两段请求报文,判断不一样的key和value,只判断d2里和d1不同的值,和全部不同的key
ok_req={ "version": "9.0.0", "is_test": True, "store": "", "urs": "", "device": { "os": "android", "imei": "99001062198893", "device_id": "CQliMWEyYTEzNTYyYzk5MzJmCTJlNmY3Zjkx", "mac": "02:00:00:00:00:00", "galaxy_tag": "CQliMWEyYTEzNTYyYzk5MzJmCTJlNmY3Zjkx", "udid": "a34b1f67dd5797df93fdd8b072f1fb8110fd0db6", "network_status": "wifi" }, "adunit": { "category": "VIDEO", "location": "1", "app": "7A16FBB6", "blacklist": "" }, "ext_param":{ "is_start" : 0, "vId":"VW0BRMTEV" }}not_ok={ "version": "9.0.0", "is_test": True, "urs": "1", "store": "", "device": { "os": "android", "imei": "99001062298893", "device_id": "CQliMWEyYTEzNTYyYzk5MzJmCTJlNmY3Zjkx", "mac": "02:00:00:00:00:00", "galaxy_tag": "CQliMWEyYTEzNTYyYzk5MzJmCTJlNmY3Zjkx", "udid": "a34b1f67dd5797da93fdd8b072f1fb8110fd0db6", "network_status": "wifi" }, "adunit": { "category": "VIDEO", "location": "1", "app": "7A16FBB6", "blacklist": "" }, "ext_param": { "is_start": 0, "vid": "VW0BRMTEV" }}
方法一的需求分析:
1. 循环d1的key,通过key去d2里取值,取不到的就是d2中不存在这个key,d2与d1里不一样的key
2. 判断通过key取值的类型,如果是dict类型的继续循环
3. 把d1和d2中的key转换成集合类型,取差集,取出的key即d1和d2中不一样的key
def compare(dic_1,dic_2): for k in dic_1: v1=dic_1.get(k) v2=dic_2.get(k,‘get不到值‘)#通过k去d2里取值,d2里如果没有这个key,返回get不到值 if type(v1)==dict: compare(v1,v2)#取值为dict类型递归 else: if v1 != v2 and v2 !=‘get不到值‘ : print (‘value不一样的:key是%s,v1是%s,v2是%s‘%(k,v1,v2)) r1 = set(dic_1.keys()) r2 = set(dic_2.keys()) res=r1.symmetric_difference(r2) print(‘两个请求报文中不一样的key是:‘,‘,‘.join(res)) compare(ok_req,not_ok)
方法二的需求分析:
1. 已知报文是二维,创建一个方法,把二维字典变成一维字典,key用特定的符号连接起来
2. 循环d1的key,取d2里面取值,如果v1==v2,就在d2里把这个键值对删除,不同的话,说明key是d1里与d2里k-v不同的数据
3. d2里剩下的全是和d1里k-v不同的数据,循环d2剩余的数据并输出
def buildDict(dict_0):#把报文的二维字典变成一维,二维字典的样式变成{一维字典key||二维key:value} dict_t = {} for key in dict_0: value = dict_0.get(key) if type(value) == dict: for k,v in value.items(): dict_t[key+‘||‘+k]=v else: dict_t[key]=value return dict_t def compare(ok_req,not_ok): dic_1 = buildDict(ok_req)#把ok_req报文变成一维的字典格式 dic_2 = buildDict(not_ok)#把not_ok报文变成一维的字典格式 for k in dic_1: v1 = dic_1.get(k) v2 = dic_2.get(k) if v1==v2: dic_2.pop(k)#把dict_2中key和value与dict_1中一样的删除 else: print(‘dic_1中数据不同的k-v,是‘ + k+‘:‘+dic_1.get(k)) for k in dic_2:#dic_2中剩余的都是和dic_1中不一样的key print(‘dic_2中不一样的k-v,是‘+k+‘:‘+dic_2.get(k)) compare(ok_req,not_ok)
原文地址:https://www.cnblogs.com/mpp0905/p/8280557.html
时间: 2024-10-12 20:02:57