无法读取未定义的“表”属性

Question

有人能帮忙吗？我不断地收到这个错误，我真的不知道还能做些什么。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手表10:37:07使用C:\Apache24\htdocs\Gulp\gulpfile.js 10:37:07开始“手表”10:37:07‘手表’在4.16毫秒10:37:07后出现错误，TypeError:无法读取未定义的手表(C:\Apache24\htdocs\Gulp\gulpfile.js:42:10) at watch的属性(C：\Apache24\htdocs\Gulp\gulpfile.js:42:10)在runBound (domain.js:422:14) at runBound (domain.js:435:12) at asyncRunner (C：\Apache24 24\htdocs\)Gulp\node_modules\异步完成\index.js:55:18)在processTicksAndRejections (内部/process/task_Quees.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

我想你的破坏有点错了。你有：

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

我认为您想用watch替换销毁过程中的watch，所以现在是：

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

在您的watch函数中，请执行：

// 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;

Answer 2

尝试一下，删除const并将其定义为一个函数。

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