Question

public class A6{

    public static void main(String[] args){
        //String personInfo[][]=new String[][] { 
        String personInfo[][]={
                {"Anderson",  "Varejao",     "1125"},
                {"Giorgi",  "Tevzadze",      "1125"},
                {"Will",      "Cherry",      "1225"},
                {"Will",      "Iams",       "12245"},
                {"Lebron",    "James",       "6025"},
                {"Kevin",     "Love",        "2525"},
                {"Kyrie",     "Livings",      "454"},
                {"Kyrie",     "Botti",       "4544"},
                {"Chris",     "Mauer",        "425"},
                {"Mot",       "Daniel",       "125"},
                {"Viktor",    "Muller",       "145"},
                {"Kiamran",   "Chris",       "1405"},
                {"Zenia",     "Vaitehovic",  "1025"},
                {"Marija",    "Grabauskaite","1471"},
                {"Milda",    "Grabauskaite", "1000"},
                {"Dion",      "Waiters",     "625" },
                {"Dion",      "Malborg",    "6250" }
        };


        System.out.println("Peoples on the list:");
        for(int i = 0; i < personInfo.length; i++) {

            for(int j = 0; j < personInfo[i].length; j++) {
                System.out.print(" ");
                System.out.print(personInfo[i][j]);
            }

            System.out.println();
        }
        System.out.println("----------------------");

        //---------v XXX  - how many people have the first name XXX----------------

        String b = "v";

        if(args[0].equals(b))

        {

            System.out.println("Persons with entered name:");

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

                for(int j = 0; j < personInfo[i].length; j++) {

                    if(personInfo[i][j].equals(args[1]))
                    {
                        System.out.println(args[1]);
                    }
                }
            }
            System.out.println();

        }
        //------vp XXX YYY - what is the salary of the person XXX YYY------
    }
}

我开始做新的自我控制练习。这里我有二维String数组，它包含这些信息{＆＃34; name＆＃34;，＆＃34; surname＆＃34;，＆＃34; salary＆＃34;}。我已经完成了我的练习的一部分，它计算了我输入命令行的名字的人数。现在我需要编写一个代码例如：我进入命令行Anderson Varejao并希望得到结果1125.换句话说，我输入一个姓名，并希望从personInfo String数组获得某个人的工资。我怎样才能达到并领取薪水？

Answer 1

for(int i = 0; i < personInfo.length; i++) {
    if(personInfo[i][0].equals(args[1]) &&
            personInfo[i][1].equals(args[2]) {
        System.out.println(personInfo[i][2]);
    }
}
System.out.println();

Answer 2

换句话说，我输入一个姓名，并希望得到一份工资 personInfo字符串数组中的某些人。我怎样才能到达并接受一个人的薪水？

名称存储在第一个索引中，工资存储在第二个维度的第三个索引中因此，您可以定义依赖于此的findSalary()方法：

public String findSalary(String personInfo[][], String name)
  for(int i = 0; i < personInfo.length; i++) {      
      if (name.equals(personInfo[i][0])){
           return personInfo[i][2];
      }        
  }
   return null;
}

并调用：

String personInfo[][]={ ...};
String salary = findSalary(personInfo, "john");

请注意，使用String来表示数字事物可能不是最好的事情。

Answer 3

按空格分割搜索条件并将结果存储在数组中，例如：result [2]

检查[i] [j] =结果[0] firstpart＆amp;＆amp; checka [i] [j + 1] =结果[1]后半部分然后打印预期值

Answer 4

如果您仍想使用这种方式，那么这就是您想要的

for (int i = 0; i < personInfo.length; i++) {
    if(personInfo[i][0] == args[0] && personInfo[i][1] == args[1]) {
       System.out.println(personInfo[i][2]);
    }
}

注意输入以空格分隔的名称和姓氏。
但是如果你想要一个好的解决方案，你应该使用面向对象的强大功能创建一个可以保存人员信息的课程喜欢：

class Person {
    String name = "";
    String surName = "";
    String salary = "";
    //setters and getters here
}

Answer 5

关于如何在数组中定位元素，因为工资总是在personInfo [2]中，我只是一次迭代一行数据并检查firstName是否对应于peopleinfo中的元素[ 0]＆amp;＆amp; lName对应于peopleinfo [1]。看起来像这样：

static Scanner sc = new Scanner(System.in);

static String personInfo[][] = {
    {"Anderson", "Varejao", "1125"},
    {"Giorgi", "Tevzadze", "1125"},
    {"Will", "Cherry", "1225"},
    {"Will", "Iams", "12245"},
    {"Lebron", "James", "6025"},
    {"Kevin", "Love", "2525"},
    {"Kyrie", "Livings", "454"},
    {"Kyrie", "Botti", "4544"},
    {"Chris", "Mauer", "425"},
    {"Mot", "Daniel", "125"},
    {"Viktor", "Muller", "145"},
    {"Kiamran", "Chris", "1405"},
    {"Zenia", "Vaitehovic", "1025"},
    {"Marija", "Grabauskaite", "1471"},
    {"Milda", "Grabauskaite", "1000"},
    {"Dion", "Waiters", "625"},
    {"Dion", "Malborg", "6250"}
};

public static void main(String[] args) {

    System.out.print("Enter First Name: ");
    String fName = sc.nextLine();
    System.out.print("Enter Last Name: ");
    String lName = sc.nextLine();
    System.out.println(fName + " " + lName + "' Salary: " + salaryLookUp(personInfo, fName, lName));

}

static String salaryLookUp(String[][] array, String fName, String lName) {
    String salary = "";

    for (String[] row : personInfo) {
        if (row[0].equalsIgnoreCase(fName) && row[1].equalsIgnoreCase(lName)) {
            salary = row[2];
            break;
        } else {
            salary = "Not Found";
        }
    }
    return salary;
}

Answer 6

这是用C＃编写的，希望它有用...

string[,] personInfo = new string[,] { 
    {"Anderson",  "Varejao",     "1125"},
    {"Giorgi",  "Tevzadze",      "1125"},
    {"Will",      "Cherry",      "1225"},
    {"Will",      "Iams",       "12245"},
    {"Lebron",    "James",       "6025"},
    {"Kevin",     "Love",        "2525"},
    {"Kyrie",     "Livings",      "454"},
    {"Kyrie",     "Botti",       "4544"},
    {"Chris",     "Mauer",        "425"},
    {"Mot",       "Daniel",       "125"},
    {"Viktor",    "Muller",       "145"},
    {"Kiamran",   "Chris",       "1405"},
    {"Zenia",     "Vaitehovic",  "1025"},
    {"Marija",    "Grabauskaite","1471"},
    {"Milda",    "Grabauskaite", "1000"},
    {"Dion",      "Waiters",     "625" },
    {"Dion",      "Malborg",    "6250" } 
};

string FirstName = "Milda";
string LastName = "Grabauskaite";

for (int i = 0; i < personInfo.GetLength(0); i++)
{
    if(personInfo[i,0] == FirstName && personInfo[i,1] == LastName)
    {
        Console.WriteLine(FirstName + " " + LastName + " Salary Is: " + personInfo[i,2]);
    }
}

如何在二维数组

6 个答案: