Question

代码：

            string[] X = Regex.Split(X, ",");
            string[] A = Regex.Split(A, ",");
            string[] B = Regex.Split(B, ",");
            string[] C= Regex.Split(C, ",");
            string[] D = Regex.Split(D, ",");



            for (int i = 0; i < splitfirsts.Length; i++)
            {
                Console.WriteLine("{0}{1}{2}{3}{4}",
                      X[i], A[i], B[i], C[i], D[i]);               
            }

string []中的信息：

X = 1,2,3

A = 4,5,6

B = 7,8,9

C =亚当，马克，汤姆

D = 16,21,88

结果将返回：

147Adam16

258Mark21

369Tom88

我想要实现的目标：

string newstring = 147Adam16258Mark21369Tom88

无论如何，通过一个简单的解决方案实现它？

Answer 1

您无需使用字符串构建器即可实现此目的。如果您只想输出147Adam16258Mark21369Tom88，则可以使用Console.Write。试试这个：

for (int i = 0; i < splitfirsts.Length; i++)
        {

            Console.Write("{0}{1}{2}{3}{4}",
                      X[i], A[i], B[i], C[i], D[i]);
        }

但是，如果您想存储它，可以使用StringBuilder：

StringBuilder sb = new StringBuilder();

        for (int i = 0; i < splitfirsts.Length; i++)
        {
            sb.AppendFormat("{0}{1}{2}{3}{4}",
              X[i], A[i], B[i], C[i], D[i]);

        }

Console.Write(sb);

希望它有所帮助！

Answer 2

string[] X = Regex.Split(X, ",");
string[] A = Regex.Split(A, ",");
string[] B = Regex.Split(B, ",");
string[] C= Regex.Split(C, ",");
string[] D = Regex.Split(D, ",");

StringBuilder sb = new StringBuilder();

for (int i = 0; i < splitfirsts.Length; i++)
{ 
  sb.AppendFormat("{0}{1}{2}{3}{4}",
          X[i], A[i], B[i], C[i], D[i]);             
}

Console.WriteLine(sb.ToString());

Array Indexer to string builder

2 个答案: