java日期的格式校验

第一种思路:


格式:

MM/dd/yyyy HH:mm:ss \d{1,2}/\d{1,2}/\d{4}(\s\d{1,2}:\d{1,2}:\d{1,2})? 2019/12/12 11:11:11 \d{4}/\d{1,2}/\d{1,2}(\s\d{1,2}:\d{1,2}:\d{1,2})? 2017-04-31 11:00:00 ^(\d{4})-([0-1]\d)-([0-3]\d)\s([0-5]\d):([0-5]\d):([0-5]\d)$ 实现: //校验时间 2019/12/12 11:11:11 String serviceReg="\\d{4}/\\d{1,2}/\\d{1,2}(\\s\\d{1,2}:\\d{1,2}:\\d{1,2})?"; if(StringUtils.isNotBlank(serviceStartTime)){ boolean checkServiceStartTime = serviceStartTime.matches(serviceReg); if(!checkServiceStartTime){ archivesSaleInfoImport.setImportFailReason("包含不合法的服务期限"); failList.add(archivesSaleInfoImport); continue; } } if(StringUtils.isNotBlank(serviceEndTime)){ boolean checkServiceEndTime = serviceEndTime.matches(serviceReg); if(!checkServiceEndTime){ archivesSaleInfoImport.setImportFailReason("包含不合法的服务期限"); failList.add(archivesSaleInfoImport); continue; } } 另一种思路: public static boolean isValidDate(String str) { boolean convertSuccess=true;      // 指定日期格式为四位年/两位月份/两位日期,注意yyyy/MM/dd区分大小写; SimpleDateFormat format = new SimpleDateFormat("yyyy/MM/dd HH:mm"); try {      // 设置lenient为false. 否则SimpleDateFormat会比较宽松地验证日期,比如2007/02/29会被接受,并转换成2007/03/01 format.setLenient(false); format.parse(str); } catch (ParseException e) { // e.printStackTrace(); // 如果throw java.text.ParseException或者NullPointerException,就说明格式不对 convertSuccess=false; } return convertSuccess; } 基于Jdk1.8线程安全的校验: try { String date = "2021/11/30 11:11:59"; final DateTimeFormatter DATE_TIME_FORMATTER = DateTimeFormatter.ofPattern("yyyy/MM/dd HH:mm:ss"); LocalDateTime dateTime = LocalDateTime.parse(date, DATE_TIME_FORMATTER); // LOCAL_DATE_FORMAT.get().parse(date); System.out.println(dateTime.toString().replaceAll("T"," ")); } catch (DateTimeParseException e) { System.out.println(e); }