c# - 使用不同的 Try 语句和相同的 Catch 语句进行 DRY

Question

所以我在方法中有以下代码块:(所有变量都是本地变量)

// ...

try
{
    if (postXml != null)
        using (StreamWriter writer = new StreamWriter(req.GetRequestStream()))
            writer.Write(postXml.ToString());
}
catch (WebException ex)
{
    HttpWebResponse response = ex.Response as HttpWebResponse;
    if (response != null)
        result = HandleOtherResponse(response, out status);
    else result = HandleBadResponse(ex.ToString(), out status);
}
catch (Exception ex)
{
    result = HandleBadResponse(ex.ToString(), out status);
}

if (result == null)
{
    try
    {
        HttpWebResponse response = req.GetResponse() as HttpWebResponse;
        result = HandleOtherResponse(response, out status);
    }
    catch (WebException ex)
    {
        HttpWebResponse response = ex.Response as HttpWebResponse;
        if (response != null)
            result = HandleOtherResponse(response, out status);
        else result = HandleBadResponse(ex.ToString(), out status);
    }
    catch (Exception ex)
    {
        result = HandleBadResponse(ex.ToString(), out status);
    }
}

// ...

可以看到，两条try语句是不同的，但是两组catch语句是完全一样的。我一直在试图想出一种可能不会在这里重复我自己的方法，但我还没有真正想到一种不会显着变慢或看起来很糟糕的方法。想知道是否有人有任何想法。

Answer 1

一种方法是编写一个“安全的”调用方法并将一个函数传递给它:

public T SafeInvocation(Func<T> myMethod)
{
    T result = default(T);

    try
    {
        // Invoke method
        result = myMethod();
    }
    catch
    {
        // Do your common catch here
    }

    return result;
}

为 Action 构建一个额外的重载，这样您就不需要返回类型了。然后你可以在别处调用它，将方法作为参数传递给你的方法(Inception？):

SafeInvocation(() => 
{
    if (postXml != null)
        using (StreamWriter writer = new StreamWriter(req.GetRequestStream()))
            writer.Write(postXml.ToString());
}