GLUP V4.0.0'gulp start'不是一个功能 [英] Glup V4.0.0 'gulp start' is not a function

问题描述

我在gulpfile.js下面有这个.当我使用"gulp start"运行应用程序时，它表明启动不是一个功能. 我正在使用V4.0.0的Glup-cli版本

I have this below gulpfile.js. When i run the app using 'gulp start' then it is showing start is not a function. Glup-cli version i'm using V4.0.0

const gulp = require('gulp');
const concat = require('gulp-concat');
const browserSync = require('browser-sync').create();

const scripts = require('./scripts');
const styles = require('./styles');

// Some pointless comments for our project.

var devMode = false;

gulp.task('css', function() {
    gulp.src(styles)
        .pipe(concat('main.css'))
        .pipe(gulp.dest('dist/css'))
        .pipe(browserSync.reload({
            stream: true
        }));
});

gulp.task('js', function() {
    gulp.src(scripts)
        .pipe(concat('scripts.js'))
        .pipe(gulp.dest('./dist/js'))
        .pipe(browserSync.reload({
            stream: true
        }));
});

gulp.task('html', function() {
    return gulp.src('./src/templates/**/*.html')
        .pipe(gulp.dest('./dist/'))
        .pipe(browserSync.reload({
            stream: true
        }));
});

gulp.task('build', function() {
    gulp.start(['css', 'js', 'html'])
});

gulp.task('browser-sync', function() {
    browserSync.init(null, {
        open: false,
        server: {
            baseDir: 'dist',
        }
    });
});

gulp.task('start', function() {
    devMode = true;
    gulp.start(['build', 'browser-sync']);
    gulp.watch(['./src/css/**/*.css'], ['css']);
    gulp.watch(['./src/js/**/*.js'], ['js']);
    gulp.watch(['./src/templates/**/*.html'], ['html']);
});

推荐答案

gulp.start在v4中已被弃用.根据您的需求，您可以改用gulp.series或gulp.parallel.

gulp.start has been deprecated in v4. Depending on your needs, you can use gulp.series or gulp.parallel instead.

- gulp.task('start', function() {
-   devMode = true;
-   gulp.start(['build', 'browser-sync']);
+ gulp.task('start', gulp.series('build', 'browser-sync'), function(done) {
+   devMode = true;
    gulp.watch(['./src/css/**/*.css'], ['css']);
    gulp.watch(['./src/js/**/*.js'], ['js']);
    gulp.watch(['./src/templates/**/*.html'], ['html']);
  });

这个问题可能是这个问题的重复，但是既然那个问题还没有被接受的答案，我就回覆马克的答案.

This question is probably a duplicated of this one, but since that question hasn't got an accepted answer I'll just echo Mark's answer.

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