How do I prevent gulp-notify from breaking gulp-wa

2020-06-28 01:02发布

I am using the gulp-notify plugin. This is an example of how I'm implementing it in a gulpfile.js ( You can see I'm using gutil and livereload as well. I don't know if they play any factors, but let me know if they do.)

gulp.task('js', function() {
  return gulp.src('./dev/scripts/*.coffee')
    .pipe(coffee({bare: true}).on('error', gutil.log))
    .pipe( gulp.dest('./build/js'))
    .pipe(notify('Coffeescript compile successful'))
    .pipe(livereload(server));
});

This plugin works on OS X and Linux. On Windows, which doesn't have a notification feature, it returns an error and breaks the gulp-watch plugin. This is an example of how I have gulp-watch setup:

gulp.task('watch', function () {
  server.listen(35729, function (err) {
    if (err) {
      return console.log(err);
    }
    gulp.watch('./dev/scripts/*.coffee',['js']);
  });
});

So, I read in the documentation the gulp-pipe plugin can help me not break gulp-watch when on Windows, but I can't find an example of how I could use it in a setup like this.

1条回答
你好瞎i
2楼-- · 2020-06-28 01:54

You can use the gulp-if plugin in combination with the os node module to determine if you are on Windows, then exclude gulp-notify, like so:

var _if = require('gulp-if');

//...

// From http://stackoverflow.com/questions/8683895/variable-to-detect-operating-system-in-node-scripts
var isWindows = /^win/.test(require('os').platform());

//...

     // use like so:
    .pipe(_if(!isWindows, notify('Coffeescript compile successful')))
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