Gulp Watch With Gulp Pug观察错误

Question

我正在尝试设置我的仓库来观看巴格的更改，但是我会收到以下错误，我不知道会发生什么：

events.js:141
      throw er; // Unhandled 'error' event
      ^
SyntaxError: Unexpected token (19:0)
    at Parser.pp$4.raise (/../node_modules/acorn/dist/acorn.js:2221:15)
    at Parser.pp.unexpected (/../node_modules/acorn/dist/acorn.js:603:10)
    at Parser.pp$3.parseExprAtom (/../node_modules/acorn/dist/acorn.js:1822:12)
    at Parser.pp$3.parseExprSubscripts (/../node_modules/acorn/dist/acorn.js:1715:21)
    at Parser.pp$3.parseMaybeUnary (/../node_modules/acorn/dist/acorn.js:1692:19)
    at Parser.pp$3.parseExprOps (/../node_modules/acorn/dist/acorn.js:1637:21)
    at Parser.pp$3.parseMaybeConditional (/../node_modules/acorn/dist/acorn.js:1620:21)
    at Parser.pp$3.parseMaybeAssign (/../node_modules/acorn/dist/acorn.js:1597:21)
    at Parser.pp$1.parseVar (/../node_modules/acorn/dist/acorn.js:1034:28)
    at Parser.pp$1.parseVarStatement (/../node_modules/acorn/dist/acorn.js:917:10)

我的gulpfile.js看起来像这样：

const gulp = require('gulp');
const pug = require('gulp-pug');    
gulp.task('pug', () =>
  gulp.src('pug/*.pug')
    .pipe(pug({ pretty: false }))
    .pipe(gulp.dest('.'))    
);
gulp.task('watch', function (){
  gulp.watch('pug/*.pug', ['pug']);
});

Answer 1

问题可能是.pipe(gulp.dest('.'))您应该指定实际文件夹名称。例如.pipe(gulp.dest('folderName'))/ .pipe(gulp.dest([__dirname])

另外将return语句添加到gulp.src('pug/*.pug')所以看起来像

return gulp.src('pug/*.pug')避免包含return语句的错误。

您可以编辑您的问题以包含哈巴狗文件吗？