• 代码的优化(二)---不要写重复代码


    如果可以的话,重复逻辑不要多次判断,代码会非常的冗余。
    这段代码没有修改之前,大概300行,优化之后只有100行左右。


    /**
    * 判断是否可以执行退还保证金操作 * 1.判断该条保证金缴纳记录是否可以退还 * 2.如果可以退还,退还的金额是多少? * * 根据时间做判断? * 不用凑整数,可以退就直接退掉 */ @Override public Map<String, Object> returnCheck(String bailPayId,Long customerId) { logger.debug("returnCheck,bailPayId {},customerId{}",bailPayId,customerId); Map<String,Object> map = null; BigDecimal returnBailPay = new BigDecimal(0); BailPay bailPay = bailPayRepository.findOne(bailPayId); if (bailPay != null) { int bailClass = bailPay.bailClass();//该条记录的保证金缴纳的类别 String agencyId = bailPay.agencyId();//缴纳该条保证金的机构id值 BigDecimal maxCommonBailLock; BigDecimal bailLockSpecialSum; if (bailClass == 1) {//通用保证金 //如果当前竞买人存在保证金冻结记录状态为"锁定"的状态,不能够执行保证金退还操作 int bailLockCount = this.bailLockCommonQuery(customerId); if (bailLockCount != 0) { map = new HashMap<String,Object>(); map.put("flag", 0);//不可以返还 map.put("returnBailPay", 0);//返还的金额为 0 return map; } maxCommonBailLock = this.checkMaxCommonFrozenBailPay(customerId); bailLockSpecialSum = new BigDecimal(0); } else {//专项保证金 maxCommonBailLock = this.bailLockSpecialQuery(customerId,agencyId); bailLockSpecialSum = this.bailLockSpecialQuery(customerId,agencyId); } BigDecimal canUseBailLockSum = new BigDecimal(0); //查询出当前该笔保证金缴纳记录之前缴纳的所有的缴纳记录的总和 Specification<BailPay> spec = (root, query, cb) -> { List<Predicate> predicates = new ArrayList<Predicate>(); Predicate predicate1 = cb.equal(root.get(BailPay_.customerId), customerId);//当前登录用户的id predicates.add(predicate1); try { SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); Predicate predicate2 = cb.lessThan(root.<Date>get(BailPay_.crtTime), sdf.parse(bailPay.crtTime().toString()));// predicates.add(predicate2); } catch (ParseException e) { e.printStackTrace(); } Predicate predicate3 = cb.equal(root.get(BailPay_.bailClass), BailPay.BAIL_CLASS_TONGYONG);//通用保证金 predicates.add(predicate3); //判断缴纳时间早于当前该条记录的缴纳记录 if (!predicates.isEmpty()) { return cb.and(predicates.toArray(new Predicate[0])); } else { return null; } }; List<BailPay> bailPayList = bailPayRepository.findAll(spec); if (bailPayList != null && bailPayList.size()>0) { for (BailPay bailPay2 : bailPayList) { if (bailPay2 != null) { canUseBailLockSum = canUseBailLockSum.add(bailPay2.remainBailPay());//每一条缴纳记录的剩余金额的总和 } } } if(canUseBailLockSum.compareTo(maxCommonBailLock.add(bailLockSpecialSum)) != -1){ //可以退还该条保证金缴纳记录的金额 map = new HashMap<String,Object>(); map.put("flag", 1);//可以返还 map.put("returnBailPay", returnBailPay);//返还的金额 return map; } } map = new HashMap<String,Object>(); map.put("flag", 0);//不可以返还 map.put("returnBailPay", 0);//返还的金额 return map; }
  • 相关阅读:
    Codeforces 912 D. Fishes (贪心、bfs)
    Codeforces 908 D.New Year and Arbitrary Arrangement (概率&期望DP)
    HDU
    HDU
    POJ-2299 Ultra-QuickSort (树状数组)
    deque!
    HDU
    乘法逆元
    Codeforces 911D. Inversion Counting (数学、思维)
    Codeforces 909E. Coprocessor (拓扑、模拟)
  • 原文地址:https://www.cnblogs.com/mr-wuxiansheng/p/6180538.html
Copyright © 2020-2023  润新知