Gulp 如何修复';必须指定任务函数';狼吞虎咽?

Gulp 如何修复';必须指定任务函数';狼吞虎咽?,gulp,gulp-watch,Gulp,Gulp Watch,我正在研究GULP,我想在修改CSS时自动更新页面 const gulp = require("gulp"); const jshint = require("gulp-jshint"); const changed = require("gulp-changed"); const watch = require("gulp-watch"); const rename = require("gulp-rename"); const minifyCSS = require("gulp-uglif

我正在研究GULP,我想在修改CSS时自动更新页面


const gulp = require("gulp");
const jshint = require("gulp-jshint");
const changed = require("gulp-changed");
const watch = require("gulp-watch");
const rename = require("gulp-rename");
const minifyCSS = require("gulp-uglifycss");
const sass = require("gulp-sass");
const cssImport = require("gulp-cssimport");

gulp.task("changed", function() {
  return gulp
    .src("src/css/*.css")
    .pipe(changed("public/assets/css/"))
    .pipe(cssImport())
    .pipe(sass())
    .pipe(minifyCSS())
    .pipe(rename({ extname: ".min.css" }))
    .pipe(gulp.dest("public/assets/css/"));
});

gulp.task("jshint", function() {
  gulp
    .src("src/css/*.css")
    .pipe(jshint())
    .pipe(jshint.reporter("default"));
});

gulp.task("watch", function() {
  watch("src/css/*.css", ["changed"]);
});

gulp.task("default", ["jshint", "watch"]);


我试图使用“gulp watch”,但它给出了错误“必须在上面的代码中指定任务函数”

问题是我试图观看的是gulp的旧版本,而不是4.0.0版

所以我把代码改成了工作代码,看起来是这样的:

const gulp = require("gulp");
const rename = require("gulp-rename");
const minifyJS = require("gulp-uglify");
const minifyCSS = require("gulp-uglifycss");
const sass = require("gulp-sass");
const babel = require("gulp-babel");
const cssImport = require("gulp-cssimport");

gulp.task(
  "base",
  gulp.series(function() {
    return gulp.src("src/templates/*.html").pipe(gulp.dest("public/"));
  })
);

gulp.task(
  "javascript",
  gulp.series(function() {
    return gulp
      .src("src/js/*.js")
      .pipe(babel({ presets: ["@babel/env"] }))
      .pipe(minifyJS())
      .pipe(rename({ extname: ".min.js" }))
      .pipe(gulp.dest("public/assets/js/"));
  })
);

gulp.task(
  "css",
  gulp.series(function() {
    return gulp
      .src("src/css/*.css")
      .pipe(cssImport())
      .pipe(sass())
      .pipe(minifyCSS())
      .pipe(rename({ extname: ".min.css" }))
      .pipe(gulp.dest("public/assets/css/"));
  })
);

gulp.task(
  "watch",
  gulp.series(function() {
    gulp.watch("src/templates/*.html", gulp.parallel(["base"]));
    gulp.watch("src/js/*.js", gulp.parallel(["javascript"]));
    gulp.watch("src/css/*.css", gulp.parallel(["css"]));
  })
);

gulp.task(
  "default",
  gulp.series(gulp.parallel("base", "javascript", "css", "watch"))
);

任务的名称是“watch”,所以请尝试使用它,而不是“gulp watch”。但是我已经使用了任务“watch”,请查看我的默认任务每当您有一个名为“default”的任务时,您甚至不需要使用该名称。您只需使用“gulp”运行默认任务。关于“监视”任务,@slotomo是对的。您只需运行“GulpWatch”,而不是“GulpWatch”。