如何在Java中获取用户输入？
我试图创建一个计算器，但我无法让它工作，因为我不知道如何获取用户输入如何在Java中获取用户输入？,java,input,Java,Input,我试图创建一个计算器，但我无法让它工作，因为我不知道如何获取用户输入 如何在Java中获取用户输入？最简单的方法之一是使用扫描仪对象，如下所示： import java.util.Scanner; Scanner reader = new Scanner(System.in); // Reading from System.in System.out.println("Enter a number: "); int n = reader.nextInt(); // Scans the nex
如何在Java中获取用户输入？最简单的方法之一是使用
扫描仪import java.util.Scanner;
Scanner reader = new Scanner(System.in); // Reading from System.in
System.out.println("Enter a number: ");
int n = reader.nextInt(); // Scans the next token of the input as an int.
//once finished
reader.close();
BufferedReader bufferReader = new BufferedReader(new InputStreamReader(System.in));
String inputLine = bufferReader.readLine();
您可以使用获取用户输入 它将在
accStrStringint以下是获取键盘输入的方法：
Scanner scanner = new Scanner (System.in);
System.out.print("Enter your name");
name = scanner.next(); // Get what the user types.
要读取行或字符串，可以将
BufferedReaderInputStreamReaderimport java.util.Scanner;
Scanner reader = new Scanner(System.in); // Reading from System.in
System.out.println("Enter a number: ");
int n = reader.nextInt(); // Scans the next token of the input as an int.
//once finished
reader.close();
BufferedReader bufferReader = new BufferedReader(new InputStreamReader(System.in));
String inputLine = bufferReader.readLine();
Console console = System.console();
String input = console.readLine("Enter input:");
在
main（）旁边添加抛出IOException
，然后
DataInputStream input = new DataInputStream(System.in);
System.out.print("Enter your name");
String name = input.readLine();
在这里，程序要求用户输入一个数字。然后，程序打印数字的位数和位数的总和
import java.util.Scanner;
public class PrintNumber {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int num = 0;
int sum = 0;
System.out.println(
"Please enter a number to show its digits");
num = scan.nextInt();
System.out.println(
"Here are the digits and the sum of the digits");
while (num > 0) {
System.out.println("==>" + num % 10);
sum += num % 10;
num = num / 10;
}
System.out.println("Sum is " + sum);
}
}
以下是使用java.util.Scanner
从问题中提取的程序：
import java.util.Scanner;
public class Example {
public static void main(String[] args) {
int input = 0;
System.out.println("The super insano calculator");
System.out.println("enter the corrosponding number:");
Scanner reader3 = new Scanner(System.in);
System.out.println(
"1. Add | 2. Subtract | 3. Divide | 4. Multiply");
input = reader3.nextInt();
int a = 0, b = 0;
Scanner reader = new Scanner(System.in);
System.out.println("Enter the first number");
// get user input for a
a = reader.nextInt();
Scanner reader1 = new Scanner(System.in);
System.out.println("Enter the scend number");
// get user input for b
b = reader1.nextInt();
switch (input){
case 1: System.out.println(a + " + " + b + " = " + add(a, b));
break;
case 2: System.out.println(a + " - " + b + " = " + subtract(a, b));
break;
case 3: System.out.println(a + " / " + b + " = " + divide(a, b));
break;
case 4: System.out.println(a + " * " + b + " = " + multiply(a, b));
break;
default: System.out.println("your input is invalid!");
break;
}
}
static int add(int lhs, int rhs) { return lhs + rhs; }
static int subtract(int lhs, int rhs) { return lhs - rhs; }
static int divide(int lhs, int rhs) { return lhs / rhs; }
static int multiply(int lhs, int rhs) { return lhs * rhs; }
}
在java中获取输入非常简单，您只需执行以下操作：
import java.util.Scanner;
class GetInputFromUser
{
public static void main(String args[])
{
int a;
float b;
String s;
Scanner in = new Scanner(System.in);
System.out.println("Enter a string");
s = in.nextLine();
System.out.println("You entered string " + s);
System.out.println("Enter an integer");
a = in.nextInt();
System.out.println("You entered integer " + a);
System.out.println("Enter a float");
b = in.nextFloat();
System.out.println("You entered float " + b);
}
}
您可以根据需求使用以下任何选项
班
和班级
班
来自DataInputStream
类的readLine
方法已被弃用。要获取字符串值，应使用BufferedReader的上一个解决方案
班
显然，这种方法在某些IDE中不起作用。这是一个使用System.in.read（）函数的简单代码。这段代码只是写出键入的内容。如果只想获取一次输入，就可以去掉while循环，如果愿意，还可以将答案存储在字符数组中
package main;
import java.io.IOException;
public class Root
{
public static void main(String[] args)
{
new Root();
}
public Root()
{
while(true)
{
try
{
for(int y = 0; y < System.in.available(); ++y)
{
System.out.print((char)System.in.read());
}
}
catch(IOException ex)
{
ex.printStackTrace(System.out);
break;
}
}
}
}
packagemain；
导入java.io.IOException；
公共类根
{
公共静态void main（字符串[]args）
{
新根（）；
}
公共根（）
{
while（true）
{
尝试
{
对于（int y=0；y
可以是这样的
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
System.out.println("Enter a number: ");
int i = reader.nextInt();
for (int j = 0; j < i; j++)
System.out.println("I love java");
}
publicstaticvoidmain（字符串[]args）{
扫描仪阅读器=新扫描仪（System.in）；
System.out.println（“输入一个数字：”）；
int i=reader.nextInt（）；
对于（int j=0；j
您可以制作一个简单的程序，询问用户名并打印回复使用的输入内容
或者让用户输入两个数字，您可以对这些数字进行加法、乘法、减法或除法，并打印用户输入的答案，就像计算器一样
所以你需要扫描器类。您必须导入java.util.Scanner
和您需要使用的代码中
Scanner input = new Scanner(System.in);
输入是一个变量名
Scanner input = new Scanner(System.in);
System.out.println("Please enter your name : ");
s = input.next(); // getting a String value
System.out.println("Please enter your age : ");
i = input.nextInt(); // getting an integer
System.out.println("Please enter your salary : ");
d = input.nextDouble(); // getting a double
看看这有什么不同：input.next（），i=input.nextInt（），d=input.nextDouble（）
根据字符串，int和double的变化方式与其他字符串相同。不要忘记代码顶部的import语句
另请参阅博文。您可以使用BufferedReader获得如下用户输入：
InputStreamReader inp = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(inp);
// you will need to import these things.
这就是应用它们的方式
String name = br.readline();
因此，当用户在控制台中输入他的名字时，“字符串名”将存储该信息
如果是要存储的数字，则代码如下所示：
int x = Integer.parseInt(br.readLine());
跳这个有帮助
import java.util.Scanner;
public class Myapplication{
public static void main(String[] args){
Scanner in = new Scanner(System.in);
int a;
System.out.println("enter:");
a = in.nextInt();
System.out.println("Number is= " + a);
}
}
扫描仪输入=新扫描仪（System.in）；
int integer=input.nextInt（）；
String String=input.next（）；
long longInteger=input.nextLong（）；
只是一个额外的细节。如果不想冒内存/资源泄漏的风险，则应在完成后关闭扫描仪流：
myScanner.close();
请注意，java 1.7及更高版本将此作为编译警告（不要问我如何知道：-）我喜欢以下内容：
public String readLine(String tPromptString) {
byte[] tBuffer = new byte[256];
int tPos = 0;
System.out.print(tPromptString);
while(true) {
byte tNextByte = readByte();
if(tNextByte == 10) {
return new String(tBuffer, 0, tPos);
}
if(tNextByte != 13) {
tBuffer[tPos] = tNextByte;
++tPos;
}
}
}
例如，我会：
String name = this.readLine("What is your name?")
以下是公认答案的更完善版本，解决了两个常见需求：
- 重复收集用户输入，直到输入退出值
- 处理无效输入值（本例中为非整数）
代码
package inputTest;
import java.util.Scanner;
import java.util.InputMismatchException;
public class InputTest {
public static void main(String args[]) {
Scanner reader = new Scanner(System.in);
System.out.println("Please enter integers. Type 0 to exit.");
boolean done = false;
while (!done) {
System.out.print("Enter an integer: ");
try {
int n = reader.nextInt();
if (n == 0) {
done = true;
}
else {
// do something with the input
System.out.println("\tThe number entered was: " + n);
}
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input type (must be an integer)");
reader.nextLine(); // Clear invalid input from scanner buffer.
