字符串格式化器可以用变量进行参数化吗?

时间:2015-09-29 12:00:37

标签: c# .net string string-formatting

实施例

以下是一个例子:

public class Program
{
    public static void Main(string[] args)
    {
        Console.WriteLine("Test {0, 10}", 1100);
        Console.WriteLine("Test {0, 10}", 2);
        Console.WriteLine("Test {0, 10}", 40);
    }
}

输出结果为:

Test       1100
Test          2
Test         40
Press any key to continue . . .

问题

是否可以在上面的示例变量中设置数字10

以下描述了意图,但没有编译,因为预期string,而不是int

public class Program
{
    public static void Main(string[] args)
    {
        int i = 10;
        Console.WriteLine("Test {0, i}", 1100);
        Console.WriteLine("Test {0, i}", 2);
        Console.WriteLine("Test {0, i}", 40);
    }
}

2 个答案:

答案 0 :(得分:6)

使用C#6,您可以使用string interpolation

Console.WriteLine($"Test {{0, {i}}}", 1100);
Console.WriteLine($"Test {{0, {i}}}", 2);
Console.WriteLine($"Test {{0, {i}}}", 40);

C#6中字符串插值的好处是它包括变量的编译时检查。为了使字符串插值工作,您需要在字符串前面加上美元符号($)。

没有字符串插值的另一个选项是:

int i = 10;
Console.WriteLine("Test {0, " + i + "}", 1100);
Console.WriteLine("Test {0, " + i + "}", 2);
Console.WriteLine("Test {0, " + i + "}", 40);

或者:

Console.WriteLine("Test " + 1100.ToString().PadLeft(i));
Console.WriteLine("Test " + 2.ToString().PadLeft(i));
Console.WriteLine("Test " + 40.ToString().PadLeft(i));

答案 1 :(得分:3)

一个简单的解决方案是:

public class Program
{
    public static void Main(string[] args)
    {
        int i = 10;
        Console.WriteLine("Test {0, " + i + "}", 1100);
        Console.WriteLine("Test {0, " + i + "}", 2);
        Console.WriteLine("Test {0, " + i + "}", 40);
    }
}