Java 根据用户输入测试字符串和整数数组
我已经为我想在程序中使用的字符串和整数创建了数组,我想使用它们而不是Java 根据用户输入测试字符串和整数数组,java,arrays,if-statement,user-input,Java,Arrays,If Statement,User Input,我已经为我想在程序中使用的字符串和整数创建了数组,我想使用它们而不是 (name.equals "barry"||"matty" 比如说 我想知道如何编写if语句来对照数组中的字符串检查用户输入 import java.util.Scanner; public class Username { public static void main (String[]args) { Scanner kb = new Scanner (System.in);
(name.equals "barry"||"matty"
比如说
我想知道如何编写if语句来对照数组中的字符串检查用户输入
import java.util.Scanner;
public class Username
{
public static void main (String[]args)
{
Scanner kb = new Scanner (System.in);
// array containing usernames
String [] name = {"barry", "matty", "olly","joey"};
System.out.println("Enter your name");
name = kb.nextLine();
if (name.equals("barry ")|| name.equals("matty" ) || name.equals("olly")||name.equals("joey"))
System.out.println("you are verified you may use the lift");
Scanner f = new Scanner(System.in);
int floor;
int [] floor = {0,1,2,3,4,5,6,7};
System.out.println("What floor do you want to go to ");
floor = f.nextInt();
if (floor >7)
System.out.println("Invalid entry");
else if (floor <= 7)
System.out.println("Entry valid");
}
}
import java.util.Scanner;
公共类用户名
{
公共静态void main(字符串[]args)
{
扫描仪kb=新扫描仪(System.in);
//包含用户名的数组
String[]name={“barry”、“matty”、“olly”、“joey”};
System.out.println(“输入您的姓名”);
name=kb.nextLine();
如果(name.equals(“巴里”)| name.equals(“马蒂”)| name.equals(“奥利”)| name.equals(“乔伊”))
System.out.println(“您已确认可以使用电梯”);
扫描器f=新扫描器(System.in);
内部楼层;
int[]地板={0,1,2,3,4,5,6,7};
System.out.println(“你想去哪一层”);
地板=f.nextInt();
如果(楼层>7)
System.out.println(“无效条目”);
否则,如果(floor您可以通过数组进行循环:
String name = kb.nextLine();
if(contains(name)) {
System.out.println("you are verified you may use the lift");
}
public boolean contains(String name) {
String [] names = {"barry", "matty", "olly","joey"};
for (int i = 0; i < names.length; i++) {
if(names[i].equals(name)) {
System.out.println("you are verified you may use the lift");
}
}
我想您只是在寻找列表。包含-但这当然需要列表而不是数组。这里有两个明显的选项
首先,您可以使用列表
从以下内容开始:
List<String> names = new ArrayList<>();
names.add("barry");
names.add("matty");
names.add("olly");
names.add("joey");
...
if (names.contains(name))
{
...
}
作为第三个选项,如果将make your names array排序(手动或调用Arrays.sort
),您可以使用来尝试查找用户输入的名称:
String[] names = {"barry", "matty", "olly", "joey"};
Arrays.sort(names);
...
if (Arrays.binarySearch(names, name) >= 0)
{
...
}
数组是非常低级的结构,不提供任何方法。您最好使用集合,集合具有contains()
方法:
Set<String> names = new HashSet<>(Arrays.asList(new String[] {"barry", "matty", "olly","joey"}));
if (names.contains(name)) {
...
}
Set name=newhashset(Arrays.asList(新字符串[]{“barry”、“matty”、“olly”、“joey”}));
if(name.contains(name)){
...
}
由于您似乎不关心名称的顺序,而只想测试集合是否包含名称,因此哈希集是最佳的数据结构:HashSet.contains()
以恒定时间(O(1))运行,而List.contains()
,例如,是O(n)
阅读。使用for循环
get input
for int i = 0, i < array size i++
if input.equals(array[i]) then do stuff
获取输入
对于int i=0,i<数组大小i++
如果input.equals(数组[i]),则执行以下操作
如果切换到arraylist
,则可以使用arraylist和List而不是字符串数组执行If(array.contains(input))
:
List<String>names = new ArrayList<String>(names);
name = kb.nextLine();
if(names.indexOf(name)>-1)System.out.println("you are verified you may use the lift");
工作
get input
for int i = 0, i < array size i++
if input.equals(array[i]) then do stuff
List<String>names = new ArrayList<String>(names);
name = kb.nextLine();
if(names.indexOf(name)>-1)System.out.println("you are verified you may use the lift");
if(names.contains(name))System.out.println("you are verified you may use the lift");