如何从缓冲的阅读器输入字符串？(How do i input a string from a buffered reader?)

如何从缓冲的阅读器输入字符串？(How do i input a string from a buffered reader?)

我也主要使用Scanner，也想尝试使用缓冲读卡器：到目前为止我所拥有的

import java.util.*; import java.io.*; public class IceCreamCone { // variables String flavour; int numScoops; Scanner flavourIceCream = new Scanner(System.in); // constructor public IceCreamCone() { } // methods public String getFlavour() throws IOexception { try{ BufferedReader keyboardInput; keyboardInput = new BufferedReader(new InputStreamReader(System.in)); System.out.println(" please enter your flavour ice cream"); flavour = keyboardInput.readLine(); return keyboardInput.readLine(); } catch (IOexception e) { e.printStackTrace(); } }

我相当肯定会得到一个你可以说的int

Integer.parseInt(keyboardInput.readLine());

但如果我想要一个字符串怎么办？

Im used too using Scanner mainly and want too try using a buffered reader: heres what i have so far

import java.util.*; import java.io.*; public class IceCreamCone { // variables String flavour; int numScoops; Scanner flavourIceCream = new Scanner(System.in); // constructor public IceCreamCone() { } // methods public String getFlavour() throws IOexception { try{ BufferedReader keyboardInput; keyboardInput = new BufferedReader(new InputStreamReader(System.in)); System.out.println(" please enter your flavour ice cream"); flavour = keyboardInput.readLine(); return keyboardInput.readLine(); } catch (IOexception e) { e.printStackTrace(); } }

im fairly sure to get an int you can say

Integer.parseInt(keyboardInput.readLine());

but what do i do if i want a String

最满意答案

keyboardInput.readLine()已经返回一个字符串，所以你应该只做：

return keyboardInput.readLine();

（更新）

readLine方法抛出IOException 。 你要么抛出异常：

public String getFlavour() throws IOException { ... }

或者你在你的方法中处理它。

public static String getFlavour() { BufferedReader keyboardInput = null; try { keyboardInput = new BufferedReader(new InputStreamReader(System.in)); System.out.println(" please enter your flavour ice cream"); // in this case, you don't need to declare this extra variable // String flavour = keyboardInput.readLine(); // return flavour; return keyboardInput.readLine(); } catch (IOException e) { // handle this e.printStackTrace(); } return null; }

keyboardInput.readLine() already returns a string so you should simply do:

return keyboardInput.readLine();

(update)

The readLine method throws an IOException. You either throw the exception:

public String getFlavour() throws IOException { ... }

or you handle it in your method.

public static String getFlavour() { BufferedReader keyboardInput = null; try { keyboardInput = new BufferedReader(new InputStreamReader(System.in)); System.out.println(" please enter your flavour ice cream"); // in this case, you don't need to declare this extra variable // String flavour = keyboardInput.readLine(); // return flavour; return keyboardInput.readLine(); } catch (IOException e) { // handle this e.printStackTrace(); } return null; }