flutter取消Future.delayed

前言

学习flutter，其中用Futere.delayed延时1s跳转到其他页面的时 如果没跳转前离开当前页面就会报

Future.delayed(const Duration(seconds: 1), () {
  Navigator.of(context).pushNamed("");
});
//报错
This widget has been unmounted, so the State no longer has a context (and should be considered defunct).
    Consider canceling any active work during "dispose" or using the "mounted" getter to determine if the State is still active.

问题分析

说明其实很清楚了，就是widget已经unmounted 不能跳转，我们需要在dispose取消下Future，所以需要看下delayed的源码：

  factory Future.delayed(Duration duration, [FutureOr<T> computation()?]) {
    _Future<T> result = new _Future<T>();
    new Timer(duration, () {
      if (computation == null) {
        result._complete(null as T);
      } else {
        try {
          result._complete(computation());
        } catch (e, s) {
          _completeWithErrorCallback(result, e, s);
        }
      }
    });
    return result;
  }

发现其实内部用是Timer实现，返回的_Future并找到对应方法取消，干脆直接用Timer实现好了，注意还是需要dispose取消

    _timer = Timer(const Duration(seconds: 1), () {
      Navigator.of(context).pushNamed("");
    });
    
	  @override
	  void dispose() {
	    super.dispose();
	    _timer?.cancel();
	  }