脑壳疼？别再写满屏的 if-else 了，看我怎么消灭它！
在实际的业务开发当中，经常会遇到复杂的业务逻辑，可能部分同学实现出来的代码并没有什么问题，但是代码的可读性很差。
本篇文章主要总结一下自己在实际开发中如何避免大面积的 if-else 代码块的问题。补充说明一点，不是说 if-else 不好，而是多层嵌套的 if-else 导致代码可读性差、维护成本高等问题。
现有如下一段示例代码，部分优化技巧是根据这段代码进行的：
public class BadCodeDemo {
private void getBadCodeBiz(Integer city, List newDataList, List oldDataList) {
if (city != null) {
if (newDataList != null && newDataList.size() > 0) {
TestCodeData newData = newDataList.stream().filter(p -> {
if (p.getIsHoliday() == 1) {
return true;
}
return false;
}).findFirst().orElse(null);
if (newData != null) {
newData.setCity(city);
}
}
} else {
if (oldDataList != null && newDataList != null) {
List oldCollect = oldDataList.stream().filter(p -> {
if (p.getIsHoliday() == 1) {
return true;
}
return false;
}).collect(Collectors.toList());
List newCollect = newDataList.stream().filter(p -> {
if (p.getIsHoliday() == 1) {
return true;
}
return false;
}).collect(Collectors.toList());
if (newCollect != null && newCollect.size() > 0 && oldCollect != null && oldCollect.size() > 0) {
for (TestCodeData newPO : newCollect) {
if (newPO.getStartTime() == 0 && newPO.getEndTime() == 12) {
TestCodeData po = oldCollect.stream().filter(p -> p.getStartTime() == 0
&& (p.getEndTime() == 12 || p.getEndTime() == 24)).findFirst().orElse(null);
if (po != null) {
newPO.setCity(po.getCity());
}
} else if (newPO.getStartTime() == 12 && newPO.getEndTime() == 24) {
TestCodeData po = oldCollect.stream().filter(
p -> (p.getStartTime() == 12 || p.getStartTime() == 0)
&& p.getEndTime() == 24).findFirst().orElse(null);
if (po != null) {
newPO.setCity(po.getCity());
}
} else if (newPO.getStartTime() == 0 && newPO.getEndTime() == 24) {
TestCodeData po = oldCollect.stream().filter(
p -> p.getStartTime() == 0 && p.getEndTime() == 24).findFirst().orElse(null);
if (po == null) {
po = oldCollect.stream().filter(
p -> p.getStartTime() == 0 && p.getEndTime() == 12).findFirst().orElse(null);
}
if (po == null) {
po = oldCollect.stream().filter(
p -> p.getStartTime() == 12 && p.getEndTime() == 24).findFirst().orElse(null);
}
if (po != null) {
newPO.setCity(po.getCity());
}
} else if (newPO.getTimeUnit().equals(Integer.valueOf(1))) {
TestCodeData po = oldCollect.stream().filter(
e -> e.getTimeUnit().equals(Integer.valueOf(1))).findFirst().orElse(null);
if (po != null) {
newPO.setCity(po.getCity());
}
}
}
}
}
}
}
}
if(null != city) {
} else {
}
private void getCityNotNull(Integer city, List newDataList ) {
if (newDataList != null && newDataList.size() > 0) {
TestCodeData newData = newDataList.stream().filter(p -> {
if (p.getIsHoliday() == 1) {
return true;
}
return false;
}).findFirst().orElse(null);
if (newData != null) {
newData.setCity(city);
}
}
}
// 合并逻辑流程
private void getBadCodeBiz(Integer city, List newDataList, List oldDataList ) {
if (city != null) {
this.getCityNull(city, newDataList);
} else {
//此处代码省略
}
}
public void getCityNotNull(Integer city, List newDataList) {
if (CollectionUtils.isEmpty(newDataList)) {
// 提前判断，返回业务逻辑
return;
}
TestCodeData newData = newDataList.stream().filter(p -> {
if (p.getIsHoliday() == 1) {
return true;
}
return false;
}).findFirst().orElse(null);
if (null != newData) {
newData.setCity(city);
}
}
public class BadCodeDemo {
public void getBadCodeBiz(Integer city, List newDataList, List oldDataList) {
if (city != null) {
this.getCityNotNull(city, newDataList);
} else {
this.getCityNull(newDataList, oldDataList);
}
}
private void getCityNotNull(Integer city, List newDataList) {
if (CollectionUtils.isEmpty(newDataList)) {
// 提前判断，返回业务逻辑
return;
}
TestCodeData newData = newDataList.stream().filter(p -> {
if (p.getIsHoliday() == 1) {
return true;
}
return false;
}).findFirst().orElse(null);
if (null != newData) {
newData.setCity(city);
}
}
private void getCityNull(List newDataList, List oldDataList) {
// 提前判断，返回业务逻辑
if (CollectionUtils.isEmpty(oldDataList) && CollectionUtils.isEmpty(newDataList)) {
return;
}
List oldCollect = oldDataList.stream().filter(p -> {
if (p.getIsHoliday() == 1) {
return true;
}
return false;
}).collect(Collectors.toList());
List newCollect = newDataList.stream().filter(p -> {
if (p.getIsHoliday() == 1) {
return true;
}
return false;
}).collect(Collectors.toList());
// 提前判断，返回业务逻辑
if (CollectionUtils.isEmpty(newCollect) && CollectionUtils.isEmpty(oldCollect)) {
return;
}
for (TestCodeData newPO : newCollect) {
if (newPO.getStartTime() == 0 && newPO.getEndTime() == 12) {
TestCodeData po = oldCollect.stream().filter(p -> p.getStartTime() == 0
&& (p.getEndTime() == 12 || p.getEndTime() == 24)).findFirst().orElse(null);
if (po != null) {
newPO.setCity(po.getCity());
}
} else if (newPO.getStartTime() == 12 && newPO.getEndTime() == 24) {
TestCodeData po = oldCollect.stream().filter(
p -> (p.getStartTime() == 12 || p.getStartTime() == 0)
&& p.getEndTime() == 24).findFirst().orElse(null);
if (po != null) {
newPO.setCity(po.getCity());
}
} else if (newPO.getStartTime() == 0 && newPO.getEndTime() == 24) {
TestCodeData po = oldCollect.stream().filter(
p -> p.getStartTime() == 0 && p.getEndTime() == 24).findFirst().orElse(null);
if (po == null) {
po = oldCollect.stream().filter(
p -> p.getStartTime() == 0 && p.getEndTime() == 12).findFirst().orElse(null);
}
if (po == null) {
po = oldCollect.stream().filter(
p -> p.getStartTime() == 12 && p.getEndTime() == 24).findFirst().orElse(null);
}
if (po != null) {
newPO.setCity(po.getCity());
}
} else if (newPO.getTimeUnit().equals(Integer.valueOf(1))) {
TestCodeData po = oldCollect.stream().filter(
e -> e.getTimeUnit().equals(Integer.valueOf(1))).findFirst().orElse(null);
if (po != null) {
newPO.setCity(po.getCity());
}
}
}
}
}
if (newPO.getStartTime() == 0 && newPO.getEndTime() == 12) {
//第一段逻辑
} else if (newPO.getStartTime() == 12 && newPO.getEndTime() == 24) {
//第二段逻辑
} else if (newPO.getStartTime() == 0 && newPO.getEndTime() == 24) {
//第三段逻辑
} else if (newPO.getTimeUnit().equals(Integer.valueOf(1))) {
//第四段逻辑
}
public enum TimeEnum {
AM("am", "上午") {
@Override
public void setCity(TestCodeData data, List oldDataList) {
TestCodeData po = oldDataList.stream().filter(p -> p.getStartTime() == 0
&& (p.getEndTime() == 12 || p.getEndTime() == 24)).findFirst().orElse(null);
if (null != po) {
data.setCity(po.getCity());
}
}
},
PM("pm", "下午") {
@Override
public void setCity(TestCodeData data, List oldCollect) {
TestCodeData po = oldCollect.stream().filter(
p -> (p.getStartTime() == 12 || p.getStartTime() == 0)
&& p.getEndTime() == 24).findFirst().orElse(null);
if (po != null) {
data.setCity(po.getCity());
}
}
},
DAY("day", "全天") {
@Override
public void setCity(TestCodeData data, List oldCollect) {
TestCodeData po = oldCollect.stream().filter(
p -> p.getStartTime() == 0 && p.getEndTime() == 24).findFirst().orElse(null);
if (po == null) {
po = oldCollect.stream().filter(
p -> p.getStartTime() == 0 && p.getEndTime() == 12).findFirst().orElse(null);
}
if (po == null) {
po = oldCollect.stream().filter(
p -> p.getStartTime() == 12 && p.getEndTime() == 24).findFirst().orElse(null);
}
if (po != null) {
data.setCity(po.getCity());
}
}
},
HOUR("hour", "小时") {
@Override
public void setCity(TestCodeData data, List oldCollect) {
TestCodeData po = oldCollect.stream().filter(
e -> e.getTimeUnit().equals(Integer.valueOf(1))).findFirst().orElse(null);
if (po != null) {
data.setCity(po.getCity());
}
}
};
public abstract void setCity(TestCodeData data, List oldCollect) ;
private String code;
private String desc;
TimeEnum(String code, String desc) {
this.code = code;
this.desc = desc;
}
public String getCode() {
return code;
}
public void setCode(String code) {
this.code = code;
}
public String getDesc() {
return desc;
}
public void setDesc(String desc) {
this.desc = desc;
}
}
for (TestCodeData data : newCollect) {
if (data.getStartTime() == 0 && data.getEndTime() == 12) {
TimeEnum.AM.setCity(data, oldCollect);
} else if (data.getStartTime() == 12 && data.getEndTime() == 24) {
TimeEnum.PM.setCity(data, oldCollect);
} else if (data.getStartTime() == 0 && data.getEndTime() == 24) {
TimeEnum.DAY.setCity(data, oldCollect);
} else if (data.getTimeUnit().equals(Integer.valueOf(1))) {
TimeEnum.HOUR.setCity(data, oldCollect);
}
}
for (TestCodeData data : newCollect) {
String code = "am"; // 这里假设 code 变量是从 data 中获取的
TimeEnum.valueOf(code).setCity(data, oldCollect);
}
业务场景描述：比如让你做一个简单的营销拉新活动，这个活动投放到不同的渠道，不同渠道过来的用户奖励不一样。另外，公众号Java精选，回复java面试获取更多面试题资料，支持在线随时随地刷题。
@RestController
@RequestMapping("/activity")
public class ActivityController {
@Resource
private AwardService awardService;
@PostMapping("/reward")
public void reward(String userId, String source) {
if ("toutiao".equals(source)) {
awardService.toutiaoReward(userId);
} else if ("wx".equals(source)) {
awardService.wxReward(userId);
}
}
}
@Service
public class AwardService {
private static final Logger log = LoggerFactory.getLogger(AwardService.class);
public Boolean toutiaoReward(String userId) {
log.info("头条渠道用户{}奖励50元红包!", userId);
return Boolean.TRUE;
}
public Boolean wxReward(String userId) {
log.info("微信渠道用户{}奖励100元红包!", userId);
return Boolean.TRUE;
}
}
看完这段代码，逻辑上是没有什么问题的。但它有一个隐藏的缺陷，如果后期又增加很多渠道的时候，你该怎么办？继续 else if 吗？
@RestController
@RequestMapping("/activity")
public class ActivityController {
@Resource
private AwardService awardService;
@PostMapping("/reward")
public void reward(String userId, String source) {
awardService.getRewardResult(userId, source);
}
}
@Service
public class AwardService {
private static final Logger log = LoggerFactory.getLogger(AwardService.class);
private MapBoolean >> sourceMap = new HashMap<>();
@PostConstruct
private void dispatcher() {
sourceMap.put("wx", (userId, source) -> this.wxReward(userId));
sourceMap.put("toutiao", (userId, source) -> this.toutiaoReward(userId));
}
public Boolean getRewardResult(String userId, String source) {
BiFunctionBoolean> result = sourceMap.get(source);
if (null != result) {
return result.apply(userId, source);
}
return Boolean.FALSE;
}
private Boolean toutiaoReward(String userId) {
log.info("头条渠道用户{}奖励50元红包!", userId);
return Boolean.TRUE;
}
private Boolean wxReward(String userId) {
log.info("微信渠道用户{}奖励100元红包!", userId);
return Boolean.TRUE;
}
}
针对一些复杂的业务场景，业务参数很多时，可以利用 @FunctionalInterface 自定义函数式接口来满足你的业务需求，使用原理和本例并无差别。
设计模式对于 if-else 的优化，我个人觉得有些重，但是也是一种优化方式。设计模式适合使用在大的业务流程和场景中使用，针对代码块中的 if-else 逻辑优化不推荐使用。
常用的设计模式有：
策略模式
模板方法
工厂模式
单例模式
还是以上面的营销拉新活动为例来说明如何使用。
| 使用技巧一：工厂模式+抽象类
public abstract class AwardAbstract {
public abstract Boolean award(String userId);
}
// 头条渠道发放奖励业务
public class TouTiaoAwardService extends AwardAbstract {
@Override
public Boolean award(String userId) {
log.info("头条渠道用户{}奖励50元红包!", userId);
return Boolean.TRUE;
}
}
// 微信渠道发放奖励业务
public class WeChatAwardService extends AwardAbstract {
@Override
public Boolean award(String userId) {
log.info("微信渠道用户{}奖励100元红包!", userId);
return Boolean.TRUE;
}
}
public class AwardFactory {
public static AwardAbstract getAwardInstance(String source) {
if ("toutiao".equals(source)) {
return new TouTiaoAwardService();
} else if ("wx".equals(source)) {
return new WeChatAwardService();
}
return null;
}
}
@PostMapping("/reward2")
public void reward2(String userId, String source) {
AwardAbstract instance = AwardFactory.getAwardInstance(source);
if (null != instance) {
instance.award(userId);
}
}
| 使用技巧二：策略模式+模板方法+工厂模式+单例模式
还是以营销拉新为业务场景来说明，这个业务流程再增加一些复杂度，比如发放奖励之前要进行身份验证、风控验证等一些列的校验，此时你的业务流程该如何实现更清晰简洁呢？
/** 策略业务接口 */
public interface AwardStrategy {
/**
* 奖励发放接口
*/
Map awardStrategy(String userId) ;
/**
* 获取策略标识,即不同渠道的来源标识
*/
String getSource();
}
public abstract class BaseAwardTemplate {
private static final Logger log = LoggerFactory.getLogger(BaseAwardTemplate.class);
//奖励发放模板方法
public Boolean awardTemplate(String userId) {
this.authentication(userId);
this.risk(userId);
return this.awardRecord(userId);
}
//身份验证
protected void authentication(String userId) {
log.info("{} 执行身份验证!", userId);
}
//风控
protected void risk(String userId) {
log.info("{} 执行风控校验!", userId);
}
//执行奖励发放
protected abstract Boolean awardRecord(String userId);
}
@Slf4j
@Service
public class ToutiaoAwardStrategyService extends BaseAwardTemplate implements AwardStrategy {
/**
* 奖励发放接口
*/
@Override
public Boolean awardStrategy(String userId) {
return super.awardTemplate(userId);
}
@Override
public String getSource() {
return "toutiao";
}
/**
* 具体的业务奖励发放实现
*/
@Override
protected Boolean awardRecord(String userId) {
log.info("头条渠道用户{}奖励50元红包!", userId);
return Boolean.TRUE;
}
}
@Slf4j
@Service
public class WeChatAwardStrategyService extends BaseAwardTemplate implements AwardStrategy {
/**
* 奖励发放接口
*/
@Override
public Boolean awardStrategy(String userId) {
return super.awardTemplate(userId);
}
@Override
public String getSource() {
return "wx";
}
/***
* 具体的业务奖励发放实现
*/
@Override
protected Boolean awardRecord(String userId) {
log.info("微信渠道用户{}奖励100元红包!", userId);
return Boolean.TRUE;
}
}
@Component
public class AwardStrategyFactory implements ApplicationContextAware {
private final static Map MAP = new HashMap<>();
@Override
public void setApplicationContext(ApplicationContext applicationContext) throws BeansException {
Map beanTypeMap = applicationContext.getBeansOfType(AwardStrategy.class);
beanTypeMap.values().forEach(strategyObj -> MAP.put(strategyObj.getSource(), strategyObj));
}
/**
* 对外统一暴露的工厂方法
*/
public Boolean getAwardResult(String userId, String source) {
AwardStrategy strategy = MAP.get(source);
if (Objects.isNull(strategy)) {
throw new RuntimeException("渠道异常!");
}
return strategy.awardStrategy(userId);
}
/**
* 静态内部类创建单例工厂对象
*/
private static class CreateFactorySingleton {
private static AwardStrategyFactory factory = new AwardStrategyFactory();
}
public static AwardStrategyFactory getInstance() {
return CreateFactorySingleton.factory;
}
}
@RestController
@RequestMapping("/activity")
public class ActivityController {
@PostMapping("/reward3")
public void reward3(String userId, String source) {
AwardStrategyFactory.getInstance().getAwardResult(userId, source);
}
}
2022-02-20 12:23:27.716 INFO 20769 --- [nio-8080-exec-1] c.a.c.e.o.c.p.s.BaseAwardTemplate : fei 执行身份验证!
2022-02-20 12:23:27.719 INFO 20769 --- [nio-8080-exec-1] c.a.c.e.o.c.p.s.BaseAwardTemplate : fei 执行风控校验!
2022-02-20 12:23:27.719 INFO 20769 --- [nio-8080-exec-1] a.c.e.o.c.p.s.WeChatAwardStrategyService : 微信渠道用户fei奖励100元红包!
其他技巧：
使用三目运算符
相同业务逻辑提取复用
不论使用那种技巧，首先是我们在业务代码开发过程中一定要多思考，将复杂的业务逻辑能通过简洁的代码表现出来，这才是你的核心能力之一，而不是一个 curd boy。与君共勉，共同进步！
作者：大飞学习笔记
https://juejin.cn/post/7066643792970907679
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