BackgroundWorker 线程必须是 STA [英] BackgroundWorker thread must be STA

问题描述

我有一个BackgroundWorker来调用一个函数在BackgroundWorker _DoWork上做一个很长的过程，当函数发生错误时我会提示一个自定义的消息框:

I have a BackgroundWorker to call a function to do a long process at BackgroundWorker _DoWork, when error occur in the function I will prompt a customized messagebox:

 WPFMessageBoxResult result = WPFMessageBox.Show("Activation Fail", "Error!!", WPFMessageBoxButtons.OK, WPFMessageBoxImage.Error);

以下异常发生在 WPFMessageBoxResult 类:

The exception below happens at WPFMessageBoxResult class :

The calling thread must be STA, because many UI components require this. 

谢谢.

推荐答案

您不应尝试从后台线程与任何 UI 组件进行交互.

You should not try to interact with any UI components from a background thread.

一种方法是在 doWork 方法中捕获异常并将其分配给后台工作器的 result 属性，然后检查该结果是否为异常类型，或者如果您不将结果用于其他任何事情，则检查该结果是否为空.然后在 backgroundWorker_completed 事件中检查它.

One way could be to catch the exception in your doWork method and assign it to the backgroundworker's result property and then check if that result is a type of exception or not null if you are not using the result for anything else. then check for it in the backgroundWorker_completed event.

BackgroundWorker_DoWork(sender, )
{
    try
    {
       // do work        
    }
    catch (Exception ex)
    {
         BackgroundWorker w = sender as BackgroundWorker;
         if (w != null)
             w.Result = ex;
    }
}

然后

BackgroundWorker_Completed()
{
    if (s.Result != null && e.Result is Exception)
    {
       Exception ex = e.Result as Exception;
       // do something with ex
    }
}

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