如何防止java.lang.NumberFormatException:对于输入字符串:"N/A"？

Question

在运行我的代码时,我得到了NumberFormatException:

java.lang.NumberFormatException: For input string: "N/A"
    at java.lang.NumberFormatException.forInputString(Unknown Source)
    at java.lang.Integer.parseInt(Unknown Source)
    at java.lang.Integer.valueOf(Unknown Source)
    at java.util.TreeMap.compare(Unknown Source)
    at java.util.TreeMap.put(Unknown Source)
    at java.util.TreeSet.add(Unknown Source)`

如何防止发生此异常？

Answer 1

"N/A"不是整数.NumberFormatException如果您尝试将其解析为整数,则必须抛出.

解析前检查.或Exception妥善处理.

1. 异常处理*

try{
    int i = Integer.parseInt(input);
} catch(NumberFormatException ex){ // handle your exception
    ...
}

或 - 整数模式匹配 -

String input=...;
String pattern ="-?\\d+";
if(input.matches("-?\\d+")){ // any positive or negetive integer or not!
 ...
}

Answer 2

“N/A”是字符串，无法转换为数字。捕获异常并处理它。例如：

    String text = "N/A";
    int intVal = 0;
    try {
        intVal = Integer.parseInt(text);
    } catch (NumberFormatException e) {
        //Log it if needed
        intVal = //default fallback value;
    }

Answer 3

显然,你不能解析N/A到int的值.你可以做以下事情来处理它NumberFormatException.

   String str="N/A";
   try {
        int val=Integer.parseInt(str);
   }catch (NumberFormatException e){
       System.out.println("not a number"); 
   } 

Answer 4

NumberFormatException如果字符串不包含可解析的整数，则Integer.parseInt(str)抛出。你可以像下面一样处理。

int a;
String str = "N/A";

try {   
   a = Integer.parseInt(str);
} catch (NumberFormatException nfe) {
  // Handle the condition when str is not a number.
}

Answer 5

制作这样的异常处理程序,

private int ConvertIntoNumeric(String xVal)
{
 try
  { 
     return Integer.parseInt(xVal);
  }
 catch(Exception ex) 
  {
     return 0; 
  }
}

.
.
.
.

int xTest = ConvertIntoNumeric("N/A");  //Will return 0