2010-06-10 96 views

回答

3

我會用一個Task

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

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

IronPython的腳本,然後在一個單獨的線程上運行。確保您的事件處理程序在更新GUI時使用適當的同步機制。

+0

我真的遇到了這個問題,並嘗試它,但注意到一個問題......如果ScriptSource.Execute()拋出異常,它不會被圍繞任務調用的try catch塊捕獲。它只是打破了這條線...有什麼辦法來處理這個例外嗎? – 2010-06-10 19:16:37

+0

我會將.Execute()調用移動到一個新函數中,並將try/catch放在該函數中,然後使用新函數ala'Task.Factory.StartNew(MyExecute)'或Task.Factory.StartNew ()=> MyExecute(...))'如果你需要關閉。 – 2010-06-10 19:47:42

3

您可以使用後臺工作器在單獨的線程上運行腳本。然後使用ProgressChanged和RunWorkerCompleted事件處理程序更新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); 
    }