用户输入不适用于 keyboard.nextLine() 和 String (Java)
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User Input not working with keyboard.nextLine() and String (Java)
提问by js9999
I recently started learning java during my spare time. So to practice, I'm making a program that takes a temperature (Celsius or Fahrenheit) and converts it to the opposite. I've already imported the keyboard scanner.
我最近在业余时间开始学习Java。所以为了练习，我正在制作一个程序，该程序需要一个温度（摄氏度或华氏度）并将其转换为相反的温度。我已经导入了键盘扫描仪。
int temp;
String opposite, type;
double product;
System.out.print("Please enter a temperature: ");
temp = keyboard.nextInt();
System.out.println("Was that in Celsius or Fahrenheit?");
System.out.print("(Enter 'C' for Celsius and 'F' for Fahrenheit) ");
type = keyboard.nextLine();
if (type == "C") // Only irrelevant temp conversion code left so I'm leaving it out
I'm new to the String and nextLinestuff and the program just skips over the user input section where you enter either C or F. Would someone explain what I can do to fix this?
我是 String 和其他nextLine东西的新手，程序只是跳过你输入 C 或 F 的用户输入部分。有人能解释一下我能做些什么来解决这个问题吗？
Thanks!
谢谢！
采纳答案by Anjula Ranasinghe
For you code Change nextLine();to next();and it will work.
对于您的代码更改nextLine();为next();，它将起作用。
System.out.println("Was that in Celsius or Fahrenheit?");
System.out.print("(Enter 'C' for Celsius and 'F' for Fahrenheit) ");
type = keyboard.next();
to get an idea for you to what happened was this:
让你知道发生了什么是这样的：
- nextLine(): Advances this scanner past the current lineand returns the input that was skipped.
- next(): Finds and returns the next complete token from this scanner.
- nextLine()：将此扫描器推进到当前行并 返回被跳过的输入。
- next()：从这个扫描器中查找并返回下一个完整的标记。
Also like the many of the answers says use equals()instead of using ==
也像许多答案说使用equals()而不是使用==
The ==checks only the references to the object are equal. .equal()compares string.
在==仅检查到对象的引用是相等的。.equal()比较字符串。
Read more Here
在这里阅读更多
回答by beatyt
.nextInt()does not read the end of line character "\n".
.nextInt()不读取行尾字符"\n"。
You need to put a keyboard.nextLine()after the .nextInt()and then it will work.
你需要把一keyboard.nextLine()后的.nextInt()，然后它会奏效。
回答by Masudul
Never use Scanner#nextLineafter Scanner#nextInt. Whenever you hit enter button after Scanner#nextIntthan it will skip the Scanner#nextLinecommand. So,
Change from
以后再也不用Scanner#nextLine了Scanner#nextInt。每当您按下回车按钮后Scanner#nextInt，它都会跳过该Scanner#nextLine命令。所以，从
int temp = keyboard.nextInt();
to
到
int temp = Integer.parseInt(keyboard.nextLine());
回答by PWC
Also, use type.equals("C")instead of if (type == "C"), the later one is comparing the reference of the value.
另外，使用type.equals("C")代替if (type == "C")，后者是比较值的引用。
回答by JClassic
Call
称呼
keyboard.nextLine();
After
后
temp = keyboard.nextInt();
Because nextInt() doesn't consume the \ncharacter.
因为 nextInt() 不消耗\n字符。
Also, compare Stringswith .equals();not ==
此外，Strings与.equals();不比较==
if(type.equals("C"));
回答by vishu9219
Use Scanner Class:
使用扫描仪类：
int temp;
java.util.Scanner s = new java.util.Scanner(System.in);
String opposite, type;
double product;
System.out.print("Please enter a temperature: ");
temp = s.nextInt();
System.out.println("Was that in Celsius or Fahrenheit?");
System.out.print("(Enter 'C' for Celsius and 'F' for Fahrenheit) ");
type = s.nextLine();
if (type.equals("C")){
do something
}
else{
do something
}
For More Input references:
更多输入参考：
Hereis a wonderful comparison on how to compare strings in java.
这是关于如何在java中比较字符串的精彩比较。
回答by STIKO
You could use keyboard.next()instead of nextLine()and as
您可以使用keyboard.next()代替nextLine()和作为
user1770155
用户1770155
mentioned to compare two strings you should use .equals()and what I would do since you're comparing to an upper letter "C" is
提到比较您应该使用的两个字符串.equals()，以及我会做的事情，因为您要与大写字母“C”进行比较
type = keyboard.next().toUpperCase();
if (type.equals("C"))
