Question

为了备份我的主linux目录，我尝试使用此命令，但现在出现错误：

var gulp = require('gulp');
gulp.task('default', function () {
  return gulp.src('./*.js') // read all
    .pipe(gulp.dest('/media/m/sdd2 ntfs/backup/home/'));
});
gulp.watch('./*.js', gulp.series('default'));

然后这个：

var gulp = require('gulp');
gulp.task('default', function () {
  return gulp.src(['./**/*.*', '!./**/*.js'])
    .pipe(gulp.dest('/media/m/sdd2 ntfs/backup/home/'));
});
gulp.watch([
  './**/*.*', 
  '!././**/*.js', 
  '!./.wine**/' ,
  '!./.npm/**/*.*'
], gulp.series('default'));

我读了一些酒，弄错了。排除葡萄酒可能是个好主意吗？或所有隐藏文件？

错误：

sudo gulp
[14:59:46] Using gulpfile /home/m/gulpfile.js
[14:59:46] Starting 'default'...
[14:59:47] The following tasks did not complete: default
[14:59:47] Did you forget to signal async completion?
events.js:174
      throw er; // Unhandled 'error' event
      ^

Error: ENOSPC: System limit for number of file watchers reached, watch '.npm/_cacache/content-v2/sha512/47/cb'
    at FSWatcher.start (internal/fs/watchers.js:165:26)
    at Object.watch (fs.js:1258:11)

gulp：达到文件查看器数量的系统限制

0 个答案: