如何判别输入的字符串符合日期规范“YYYYMMDDHH”,麻烦各位大虾了。在线等

Question

you can use calendar class to check itint yy = Integer.parseInt(your_string.substring(0,4));
int mm = Integer.parseInt(your_string.substring(4,6));
int dd = Integer.parseInt(your_string.substring(6,8));
int hh = Integer.parseInt(your_string.substring(8,10));
Calendar c = new GregorianCalendar(yy,mm-1,dd,hh,0,0);
if (c.get(Calendar.YEAR) != yy || c.get(Calendar.MONTH) != mm-1 || c.get(Calendar.DAY) != dd || c.get(Calendar.HOUR) != hh) {
    return false;
}
return true;

Answer 1

String date = "2005020313";
SimpleDateFormat fmt = new SimpleDateFormat("yyyyMMddHH");
fmt.setLenient(false);
try {
    fmt.parse(date);
} catch (ParseException e) {
    //不符合日期格式
    e.printStackTrace();
}

Answer 2

String dateString = "2005010213";
SimpleDateFormat sdf = new SimpleDateFormat("yyyyMMddHH");
Date date = sdf.parse(dateString);//在此捕捉异常，如果有，则字符串格式不对

Answer 3

正则表达式
 function validateDate(form) {
               var bValid = true;
               var focusField = null;
               var i = 0;
               var fields = new Array();
               oDate = new DateValidations();
               for (x in oDate) {
                   var value = form[oDate[x][0]].value;
                   var datePattern = oDate[x][2]("datePatternStrict");
                   if ((form[oDate[x][0]].type == 'text' ||
                        form[oDate[x][0]].type == 'textarea') &&
                       (value.length > 0) &&
                       (datePattern.length > 0)) {
                     var MONTH = "MM";
                     var DAY = "dd";
                     var YEAR = "yyyy";
                     var orderMonth = datePattern.indexOf(MONTH);
                     var orderDay = datePattern.indexOf(DAY);
                     var orderYear = datePattern.indexOf(YEAR);
                     if ((orderDay < orderYear && orderDay > orderMonth)) {
                         var iDelim1 = orderMonth + MONTH.length;
                         var iDelim2 = orderDay + DAY.length;
                         var delim1 = datePattern.substring(iDelim1, iDelim1 + 1);
                         var delim2 = datePattern.substring(iDelim2, iDelim2 + 1);
                         if (iDelim1 == orderDay && iDelim2 == orderYear) {
                            dateRegexp = new RegExp("^(\\d{2})(\\d{2})(\\d{4})$");
                         } else if (iDelim1 == orderDay) {
                            dateRegexp = new RegExp("^(\\d{2})(\\d{2})[" + delim2 + "](\\d{4})$");
                         } else if (iDelim2 == orderYear) {
                            dateRegexp = new RegExp("^(\\d{2})[" + delim1 + "](\\d{2})(\\d{4})$");
                         } else {
                            dateRegexp = new RegExp("^(\\d{2})[" + delim1 + "](\\d{2})[" + delim2 + "](\\d{4})$");
                         }
                         var matched = dateRegexp.exec(value);
                         if(matched != null) {
                            if (!isValidDate(matched[2], matched[1], matched[3])) {
                               if (i == 0) {
                                   focusField = form[oDate[x][0]];
                               }
                               fields[i++] = oDate[x][1];
                               bValid =  false;
                            }
                         } else {
                            if (i == 0) {
                                focusField = form[oDate[x][0]];
                            }
                            fields[i++] = oDate[x][1];
                            bValid =  false;
                         }
                     } else if ((orderMonth < orderYear && orderMonth > orderDay)) {
                         var iDelim1 = orderDay + DAY.length;
                         var iDelim2 = orderMonth + MONTH.length;
                         var delim1 = datePattern.substring(iDelim1, iDelim1 + 1);
                         var delim2 = datePattern.substring(iDelim2, iDelim2 + 1);
                         if (iDelim1 == orderMonth && iDelim2 == orderYear) {
                             dateRegexp = new RegExp("^(\\d{2})(\\d{2})(\\d{4})$");
                         } else if (iDelim1 == orderMonth) {
                             dateRegexp = new RegExp("^(\\d{2})(\\d{2})[" + delim2 + "](\\d{4})$");
                         } else if (iDelim2 == orderYear) {
                             dateRegexp = new RegExp("^(\\d{2})[" + delim1 + "](\\d{2})(\\d{4})$");
                         } else {
                             dateRegexp = new RegExp("^(\\d{2})[" + delim1 + "](\\d{2})[" + delim2 + "](\\d{4})$");
                         }
                         var matched = dateRegexp.exec(value);
                         if(matched != null) {
                             if (!isValidDate(matched[1], matched[2], matched[3])) {
                                 if (i == 0) {
                                     focusField = form[oDate[x][0]];
                                 }
                                 fields[i++] = oDate[x][1];
                                 bValid =  false;
                              }
                         } else {
                             if (i == 0) {
                                 focusField = form[oDate[x][0]];
                             }
                             fields[i++] = oDate[x][1];
                             bValid =  false;
                         }
                     } else if ((orderMonth > orderYear && orderMonth < orderDay)) {
                         var iDelim1 = orderYear + YEAR.length;
                         var iDelim2 = orderMonth + MONTH.length;
                         var delim1 = datePattern.substring(iDelim1, iDelim1 + 1);
                         var delim2 = datePattern.substring(iDelim2, iDelim2 + 1);
                         if (iDelim1 == orderMonth && iDelim2 == orderDay) {
                             dateRegexp = new RegExp("^(\\d{4})(\\d{2})(\\d{2})$");
                         } else if (iDelim1 == orderMonth) {
                             dateRegexp = new RegExp("^(\\d{4})(\\d{2})[" + delim2 + "](\\d{2})$");
                         } else if (iDelim2 == orderDay) {
                             dateRegexp = new RegExp("^(\\d{4})[" + delim1 + "](\\d{2})(\\d{2})$");
                         } else {
                             dateRegexp = new RegExp("^(\\d{4})[" + delim1 + "](\\d{2})[" + delim2 + "](\\d{2})$");
                         }
                         var matched = dateRegexp.exec(value);
                         if(matched != null) {
                             if (!isValidDate(matched[3], matched[2], matched[1])) {
                                 if (i == 0) {
                                     focusField = form[oDate[x][0]];
                                  }
                                  fields[i++] = oDate[x][1];
                                  bValid =  false;
                              }
                          } else {
                              if (i == 0) {
                                  focusField = form[oDate[x][0]];
                              }
                              fields[i++] = oDate[x][1];
                              bValid =  false;
                          }
                     } else {
                         if (i == 0) {
                             focusField = form[oDate[x][0]];
                         }
                         fields[i++] = oDate[x][1];
                         bValid =  false;
                     }
                  }
               }
               if (fields.length > 0) {
                  focusField.focus();
                  alert(fields.join('\n'));
               }
               return bValid;
            }

