Question

我一直在尝试这个问题但是我仍然无法找到解决方案。任何帮助表示赞赏！感谢

问题 - 声明并创建一个数组来存储四个玩家的名字。调用数组pNames。

要求用户输入四个播放器的名称并将其名称存储在pNames数组中。

private DataTable GenerateTransposedTable(DataTable inputTable)
    {
        DataTable outputTable = new DataTable(inputTable.TableName);
        outputTable.Columns.Add(inputTable.Columns[0].ColumnName);


        foreach (DataRow inRow in inputTable.Rows)
        {
            string newColName = inRow[0].ToString();
            outputTable.Columns.Add(newColName);
        }


        for (int rCount = 1; rCount <= inputTable.Columns.Count - 1; rCount++)
        {
            DataRow newRow = outputTable.NewRow();


            newRow[0] = inputTable.Columns[rCount].ColumnName;
            for (int cCount = 0; cCount <= inputTable.Rows.Count - 1; cCount++)
            {
                string colValue = inputTable.Rows[cCount][rCount].ToString();
                newRow[cCount + 1] = colValue;
            }
            outputTable.Rows.Add(newRow);
        }

        return outputTable;
    }

输出问候消息以迎接每个玩家。

  Enter Name of Player 1 > Johnny
  Enter Name of Player 2 > Jackie
  Enter Name of Player 3 > Jessie
  Enter Name of Player 4 > Jeremy

答案：

Hello Johnny
Hello Jackie
Hello Jessie
Hello Jeremy

}

Answer 1

数组基于0索引。

在循环中，您使用的是1索引库。

更改

for(int i = 1; i <= pNames.length; i++) {

    System.out.print("Enter name of Player" + " " + i + " > ");

    pNames[i] = sc.nextLine();
}

到

for(int i = 1; i <= pNames.length; i++) {

    System.out.print("Enter name of Player" + " " + i + " > ");

    // The index of pNames must start from 0!
    pNames[i-1] = sc.nextLine();
}

更接近程序员推理模式的解决方案是使用从0开始的循环并更改System.out，如下所示：

// Loop start from 0 and goes to 4 excluded
for (int i = 0; i < pNames.length; i++) {

    // To print Enter name of Player 1 change i to (i + 1)
    System.out.print("Enter name of Player" + " " + (i + 1) + " > ");

    pNames[i] = sc.nextLine();
}

Answer 2

不是从1开始索引，而是需要将数组索引启动为0并在显示“输入播放器名称”时，只显示i的递增值1

 import java.util.Scanner;
  public class Q2 {
  public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);

    String [] pNames = new String[4];

    for(int i = 0; i < pNames.length; i++) {

        System.out.print("Enter name of Player" + " " + (i+1) + " > ");

        pNames[i] = sc.nextLine();
    }

    System.out.println("Hello" + " " + pNames[0]);
    System.out.println("Hello" + " " + pNames[1]);
    System.out.println("Hello" + " " + pNames[2]);
    System.out.println("Hello" + " " + pNames[3]);
}

Answer 3

请改为尝试：

从0开始，使用sc.Next();代替sc.nextLine();：

  import java.util.Scanner;
  public class Q2 {
  public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

String [] pNames = new String[4];

for(int i = 0; i < pNames.length; i++) {

    int a = i+1;
    System.out.print("Enter name of Player" + " " + a + " > ");

    pNames[i] = sc.next();
}

System.out.println("Hello" + " " + pNames[0]);
System.out.println("Hello" + " " + pNames[1]);
System.out.println("Hello" + " " + pNames[2]);
System.out.println("Hello" + " " + pNames[3]);
}

}

Answer 4

只需更改此行：

pNames[i] = sc.nextLine();

到此：

pNames[i-1] = sc.nextLine();

由于您正在访问1到4，因此您的数组实际上在0-3范围内，因此您需要减去一个。

在数组中输入和存储名称 - Java

4 个答案: