Gulp:如何将参数从手表传递到任务 [英] Gulp: how to pass parameters from watch to tasks

问题描述

使用 gulp 你经常会看到这样的模式:

With gulp you often see patterns like this:

gulp.watch('src/*.jade',['templates']);

gulp.task('templates', function() {
  return gulp.src('src/*.jade')
    .pipe(jade({
      pretty: true
    }))
    .pipe(gulp.dest('dist/'))
    .pipe( livereload( server ));
});

这是否实际上将监视的文件传递到模板任务中?这些如何覆盖/扩展/过滤 src'ed 任务?

Does this actually pass the watch'ed files into the templates task? How do these overwrite/extend/filter the src'ed tasks?

推荐答案

我前段时间也有同样的问题，经过一番挖掘得出以下结论.

I had the same question some time ago and came to the following conclusion after digging for a bit.

gulp.watch 是一个发出 change 事件的 eventEmitter，因此您可以这样做:

gulp.watch is an eventEmitter that emits a change event, and so you can do this:

var watcher = gulp.watch('src/*.jade',['templates']);

watcher.on('change', function(f) {
  console.log('Change Event:', f);
});

你会看到这个:

Change Event: { type: 'changed',
  path: '/Users/developer/Sites/stackoverflow/src/touch.jade' }

这个信息大概可以通过它的任务函数或 gulp.src 的行为传递给 template 任务.

This information could presumably be passed to the template task either via its task function, or the behavior of gulp.src.

任务函数本身只能接收回调(https://github.com/gulpjs/gulp/blob/master/docs/API.md#fn)并且无法接收有关乙烯基文件的任何信息(https://github.com/wearefractal/vinyl-fs).

The task function itself can only receive a callback (https://github.com/gulpjs/gulp/blob/master/docs/API.md#fn) and cannot receive any information about vinyl files (https://github.com/wearefractal/vinyl-fs) that are used by gulp.

启动任务的源(在这种情况下为 .watch，或 gulp 命令行)对 gulp.src('src-glob', [options] 的行为没有影响).'src-glob' 是一个字符串(或字符串数​​组)和 options (https://github.com/isaacs/node-glob#options) 没有任何文件更改.

The source starting a task (.watch in this case, or gulp command line) has no effect on the behavior of gulp.src('src-glob', [options]). 'src-glob' is a string (or array of strings) and options (https://github.com/isaacs/node-glob#options) has nothing about any file changes.

因此，我看不到 .watch 可以直接影响它触发的任务的行为.

Hence, I don't see any way in which .watch could directly affect the behavior of a task it triggers.

如果您只想处理更改的文件，可以使用 gulp-changed (https://www.npmjs.com/package/gulp-changed) 如果你想使用 gulp.watch，或者你冷使用 gulp-watch.

If you want to process only the changed files, you can use gulp-changed (https://www.npmjs.com/package/gulp-changed) if you want to use gulp.watch, or you cold use gulp-watch.

或者，您也可以这样做:

Alternatively, you could do this as well:

var gulp = require('gulp');
var jade = require('gulp-jade');
var livereload = require('gulp-livereload');

gulp.watch('src/*.jade', function(event){
  template(event.path);
});

gulp.task('templates', function() {
  template('src/*.jade');
});

function template(files) {
  return gulp.src(files)
    .pipe(jade({
      pretty: true
    }))
    .pipe(gulp.dest('dist/'))
}

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