Question

private void Include(IList<string> includes, Action action)
{
    if (includes != null)
    {
        foreach (var include in includes)
            action(<add include here>);
    }
}

我想称之为

this.Include(includes, _context.Cars.Include(<NEED TO PASS each include to here>));

这个想法是将每个include传递给方法。

Answer 1

如果您知道要传递的参数，请为该类型选择Action<T>。例如：

void LoopMethod (Action<int> code, int count) {
     for (int i = 0; i < count; i++) {
         code(i);
     }
}

如果要将参数传递给方法，请将方法设为通用：

void LoopMethod<T> (Action<T> code, int count, T paramater) {
     for (int i = 0; i < count; i++) {
         code(paramater);
     }
}

来电者代码：

Action<string> s = Console.WriteLine;
LoopMethod(s, 10, "Hello World");

更新。您的代码应如下所示：

private void Include(IList<string> includes, Action<string> action)
{
    if (includes != null)
    {
         foreach (var include in includes)
             action(include);
    }
}

public void test()
{
    Action<string> dg = (s) => {
        _context.Cars.Include(s);
    };
    this.Include(includes, dg);
}

Answer 2

您正在寻找带有参数的Action<T>。

Answer 3

肮脏的技巧：您也可以使用lambda表达式传递您想要的任何代码，包括带参数的调用。

this.Include(includes, () =>
{
    _context.Cars.Include(<parameters>);
});

如何在Action中传递参数？

3 个答案: