如何在Gulp任务错误回调中不显示堆栈跟踪

Question

当我在一个任务中出错时:

gulp.task('foo', function(onDone) {
    onDone('It did not work');
});

我得到一个大的不相关的堆栈跟踪:

[13:10:25] 'foo' errored after 4.27 s
[13:10:25] Error: It did not work
    at formatError (C:npmnode_modulesgulpbingulp.js:169:10)
    at Gulp.<anonymous> (C:npmnode_modulesgulpbingulp.js:195:15)
    at emitOne (events.js:96:13)
    at Gulp.emit (events.js:188:7)
    at Gulp.Orchestrator._emitTaskDone (D:Codemy-projectnode_modulesorchestratorindex.js:264:8)
    at D:Codemy-projectnode_modulesorchestratorindex.js:275:23
    at finish (D:Codemy-projectnode_modulesorchestratorlibrunTask.js:21:8)
    at cb (D:Codemy-projectnode_modulesorchestratorlibrunTask.js:29:3)
    at D:Codemy-projectgulp-tasksinsert-docs.js:116:25
    at Request._callback (D:Codemy-projectgulpfile.js:112:17)

我怎么能不显示呢?我更喜欢使用内置的gulp功能，而不是第三方模块或console.error()。

Answer 1

Gulp检查传递给回调的Error对象上是否存在showStack属性。这可能是为了在gutil.PluginError上支持该选项而引入的，但是可以与任何Error对象一起使用:

function withError(msg) {
  var err = new Error(msg);
  err.showStack = false;
  return err;
}
gulp.task('foo', function(done) {
  done(withError('It did not work'));
});