gulp task for(... in ...) 创建正确吗？

Question

我做错了什么？

gulpfile.js

'use strict';

// Modules & Plugins
var gulp = require('gulp');
const image = require('gulp-image');
var $ = require('gulp-load-plugins')();

let directories = [
  '2006/Art1',
  '2006/Art2',
  '2010/Art1'
];

function gulpTask(directoryName) {
  gulp.task('blogmotionAllMedia', function () {
    return gulp
      .src('src/content/' + directoryName + '/media/*.{jpg,jpeg,png,gif}')
      .pipe($.responsive({
        '*.jpg': [
          { width: 400, rename: { suffix: '-4' } },
          { rename: { suffix: '-6' } }
        ],
      }))
      .pipe(image({
        jpegRecompress: true
      }))
      .pipe(gulp.dest('dist/content/' + directoryName + '/media'))
  });
}

var directory;

for (directory in directories) {
  gulpTask(directory);
}

运行任务 npx gulp gulpTask

意外

我收到以下错误消息：

Task never defined: gulpTask

我仍然依附于我的past question 和来自Miles 的answer（想法）

Answer 1

首先，gulp 任务应注册为您要调用它的名称gulp.task('blogmotionAllMedia', ... 应为gulp.task('gulpTask', ...。 此外，我认为您应该遍历任务内部的目录。

此外，我认为您需要使用 merge-stream by const merge = require('merge-stream') 和 return merge(directories.map(singleDirectoryGulpTask)); 合并结果流

结果应如下所示： （免责声明：多年来没有做过 gulp 的东西。仍然希望它有所帮助）

'use strict';

// Modules & Plugins
var gulp = require('gulp');
const image = require('gulp-image');
var $ = require('gulp-load-plugins')();
const merge = require('merge-stream')

let directories = [
    '2006/Art1',
    '2006/Art2',
    '2010/Art1'
];

gulp.task('gulpTask', function () {
    function singleDirectoryGulpTask(directoryName) {
        return gulp
            .src('src/content/' + directoryName + '/media/*.{jpg,jpeg,png,gif}')
            .pipe($.responsive({
                '*.jpg': [
                    { width: 400, rename: { suffix: '-4' } },
                    { rename: { suffix: '-6' } }
                ],
            }))
            .pipe(image({
                jpegRecompress: true
            }))
            .pipe(gulp.dest('dist/content/' + directoryName + '/media'));
    }

    return merge(directories.map(singleDirectoryGulpTask));
});

Answer 2

变化：

for (directory in directories) {
  gulpTask(directory);
}

到：

for (directory of directories) {
  gulpTask(directory);
}