決,是這個問題.以 ./ 開頭的 Glob(即 src 的值)似乎無法在 ATM 上工作.

                  EDIT: Solved, it was this issue. Globs starting with ./ (that was the value of src) seems not working ATM.

                  推薦答案

                  顯然 gulp.watch 現(xiàn)在可以處理新文件或已刪除文件.問問題時沒有.

                  Apparently gulp.watch does work with new or deleted files now. It did not when the question was asked.

                  我的回答仍然有效:gulp-watch 通常是更好的解決方案,因為它允許您僅對已修改的文件執(zhí)行特定操作,而 gulp.watch 只允許您運行完整的任務(wù).對于一個合理規(guī)模的項目,這很快就會變得太慢而無法使用.

                  The rest of my answer still stands: gulp-watch is usually a better solution because it lets you perform specific actions only on the files that have been modified, while gulp.watch only lets you run complete tasks. For a project of a reasonable size, this will quickly become too slow to be useful.

                  你沒有錯過任何東西.gulp.watch 不適用于新文件或已刪除文件.這是一個為簡單項目設(shè)計的簡單解決方案.

                  You aren't missing anything. gulp.watch does not work with new or deleted files. It's a simple solution designed for simple projects.

                  要獲得可以查找新文件的文件監(jiān)視,請使用 gulp-watch 插件,更強大.用法如下所示:

                  To get file watching that can look for new files, use the gulp-watch plugin, which is much more powerful. Usage looks like this:

                  var watch = require('gulp-watch');
                  
                  // in a task
                  watch({glob: <<glob or array of globs>> })
                          .pipe( << add per-file tasks here>> );
                  
                  // if you'd rather rerun the whole task, you can do this:
                  watch({glob: <<glob or array of globs>>}, function() {
                      gulp.start( <<task name>> );
                  });
                  

                  我個人推薦第一個選項.這允許更快的每個文件進程.只要您不連接任何文件,它就可以在使用 livereload 的開發(fā)過程中運行良好.

                  Personally, I recommend the first option. This allows for much faster, per-file processes. It works great during development with livereload as long as you aren't concatenating any files.

                  您可以使用 我的 lazypipe 庫 或簡單地使用函數(shù)和 stream-combiner 像這樣:

                  You can wrap up your streams either using my lazypipe library, or simply using a function and stream-combiner like this:

                  var combine = require('stream-combiner');
                  
                  function scriptsPipeline() {
                      return combine(coffeeescript(), uglify(), gulp.dest('/path/to/dest'));
                  }
                  
                  watch({glob: 'src/scripts/**/*.js' })
                          .pipe(scriptsPipeline());
                  

                  <小時>

                  更新 2014 年 10 月 15 日


                  UPDATE October 15, 2014

                  正如下面@pkyeck 所指出的,顯然 gulp-watch 的 1.0 版本稍微改變了格式,所以上面的例子現(xiàn)在應(yīng)該是:

                  As pointed out by @pkyeck below, apparently the 1.0 release of gulp-watch changed the format slightly, so the above examples should now be:

                  var watch = require('gulp-watch');
                  
                  // in a task
                  watch(<<glob or array of globs>>)
                          .pipe( << add per-file tasks here>> );
                  
                  // if you'd rather rerun the whole task, you can do this:
                  watch(<<glob or array of globs>>, function() {
                      gulp.start( <<task name>> );
                  });
                  

                  var combine = require('stream-combiner');
                  
                  function scriptsPipeline() {
                      return combine(coffeeescript(), uglify(), gulp.dest('/path/to/dest'));
                  }
                  
                  watch('src/scripts/**/*.js')
                          .pipe(scriptsPipeline());
                  

                  這篇關(guān)于Gulps gulp.watch 不會為新文件或已刪除文件觸發(fā)?的文章就介紹到這了,希望我們推薦的答案對大家有所幫助,也希望大家多多支持html5模板網(wǎng)!

                  【網(wǎng)站聲明】本站部分內(nèi)容來源于互聯(lián)網(wǎng),旨在幫助大家更快的解決問題,如果有圖片或者內(nèi)容侵犯了您的權(quán)益,請聯(lián)系我們刪除處理,感謝您的支持!

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