2010-06-10 64 views

回答

3

我会用一个Task

ScriptEngine engine = ...; 
// initialize your script and events 

Task.Factory.StartNew(() => engine.Execute(...)); 

IronPython的脚本,然后在一个单独的线程上运行。确保您的事件处理程序在更新GUI时使用适当的同步机制。

+0

我真的遇到了这个问题,并尝试它,但注意到一个问题......如果ScriptSource.Execute()抛出异常,它不会被围绕任务调用的try catch块捕获。它只是打破了这条线...有什么办法来处理这个例外吗? – 2010-06-10 19:16:37

+0

我会将.Execute()调用移动到一个新函数中,并将try/catch放在该函数中,然后使用新函数ala'Task.Factory.StartNew(MyExecute)'或Task.Factory.StartNew ()=> MyExecute(...))'如果你需要关闭。 – 2010-06-10 19:47:42

3

您可以使用后台工作器在单独的线程上运行脚本。然后使用ProgressChanged和RunWorkerCompleted事件处理程序更新ui。

BackgroundWorker worker; 
    private void RunScriptBackground() 
    { 
    string path = "c:\\myscript.py"; 
    if (File.Exists(path)) 
    {    
     worker = new BackgroundWorker(); 
     worker.DoWork += new DoWorkEventHandler(bw_DoWork); 
     worker.ProgressChanged += new ProgressChangedEventHandler(bw_ProgressChanged); 
     worker.RunWorkerCompleted += new RunWorkerCompletedEventHandler(bw_RunWorkerCompleted); 
     worker.RunWorkerAsync(); 
    } 
    } 

    private void bw_RunWorkerCompleted(object sender, RunWorkerCompletedEventArgs e) 
    { 
    // handle completion here 
    } 

    private void bw_ProgressChanged(object sender, ProgressChangedEventArgs e) 
    { 
    // handle progress updates here 
    } 

    private void bw_DoWork(object sender, DoWorkEventArgs e) 
    { 
    // following assumes you have setup IPy engine and scope already 
    ScriptSource source = engine.CreateScriptSourceFromFile(path); 
    var result = source.Execute(scope); 
    }