}
}
System.out.println("Exiting...");
reader.close();
}
}
Please enter integers. Type 0 to exit.
Enter an integer: 12
The number entered was: 12
Enter an integer: -56
The number entered was: -56
Enter an integer: 4.2
Invalid input type (must be an integer)
Enter an integer: but i hate integers
Invalid input type (must be an integer)
Enter an integer: 3
The number entered was: 3
Enter an integer: 0
Exiting...
示例
package inputTest;
import java.util.Scanner;
import java.util.InputMismatchException;
public class InputTest {
public static void main(String args[]) {
Scanner reader = new Scanner(System.in);
System.out.println("Please enter integers. Type 0 to exit.");
boolean done = false;
while (!done) {
System.out.print("Enter an integer: ");
try {
int n = reader.nextInt();
if (n == 0) {
done = true;
}
else {
// do something with the input
System.out.println("\tThe number entered was: " + n);
}
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input type (must be an integer)");
reader.nextLine(); // Clear invalid input from scanner buffer.
}
}
System.out.println("Exiting...");
reader.close();
}
}
Please enter integers. Type 0 to exit.
Enter an integer: 12
The number entered was: 12
Enter an integer: -56
The number entered was: -56
Enter an integer: 4.2
Invalid input type (must be an integer)
Enter an integer: but i hate integers
Invalid input type (must be an integer)
Enter an integer: 3
The number entered was: 3
Enter an integer: 0
Exiting...
请注意，如果没有nextLine（）
，错误输入将在无限循环中重复触发相同的异常。根据具体情况，您可能希望使用next（）
，但要知道，像这样的输入有空格
会产生多个异常。使用扫描仪输入键盘是可能的，正如其他人所说。但是在这个高度图形化的时代，没有图形用户界面（GUI）的计算器是毫无意义的
在现代Java中，这意味着使用JavaFX拖放工具，如布局类似计算器控制台的GUI。
请注意，使用Scene Builder直观上很容易，并且不需要您已经拥有的事件处理程序的额外Java技能
对于用户输入，您应该在GUI控制台的顶部有一个宽文本字段。
这是用户输入他们想要在其上执行功能的数字的地方。
在文本字段下方，您将有一组功能按钮，执行基本（即加法/减法/乘法/除法和记忆/调用/清除）功能。
一旦GUI被放置出来，您就可以添加将每个按钮函数链接到其Java实现的“控制器”引用，例如在项目的控制器类中调用方法
有点旧，但仍然显示了场景生成器的易用性。您可以使用扫描仪获取用户输入。对于不同的数据类型，您可以使用适当的方法进行适当的输入验证，例如next（）
用于String
或nextInt（）
用于Integer
import java.util.Scanner;
Scanner scanner = new Scanner(System.in);
//reads the input until it reaches the space
System.out.println("Enter a string: ");
String str = scanner.next();
System.out.println("str = " + str);
//reads until the end of line
String aLine = scanner.nextLine();
//reads the integer
System.out.println("Enter an integer num: ");
int num = scanner.nextInt();
System.out.println("num = " + num);
//reads the double value
System.out.println("Enter a double: ");
double aDouble = scanner.nextDouble();
System.out.println("double = " + aDouble);
//reads the float value, long value, boolean value, byte and short
double aFloat = scanner.nextFloat();
long aLong = scanner.nextLong();
boolean aBoolean = scanner.nextBoolean();
byte aByte = scanner.nextByte();
short aShort = scanner.nextShort();
scanner.close();
最好的两个选项是Buffer
Scanner input=new Scanner(System.in);
int integer=input.nextInt();
String string=input.next();
long longInteger=input.nextLong();
myScanner.close();
public String readLine(String tPromptString) {
byte[] tBuffer = new byte[256];
int tPos = 0;
System.out.print(tPromptString);
while(true) {
byte tNextByte = readByte();
if(tNextByte == 10) {
return new String(tBuffer, 0, tPos);
}
if(tNextByte != 13) {
tBuffer[tPos] = tNextByte;
++tPos;
}
}
}
String name = this.readLine("What is your name?")
