Question

我写2 ArrayList类型String包含日期和可能的时间，我希望用户输入input，然后检查输入是否不是来自数组，它将显示一条消息，提示输入无效，用户再次输入。但是我的代码的结果给了我相反的:(当我在数组外输入内容时，它将接受它我的代码怎么了？请给我看看正确的代码：（

package javaapplication19;

import java.util.ArrayList;
import java.util.Scanner;

public class JavaApplication19 {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        ArrayList<String> dayArray = new ArrayList<>();
        ArrayList<String> timeArray = new ArrayList<>();
        dayArray.add("sunday");
        dayArray.add("monday");
        dayArray.add("tuesday");
        dayArray.add("wednesday");
        dayArray.add("thursday");
        timeArray.add("8am");
        timeArray.add("9am");
        timeArray.add("10am");
        timeArray.add("11am");
        timeArray.add("12pm");
        timeArray.add("1pm");
        timeArray.add("2pm");
        timeArray.add("3pm");
        timeArray.add("4pm");

        System.out.println("please enter day :");
        String a1 = input.nextLine();
        for (int g = 0; g < dayArray.size(); g++){
            if (dayArray.get(g).equals(a1))
                System.out.println("invalid day , please enter another day : ");
        a1 = input.nextLine();
}

        System.out.println("please enter time : ");
        String a2 = input.nextLine();
        for (int s = 0; s < timeArray.size(); s++) {
            if (timeArray.get(s).equals(a2))
                System.out.println("invalid time , please enter another time : ");
            a2 = input.nextLine();

        }
    }
}

Answer 1

您可以这样：

XAxis

不需要遍历数组。方法contains（）为您完成了

Answer 2

您可以使用以下代码查找if条件的变化。

System.out.println("please enter time : ");
String a2 = input.nextLine();
for (int s = 0; s < timeArray.size(); s++) {
if (!timeArray.get(s).equals(a2))
System.out.println("invalid time , please enter another time : ");
a2 = input.nextLine();
}

Answer 3

您的代码应该是这样的。

System.out.println("please enter day :");

boolean validDate = false;

while(!validDate){
    String a1 = input.nextLine();
    if(dayArray.contains(a1)){
        System.out.println("invalid day , please enter another day : ");
    }else{
        validDate=true;
    }
}

boolean validHour = false;
System.out.println("please enter a time : ");
while(!validHour){
    String a2 = input.nextLine();
    if(timeArray.contains(a2)){
        System.out.println("invalid time , please enter another time : ");
    }else{
        validHour=true;
    }
}

Answer 4

我确实重构了您的代码，并添加了一些注释，以使其更容易理解

Scanner input = new Scanner(System.in);
List<String> dayArray = Arrays.asList("sunday","monday","tuesday","wednesday","thursday");
List<String> timeArray = Arrays.asList("8am","9am","10am","11am","12pm","1pm","2pm","3pm","4pm");

String a1 = "";
String a2 = ""; //Store user input

boolean nextInput = false; // flag indicate we continue or we force user re-enter information
//try to read the day input
while (!nextInput) {
    System.out.println("please enter day :");
    a1 = input.nextLine();
    nextInput = dayArray.contains(a1);
}

//try to read the time input
nextInput = false; //<= reset flag
while(!nextInput) {
    System.out.println("please enter time : ");
    a2 = input.nextLine();
    nextInput = timeArray.contains(a2);
}

System.out.println("day :" + a1);
System.out.println("time :" + a2);
System.out.println("program finished");

input.close(); //close scanner

Answer 5

这样写

WARNING:socketIO-client:insta-menorah.herokuapp.com:39203//socket.io [engine.io waiting for connection] HTTPConnectionPool(host='insta-menorah.herokuapp.com', port=39203): Max retries exceeded with url: //socket.io/?EIO=3&transport=polling&t=1543479369533-0 (Caused by NewConnectionError('<urllib3.connection.HTTPConnection object at 0x1027c3a90>: Failed to establish a new connection: [Errno 60] Operation timed out',))

检查输入是否在ArrayList中

5 个答案: