Java JOptionPane 输入到 int
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JOptionPane Input to int
提问by kxk
I am trying to make a JOptionPane get an input and assign it to an int but I am getting some problems with the variable types.
我试图让 JOptionPane 获得一个输入并将它分配给一个 int 但我在变量类型方面遇到了一些问题。
I am trying something like this:
我正在尝试这样的事情：
Int ans = (Integer) JOptionPane.showInputDialog(frame,
"Text",
JOptionPane.INFORMATION_MESSAGE,
null,
null,
"[sample text to help input]");
But I am getting:
但我得到：
Exception in thread "main" java.lang.ClassCastException: java.lang.String cannot
be cast to java.lang.Integer
Which sounds logical yet, I cannot think of another way to make this happen.
这听起来合乎逻辑，但我想不出另一种方法来实现这一目标。
Thanks in advance
提前致谢
采纳答案by jjnguy
Simply use:
只需使用：
int ans = Integer.parseInt( JOptionPane.showInputDialog(frame,
"Text",
JOptionPane.INFORMATION_MESSAGE,
null,
null,
"[sample text to help input]"));
You cannot cast a Stringto an int, but you can convert it using Integer.parseInt(string).
您不能将 a转换String为 an int，但可以使用Integer.parseInt(string).
回答by Hyman
This because the input that the user inserts into the JOptionPaneis a Stringand it is stored and returned as a String.
这是因为用户插入的输入JOptionPane是 aString并且它被存储并作为 a 返回String。
Java cannot convert between strings and number by itself, you have to use specific functions, just use:
Java本身无法在字符串和数字之间进行转换，您必须使用特定的函数，只需使用：
int ans = Integer.parseInt(JOptionPane.showInputDialog(...))
回答by Davy Meers
Please note that Integer.parseInt throws an NumberFormatException if the passed string doesn't contain a parsable string.
请注意，如果传递的字符串不包含可解析的字符串，则 Integer.parseInt 会抛出 NumberFormatException。
回答by MyStack
// sample code for addition using JOptionPane
import javax.swing.JOptionPane;
public class Addition {
public static void main(String[] args) {
String firstNumber = JOptionPane.showInputDialog("Input <First Integer>");
String secondNumber = JOptionPane.showInputDialog("Input <Second Integer>");
int num1 = Integer.parseInt(firstNumber);
int num2 = Integer.parseInt(secondNumber);
int sum = num1 + num2;
JOptionPane.showMessageDialog(null, "Sum is" + sum, "Sum of two Integers", JOptionPane.PLAIN_MESSAGE);
}
}
回答by mubarak baloch
String String_firstNumber = JOptionPane.showInputDialog("Input Semisecond");
int Int_firstNumber = Integer.parseInt(firstNumber);
Now your Int_firstnumbercontains integer value of String_fristNumber.
现在您Int_firstnumber包含 的整数值String_fristNumber。
hope it helped
希望它有帮助
