Gulp错误:监视任务必须是一个函数 [英] Gulp error: watch task has to be a function

问题描述

这是我的gulpfile:

Here is my gulpfile:

// Modules & Plugins
var gulp = require('gulp');
var concat = require('gulp-concat');
var myth = require('gulp-myth');
var uglify = require('gulp-uglify');
var jshint = require('gulp-jshint');
var imagemin = require('gulp-imagemin');

// Styles Task
gulp.task('styles', function() {
    return gulp.src('app/css/*.css')
        .pipe(concat('all.css'))
        .pipe(myth())
        .pipe(gulp.dest('dist'));
});

// Scripts Task
gulp.task('scripts', function() {
    return gulp.src('app/js/*.js')
        .pipe(jshint())
        .pipe(jshint.reporter('default'))
        .pipe(concat('all.js'))
        .pipe(uglify())
        .pipe(gulp.dest('dist'));
});

// Images Task
gulp.task('images', function() {
    return gulp.src('app/img/*')
        .pipe(imagemin())
        .pipe(gulp.dest('dist/img'));
});

// Watch Task
gulp.task('watch', function() {
    gulp.watch('app/css/*.css', 'styles');
    gulp.watch('app/js/*.js', 'scripts');
    gulp.watch('app/img/*', 'images');
});

// Default Task
gulp.task('default', gulp.parallel('styles', 'scripts', 'images', 'watch'));

如果我单独运行images，scripts或css任务，它将起作用.我必须在任务中添加return-这不在本书中，但谷歌搜索显示这是必需的.

If I run the images, scripts or css task alone it works. I had to add the return in the tasks - this wasn't in the book but googling showed me this was required.

我遇到的问题是default任务错误:

The problem I have is that the default task errors:

[18:41:59] Error: watching app/css/*.css: watch task has to be a function (optionally generated by using gulp.parallel or gulp.series)
    at Gulp.watch (/media/sf_VM_Shared_Dev/webdevadvlocal/gulp/public_html/gulp-book/node_modules/gulp/index.js:28:11)
    at /media/sf_VM_Shared_Dev/webdevadvlocal/gulp/public_html/gulp-book/gulpfile.js:36:10
    at taskWrapper (/media/sf_VM_Shared_Dev/webdevadvlocal/gulp/public_html/gulp-book/node_modules/undertaker/lib/set-task.js:13:15)
    at bound (domain.js:287:14)
    at runBound (domain.js:300:12)
    at asyncRunner (/media/sf_VM_Shared_Dev/webdevadvlocal/gulp/public_html/gulp-book/node_modules/async-done/index.js:36:18)
    at nextTickCallbackWith0Args (node.js:419:9)
    at process._tickCallback (node.js:348:13)
    at Function.Module.runMain (module.js:444:11)
    at startup (node.js:136:18)

我认为是因为监视任务中也没有return.错误消息也不清楚-至少对我来说.我尝试在最后一个gulp.watch()之后添加一个return，但是那也不起作用.

I think it is because there is also no return in the watch task. Also the error message isn't clear - at least to me. I tried adding a return after the last gulp.watch() but that didn't work either.

推荐答案

在gulp 3.x中，您可以像这样将任务的名称传递给gulp.watch():

In gulp 3.x you could just pass the name of a task to gulp.watch() like this:

gulp.task('watch', function() {
  gulp.watch('app/css/*.css', ['styles']);
  gulp.watch('app/js/*.js', ['scripts']);
  gulp.watch('app/img/*', ['images']);
});

在gulp 4.x中不再是这种情况.您必须传递功能.在gulp 4.x中执行此操作的惯用方式是仅使用一个任务名称传递gulp.series()调用.这将返回一个仅执行指定任务的函数:

In gulp 4.x this is no longer the case. You have to pass a function. The customary way of doing this in gulp 4.x is to pass a gulp.series() invocation with only one task name. This returns a function that only executes the specified task:

gulp.task('watch', function() {
  gulp.watch('app/css/*.css', gulp.series('styles'));
  gulp.watch('app/js/*.js', gulp.series('scripts'));
  gulp.watch('app/img/*', gulp.series('images'));
});

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