package inputTest;
import java.util.Scanner;
import java.util.InputMismatchException;
public class InputTest {
public static void main(String args[]) {
Scanner reader = new Scanner(System.in);
System.out.println("Please enter integers. Type 0 to exit.");
boolean done = false;
while (!done) {
System.out.print("Enter an integer: ");
try {
int n = reader.nextInt();
if (n == 0) {
done = true;
}
else {
// do something with the input
System.out.println("\tThe number entered was: " + n);
}
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input type (must be an integer)");
reader.nextLine(); // Clear invalid input from scanner buffer.
}
}
System.out.println("Exiting...");
reader.close();
}
}
Please enter integers. Type 0 to exit.
Enter an integer: 12
The number entered was: 12
Enter an integer: -56
The number entered was: -56
Enter an integer: 4.2
Invalid input type (must be an integer)
Enter an integer: but i hate integers
Invalid input type (must be an integer)
Enter an integer: 3
The number entered was: 3
Enter an integer: 0
Exiting...
import java.util.Scanner;
Scanner scanner = new Scanner(System.in);
//reads the input until it reaches the space
System.out.println("Enter a string: ");
String str = scanner.next();
System.out.println("str = " + str);
//reads until the end of line
String aLine = scanner.nextLine();
//reads the integer
System.out.println("Enter an integer num: ");
int num = scanner.nextInt();
System.out.println("num = " + num);
//reads the double value
System.out.println("Enter a double: ");
double aDouble = scanner.nextDouble();
System.out.println("double = " + aDouble);
//reads the float value, long value, boolean value, byte and short
double aFloat = scanner.nextFloat();
long aLong = scanner.nextLong();
boolean aBoolean = scanner.nextBoolean();
byte aByte = scanner.nextByte();
short aShort = scanner.nextShort();
scanner.close();
Scanner scanner = new Scanner (System.in); // create scanner
System.out.print("Enter your name"); // prompt user
name = scanner.next(); // get user input
import java.util.InputMismatchException; // import the exception catching class
import java.util.Scanner; // import the scanner class
public class RunScanner {
// main method which will run your program
public static void main(String args[]) {
// create your new scanner
// Note: since scanner is opened to "System.in" closing it will close "System.in".
// Do not close scanner until you no longer want to use it at all.
Scanner scanner = new Scanner(System.in);
// PROMPT THE USER
// Note: when using scanner it is recommended to prompt the user with "System.out.print" or "System.out.println"
System.out.println("Please enter a number");
// use "try" to catch invalid inputs
try {
// get integer with "nextInt()"
int n = scanner.nextInt();
System.out.println("Please enter a decimal"); // PROMPT
// get decimal with "nextFloat()"
float f = scanner.nextFloat();
System.out.println("Please enter a word"); // PROMPT
// get single word with "next()"
String s = scanner.next();
// ---- Note: Scanner.nextInt() does not consume a nextLine character /n
// ---- In order to read a new line we first need to clear the current nextLine by reading it:
scanner.nextLine();
// ----
System.out.println("Please enter a line"); // PROMPT
// get line with "nextLine()"
String l = scanner.nextLine();
// do something with the input
System.out.println("The number entered was: " + n);
System.out.println("The decimal entered was: " + f);
System.out.println("The word entered was: " + s);
System.out.println("The line entered was: " + l);
}
catch (InputMismatchException e) {
System.out.println("\tInvalid input entered. Please enter the specified input");
}
scanner.close(); // close the scanner so it doesn't leak
}
}
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
System.out.println("Welcome to the best program in the world! ");
while (true) {
System.out.print("Enter a query: ");
Scanner scan = new Scanner(System.in);
String s = scan.nextLine();
if (s.equals("q")) {
System.out.println("The program is ending now ....");
break;
} else {
System.out.println("The program is running...");
}
}
}
}
import java.util.Scanner;
public class main {
public static void main(String[]args) {
Scanner sc=new Scanner(System.in);
int a;
String b;
System.out.println("Type an integer here: ");
a=sc.nextInt();
System.out.println("Type anything here:");
b=sc.nextLine();
Int a =JOptionPane.showInputDialog(null,"Enter number:");