Java异常处理

Question

您好我想使用下面的一段代码，我得到了一个分析异常，需要知道如何处理

DateFormat formatter1 = new SimpleDateFormat("dd/MM/yyyy"); 
Date date1 = (Date)formatter1.parse(startDate); 
Date date2 = (Date)formatter1.parse(endDate);    
DateFormat formatter2=new SimpleDateFormat("yyyy-MM-dd"); 
String startDate1=formatter2.format(date1); 
String endDate1=formatter2.format(date2); 

在此先感谢

Answer 1

尝试使用getErrorOffset()：

DateFormat formatter1 = new SimpleDateFormat("dd/MM/yyyy"); 
try 
{ 
    Date date1 = (Date)formatter1.parse(startDate); 
    Date date2 = (Date)formatter1.parse(endDate);    
}catch(ParseException e) 
{ 
    // Handle Exception, take a look at e.ErrorOffset to know where the problem occurs 
    // if the Message doesnt give enough information 
}    
DateFormat formatter2=new SimpleDateFormat("yyyy-MM-dd");    
String startDate1=formatter2.format(date1);    
String endDate1=formatter2.format(date2); 

Answer 2

你需要把把你的例外味精，

Date date1 = null, date2= null; 
    DateFormat formatter1 = new SimpleDateFormat("dd/MM/yyyy"); 


    try { 
     date1 = (Date)formatter1.parse("08/01/1988"); 
     date2 = (Date)formatter1.parse("08/01/2000"); 
    } catch (ParseException ex) { 
     //Handle Exception here 
    } 
    DateFormat formatter2=new SimpleDateFormat("yyyy-MM-dd"); 
    String startDate1=formatter2.format(date1); 
    String endDate1=formatter2.format(date2); 

    System.out.println(startDate1); 
    System.out.println(endDate1);