java.lang.NumberFormatException：对于输入字符串：“10.0”
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java.lang.NumberFormatException: For input string: "10.0"
提问by George
This code must validate input data from the findActions() method:
此代码必须验证来自 findActions() 方法的输入数据：
try {
System.out.println(findActions(lookingArea.substring(0, right)));// always printing valid number string
Integer.parseInt(findActions(lookingArea.substring(0, right)));// checking for number format
}
catch(NumberFormatException exc) {
System.out.println(exc);
}
But I always have java.lang.NumberFormatException: For input string: "*number*"that is so strange, because checking with System.out.println(findActions(lookingArea.substring(0, right)));,
但我总是java.lang.NumberFormatException: For input string: "*number*"觉得这很奇怪，因为检查System.out.println(findActions(lookingArea.substring(0, right)));，
I get *number*like 10.0
我得到*number*像 10.0
采纳答案by rgettman
Integer.parseIntdoesn't expect the .character. If you're sure it can be converted to an int, then do one of the following:
Integer.parseInt不期待这个.角色。如果您确定可以将其转换为int，请执行以下操作之一：
- Eliminate the
".0"off the end of the string before parsing it, or - Call
Double.parseDouble, and cast the result toint.
".0"在解析之前消除字符串的末尾，或者- 调用
Double.parseDouble，并将结果转换为int。
Quoting the linked Javadocs above:
引用上面链接的 Javadocs：
The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign '-' ('\u002D') to indicate a negative value or an ASCII plus sign '+' ('\u002B') to indicate a positive value.
字符串中的字符必须都是十进制数字，除了第一个字符可以是 ASCII 减号 '-' ('\u002D') 表示负值或 ASCII 加号 '+' ('\u002B')表示正值。
回答by Eng.Fouad
10.0is not an integer number. Instead, you can use:
10.0不是整数。相反，您可以使用：
int num = (int) Double.parseDouble(...);