Answer 4

路上的怎么把javascript也搬出来了？

Answer 5

SimpleDateFormat的parse方法最方便，但是如何使yyyyMMddHH这里面的HH必须是2位数，
比如2005020401可以，但200502041就不行呢？

Answer 6

最好用捕捉异常
import java.util.*;public class formular {   public static void main(String[] args) {
      System.out.println(isvaliedate("2004022918"));
   }   public static boolean isvaliedate(String timestr) {
      if (timestr.length() != 10)
         return false;
      int y, m, d, h, yy, mm, dd, hh;
      try {
         y = Integer.parseInt(timestr.substring(0, 4));
         m = Integer.parseInt(timestr.substring(4, 6)) - 1;
         d = Integer.parseInt(timestr.substring(6, 8));
         h = Integer.parseInt(timestr.substring(8, 10));
      } catch (NumberFormatException e) {
         return false;
      }
      Calendar dt = Calendar.getInstance();
      dt.clear();
      dt.set(y, m, d, h, 0, 0);
      yy = dt.get(Calendar.YEAR);
      mm = dt.get(Calendar.MONTH);
      dd = dt.get(Calendar.DATE);
      hh = dt.get(Calendar.HOUR);
      if (dt.get(Calendar.AM_PM) == Calendar.PM)
         hh += 12;
      if (yy != y || mm != m || dd != d || hh != h) {
         return false;
      }
      return true;
   }
}

Answer 7

建议不使用SimpleDateFormat，使用Calendar，很好用

Answer 8

回复人： labrun(labrun) ( ) 信誉：100  2005-02-04 17:43:00  得分: 0  
 
 
   SimpleDateFormat的parse方法最方便，但是如何使yyyyMMddHH这里面的HH必须是2位数，
比如2005020401可以，但200502041就不行呢？
  
 
－－－－－－－－－－－－－－－－－－－－－－－
我前面的回复中有啊，加上这么一句：
fmt.setLenient(false);

Answer 9

回复人： fogs(菜鸟想飞) ( ) 信誉：100  2005-02-04 21:16:00  得分: 0  
 
 
   建议不使用SimpleDateFormat，使用Calendar，很好用
  
 
－－－－－－－－－－－－－－－－－－－－－－－－
why?请问SimpleDateFormat有什么不妥么？

Answer 10

回复人： labrun(labrun) ( ) 信誉：100  2005-02-04 17:43:00  得分: 0  
 
 
   SimpleDateFormat的parse方法最方便，但是如何使yyyyMMddHH这里面的HH必须是2位数，
比如2005020401可以，但200502041就不行呢？
  
 
－－－－－－－－－－－－－－－－－－－－－－－
最近老眼昏花，老是看错，真不好意思
对于你的要求，setLenient()方法也不行

Answer 11

捕获异常是极其消耗资源的
用捕获异常来进行判断是异常的一种错误用法，在这里正确的方法是用正则表达式或自己分析(同样不能用捕获Integer.parseInt()的方式进行，而是应该先看这个字符串是不是全是数字s.char(i)在字符'0'到'9'之间)

Answer 12

如果是文本框的话，可以用JFormattedTextField一开始就阻止用户输入非法的数值啊！