为什么'；python补丁请求不能用于循环吗？
我有一个代码，我想在其中为每个用户发出补丁请求（请参阅“for user in users”中的）。循环内的打印输出如下： 5ee5eee85cb7558196e1f742 101 5ee5eee85cb7558196e1f742 108 5EF9A4 EC59BD2A2ABDD13D 242 5ee5eee85cb7558196e1f742 122 5ee5eee85cb7558196e1f742 129 5EF9A4 EC59BD2A2ABDD13D 263 5ee5eee85cb7558196e1f742 143 5ee5eee85cb7558196e1f742 一百五十 5EF9A4 EC59BD2A2ABDD13D 284 但是，该补丁只制作了一次，并且仅用于第一个“5ee5eee85cb7558196e1f742”和第一个“5EF9A4 EC59BD2A2AABD13D”，而不是循环中的所有其他时间。有人能帮我吗为什么'；python补丁请求不能用于循环吗？,python,python-requests,patch,Python,Python Requests,Patch,我有一个代码，我想在其中为每个用户发出补丁请求（请参阅“for user in users”中的）。循环内的打印输出如下： 5ee5eee85cb7558196e1f742 101 5ee5eee85cb7558196e1f742 108 5EF9A4 EC59BD2A2ABDD13D 242 5ee5eee85cb7558196e1f742 122 5ee5eee85cb7558196e1f742 129 5EF9A4 EC59BD2A2ABDD13D 263 5ee5eee85cb75581
extrapoints=0
for position in positions:
position_type = position['position_type']
#an kapoiois user stathmeusi
if position_type == "parking":
#kratame tis suntetagmenes thesis gia na doume an thn exei anaferei kapoios reporter
coordinatesx = position['location']['coordinates'][0]
coordinatesy = position['location']['coordinates'][1]
#pame sti sullogi reports na doume an uparxoun oi suntetagmenes se kapoia anafora
for report in reports:
#rid = report['reporter_id']
rcoordinatesy = report['location']['coordinates'][1]
rcoordinatesx = report['location']['coordinates'][0]
#extrapoints = 0
if rcoordinatesx==coordinatesx and rcoordinatesy==coordinatesy:
#an kapoia anafora exei tetoies suntetagmenes prosthetoume 7 pontous ston reporter
#rid = report['reporter_id']
rid= report['reporter_id']
extrapoints = extrapoints + 7
for user in users:
userid = user['_id']
rewarding_p = user['rewarding_points']
if userid == rid:
etag = user['_etag']
print (etag)
rewarding_p = rewarding_p + extrapoints
print(userid)
print(rewarding_p)
data={"rewarding_points":rewarding_p}
headers = {'Authorization':'Bearer 8b59cc74f15befc05b48f90d5e09866028b2b624','If-Match': etag}
url = 'https://socialpark.iti.gr:8006/users'
new_url = "{}/{}".format(url, userid)
r = requests.patch(new_url, data, headers=headers)
time.sleep(2)