Question

如何使用字符串搜索数组以显示相应的数字？我的任务要求我们让用户输入一个名字，然后让他们相应的工资得到它，这是我到目前为止的代码，我不知道该怎么用来做这个

import java.io.*;

public class Employees {

    public static void main(String agrs[]) throws IOException {

        BufferedReader keyboardInput = new BufferedReader(new InputStreamReader(System.in));

        double pay;
        String name;
        String answer;

        //Arrays
        double[] employeePay = new double[3];
        String[] employeeName = new String[3];

        pay = employeePay.length;

        //for loops
        for (int x = 0; x <= employeePay.length - 1; x++) {
            System.out.print("Please enter the employee name:");
            employeeName[x] = (keyboardInput.readLine());
            System.out.print("Please enter the employe pay :");
            employeePay[x] = Double.parseDouble(keyboardInput.readLine());

        }

        //requesting a name
        System.out.println("Please enter an employee name to search for:");
        answer = (keyboardInput.readLine());

        for (int x = 0; x <= employeePay.length - 1; x++) {
            if (employeeName[x].equalsIgnoreCase(answer)) {
                System.out.print(employeeName + " pay rate is " + pay);

            } else {
                System.out.print(" The employee name hasnt been entered");
            }

        }

    }
}

Answer 1

考虑使用HashMap而不是使用2个单维数组

int employeeCount = 3;
HashMap<String, Double> employeePayMap = new HashMap<String, Double>();
// ... code to capture input
for (int x = 0; x < employeeCount; x++) {
        System.out.print("Please enter the employee name:");
        String name = (keyboardInput.readLine());
        System.out.print("Please enter the employe pay :");
        Double pay = Double.parseDouble(keyboardInput.readLine());
        employeePayMap.put( name, pay );

    }

然后将付款检索为

 Double payFound = employeePayMap.get(answer);

将payFound检查为null - 如果为null，则名称不匹配。

Answer 2

你做错了什么：

if (employeeName[x].equalsIgnoreCase(answer)) {
    System.out.print(employeeName + " pay rate is " + pay);
}

在这里打印整个employeeName，而不是预期的索引。它应该是employeeName[x]。 pay也只是employeePay.length。我不认为你在这里打印employeePay数组的长度。它应该是employeePay[x]。

if (employeeName[x].equalsIgnoreCase(answer)) {
    System.out.print(employeeName[x] + " pay rate is = " + employeePay[x]);
}

更好的方法：

如果您有配对的内容（例如员工姓名和付款），请更好地使用Map而不是两个离散数组。这是更好，更清洁的方法。

您不仅可以轻松地将数据插入到地图中，还可以从地图中搜索数据，而无需在代码中进行迭代。

如何使用字符串搜索数组以显示相应的数字

2 个答案: