对于过多if-else分支的优化

场景1：equals判断

int code;
if("Name".equals(str))
    code = 0;
else if("Age".equals(str))
    code = 1;
else if("Address".equals(str))
    code = 2;
...

可以用Map进行优化:

Map typeCodeMap = new HashMap();
typeCodeMap.put("Name", 0);
typeCodeMap.put("Age", 1);
typeCodeMap.put("Address", 2);
...
int code = typeCode.get(type);

场景2： 判断多个变量是否为空

        if (StringUtils.isBlank(a)) {
            response.setSuccess(false);
            response.setMessage("a can not be null");
        } else if (StringUtils.isBlank(b)) {
            response.setSuccess(false);
            response.setMessage("b can not be null");
        } else if (StringUtils.isBlank(c)) {
            response.setSuccess(false);
            response.setMessage("c can not be null");
        } else if (StringUtils.isBlank(d)) {
            response.setSuccess(false);
            response.setMessage("d can not be null");
        } else if (StringUtils.isBlank(e)) {
            response.setSuccess(false);
            response.setMessage("e can not be null");
        } 

用Map进行优化:

        Map<String, String> paramMap = new HashMap<>(5);
        paramMap.put("a", a);
        paramMap.put("b", b);
        paramMap.put("c", c);
        paramMap.put("d", d);
        paramMap.put("e", e);
        for (Map.Entry<String, String> entry : paramMap.entrySet()) {
            if (StringUtils.isBlank(entry.getValue())) {
                response.setSuccess(false);
                response.setMessage(entry.getKey() + " can not be null");
                return response;
            }
        }