我不应该像我想的那样抓住异步的东西,因为这让我头疼了几个小时。
我想要完成的是:
在我看来,我的WPF按钮有以下回调
private void Button_Click(object sender, RoutedEventArgs e)
{
GeneratorButton.IsEnabled = false;
ResultMsg.Visibility = Visibility.Visible; //default text of "Fetching data..."
string tickersString = Tickers.Text;
if (!string.IsNullOrEmpty(tickersString))
{
tickersString = tickersString.ToUpper();
string[] tickers = tickersString.Split(',', ' ');
var stockGeneratorTask = Task.Factory.StartNew<bool>(() => GetStockDataAndGenerateCSV(tickers).Result);
var allTasks = new Task[] { stockGeneratorTask };
var tasksHandler = Task.Factory.ContinueWhenAll(allTasks, completedTasks =>
{
return stockGeneratorTask.Result;
});
if (tasksHandler.Result)
{
ResultMsg.Foreground = Brushes.Green;
ResultMsg.Text = "Your report was successfully generated.";
}
else
{
ResultMsg.Text = "There was an error while generating your report. Please try again later.";
ResultMsg.Foreground = Brushes.Red;
}
}
GeneratorButton.IsEnabled = true;
}
我的期望是,只要用户点击按钮,他们就会看到由于可见性属性开关而导致的“获取数据...”的默认文本。根据我的理解,正在发生的事情是程序直接进入异步任务,当它返回时它会显示成功(或错误)消息,但它会完全跳过“获取数据......”,除非我发表评论从stockGeneratorTask声明中删除代码(没有异步调用发生)。
现在,如果在异步任务被触发之前发生了这种可见性切换,那么在实际进行异步调用之前,我不应该看到这种变化吗?为什么我回来后才看到它?
事件处理程序是允许async void
的一个例外。考虑使事件处理程序始终异步,而不是混合阻塞调用,如.Result
,这可能导致死锁。假设GetStockDataAndGenerateCSV
返回Task<bool>
然后它应该等待。
private async void Button_Click(object sender, RoutedEventArgs e) {
GeneratorButton.IsEnabled = false;
ResultMsg.Visibility = Visibility.Visible; //default text of "Fetching data..."
var tickersString = Tickers.Text;
if (!string.IsNullOrEmpty(tickersString)) {
tickersString = tickersString.ToUpper();
string[] tickers = tickersString.Split(',', ' ');
if (await GetStockDataAndGenerateCSV(tickers)) {
ResultMsg.Foreground = Brushes.Green;
ResultMsg.Text = "Your report was successfully generated.";
} else {
ResultMsg.Foreground = Brushes.Red;
ResultMsg.Text = "There was an error while generating your report. Please try again later.";
}
}
GeneratorButton.IsEnabled = true;
}