下面是我现在所拥有的。我只需要知道的主要事情是我怎么能检查一个字符串的字符数，阵列中，然后检查，对所有其他的条目，只拉最大的一个。我处理这样的问题：
编写一个程序，直到进入一个空行从用户读取姓名和出生年份。这个名字和出生年份是用逗号分隔。
此后，程序打印名字最长，平均诞生年。如果有多个名字同样最长的，你可以打印任何人。你可以假设用户进入至少一个人。示例输出
塞巴斯蒂安，2017年卢卡斯，2017年百合，2017年汉娜2014加布里埃尔，2009年
最长的名字：出生年的塞巴斯蒂安平均：2014.8
import java.util.Scanner;
public class Test112 {
//Asks the user for input and will print out who is the oldest
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int checkVal = 0;
String checkName = null;
System.out.println("Provide a name and age in this format: name,#");
while (true) {
//The unique thing about this code, is each input is a separate array that is not put in memory
//You just use the array system to pull data of the values entered at that specific time
String userIn = scanner.nextLine();
if (userIn.equals("")) {
break;
}
//Here we get the info from the user, then split it
String[] valArray = userIn.split(",");
//This now uses the variable to check which value of arguments entered is the greatest
//It'll then assign that value to the variable we created
**if (checkVal < Integer.valueOf((int)valArray[0].length) {**
//We need the checkVal variable to be a constant comparison against new entries
checkVal = Integer.valueOf(valArray[1]);
//This pulls the name of the value that satisfies the argument above
checkName = valArray[0];
}
}
System.out.println("Name of the oldest: " + checkName);
scanner.close();
}
}
我怎么能算的字符数的字符串值，在阵列内，然后把它比作他人找到最长的一个？
做到这一点，如下所示：
if (checkVal < valArray[0].length()) {
checkVal = valArray[0].length();
checkName = valArray[0];
}
请注意，length()是用来寻找的长度String。
其他一些重要的点：
Scanner了System.in。下面给出的是结合了笔记完整的程序：
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int checkVal = 0;
String checkName = "", name, birthYearStr;
int sum = 0, count = 0;
while (true) {
System.out.print("Provide a name and age in this format: name,#");
String userIn = scanner.nextLine();
if (userIn.equals("")) {
break;
}
String[] valArray = userIn.split(",");
name = valArray[0].trim();
birthYearStr = valArray[1].trim();
if (valArray.length == 2 && birthYearStr.length() == 4) { // Birth year should be of length 4
try {
sum += Integer.parseInt(birthYearStr);
if (checkVal < name.length()) {
checkVal = name.length();
checkName = name;
}
count++;
} catch (NumberFormatException e) {
System.out.println("Birth year should be an integer. Please try again.");
}
} else {
System.out.println("This is an invalid entry. Please try again.");
}
}
System.out.println("Longest name: " + checkName);
System.out.println("Average of birth years: " + (sum / count));
}
}
一个运行示例：
Provide a name and age in this format: name,#sebastian,2017
Provide a name and age in this format: name,#lucas,2017
Provide a name and age in this format: name,#lily,2017
Provide a name and age in this format: name,#hanna,2014
Provide a name and age in this format: name,#gabriel,2009
Provide a name and age in this format: name,#
Longest name: sebastian
Average of birth years: 2014
另一个样品运行：
Provide a name and age in this format: name,#sebastian,201
This is an invalid entry. Please try again.
Provide a name and age in this format: name,#sebastian,hello
This is an invalid entry. Please try again.
Provide a name and age in this format: name,#sebastian,2017
Provide a name and age in this format: name,#lucas,2017
Provide a name and age in this format: name,#lily,2017
Provide a name and age in this format: name,#hanna,2014
Provide a name and age in this format: name,#gabriel,2009
Provide a name and age in this format: name,#
Longest name: sebastian
Average of birth years: 2014
另一个样品运行：
Provide a name and age in this format: name,#hello,2018,sebastian,2017
This is an invalid entry. Please try again.
Provide a name and age in this format: name,#hello,2018
Provide a name and age in this format: name,#sebastian,2017
Provide a name and age in this format: name,#
Longest name: sebastian
Average of birth years: 2017
另一个样品运行：
Provide a name and age in this format: name,#sebastian,rama
Birth year should be an integer. Please try again.
Provide a name and age in this format: name,#sebastian,12.5
Birth year should be an integer. Please try again.
Provide a name and age in this format: name,#sebastian,2017
Provide a name and age in this format: name,#abcdefghi,2018
Provide a name and age in this format: name,#rama,2009
Provide a name and age in this format: name,#
Longest name: sebastian
Average of birth years: 2014
另一个样品运行：
Provide a name and age in this format: name,#sebastian,2017
Provide a name and age in this format: name,#lucas,2017
Provide a name and age in this format: name,#lily , 2017
Provide a name and age in this format: name,#
Longest name: sebastian
Average of birth years: 2017
本文收集自互联网，转载请注明来源。
如有侵权，请联系 [email protected] 删除。
我来说两句