无法读取 gulp 中未定义的属性 'watch'

Question

有人可以帮忙吗？我一直收到这个错误，我真的不知道还能做什么。当我在 cmd 中触发命令 gulp watch 时，我收到以下错误 TypeError: Cannot read property 'watch' of undefined

C:\Apache24\htdocs\Gulp>gulp watch
[10:37:07] Using gulpfile C:\Apache24\htdocs\Gulp\gulpfile.js
[10:37:07] Starting 'watch'...
[10:37:07] 'watch' errored after 4.16 ms
[10:37:07] TypeError: Cannot read property 'watch' of undefined
   at watch (C:\Apache24\htdocs\Gulp\gulpfile.js:42:10)
   at watch (C:\Apache24\htdocs\Gulp\node_modules\undertaker\lib\set-task.js:13:15)
   at bound (domain.js:422:14)
   at runBound (domain.js:435:12)
   at asyncRunner (C:\Apache24\htdocs\Gulp\node_modules\async-done\index.js:55:18)
   at processTicksAndRejections (internal/process/task_queues.js:75:11)

const {src, dest, series, gulp, parallel} = require('gulp');
var rename = require('gulp-rename');
var sass = require('gulp-sass');
var autoprefixer = require('gulp-autoprefixer');
var sourcemaps = require('gulp-sourcemaps');

var styleSRC = './src/scss/style.scss';
var styleDIST = './dist/css/';


var jsSRC = './src/js/script.js';
var jsDIST = './dist/js/';


function style() {
    "use strict";
    return src(styleSRC)
        .pipe(sourcemaps.init())
        .pipe(sass({
            errorLogToConsole: true,
            outputStyle: 'compressed'
        }))
        .on('error', console.error.bind(console))
        .pipe(autoprefixer({
            overrideBrowserslist: ["defaults"],
            cascade: false
        }))
        .pipe(rename({basename: 'style', suffix: '.min'}))
        .pipe(sourcemaps.write('./'))
        .pipe(dest(styleDIST));
}


function js() {
    "use strict";
    return src(jsSRC)
        .pipe(dest(jsDIST));
}

const watch = function () {
    "use strict";
    gulp.watch('./src/scss/**/*.scss', {usePolling: true}, gulp.series(style));
    gulp.watch('./src/js/**/*.js', {usePolling: true}, gulp.series(js));
};
exports.default = series(
    parallel(style, js),
    watch
);

exports.watch = watch;
exports.js = js;
exports.style = style;

Answer 1

试试这个，删除 const 并将其定义为函数。

function watch() {
    "use strict";
    gulp.watch('./src/scss/**/*.scss', {usePolling: true}, gulp.series(style));
    gulp.watch('./src/js/**/*.js', {usePolling: true}, gulp.series(js));
};

Answer 2

我认为你的解构有点错误。你有：

const {src, dest, series, gulp, parallel} = require('gulp');
//                          ^ this does not exist - hence the undefined property

我想你想用watch替换解构中的gulp，所以现在是：

const {src, dest, series, watch, parallel} = require('gulp');

并在您的 watch 函数中执行：

// renamed the function to avoid conflict
const watchTask = function () {
    "use strict";
    watch('./src/scss/**/*.scss', {usePolling: true}, series(style));
    watch('./src/js/**/*.js', {usePolling: true}, series(js));
};

exports.default = series(
    parallel(style, js),
    watchTask
);

exports.watch = watchTask;

无法读取 gulp 中未定义的属性 'watch'

Cannot read property 'watch' of undefined in gulp

javascript

node.js

gulp

无法读取 gulp 中未定义的 属性 'watch'

Cannot read property 'watch' of undefined in gulp

javascript

node.js

gulp

无法读取 gulp 中未定义的属性 'watch'