python - 无法从 slack 打印用户名 "@John"以及如何在不手动输入 cntrl+c 的情况下关闭 Linux 中的 python 文件

Question

Q1) Unable to print username from slack as "@John" instead its printing "WLCLZ2LDP"

So for variable "username" it should print slack username ( for example : JohnDoe) but instead printing randomized letter as WLCLZ2LDP

代码

@slack.RTMClient.run_on(event='message')
def say_hello(**payload):
    try:
        ##0 get clients and payload
        logging.info('msg received')

        data = payload['data']
        web_client = payload['web_client']
        rtm_client = payload['rtm_client']

        ##0 - 1 Check if it is the first msg, not replied msg by me
        # print(data)
        if data.get('text') == None:
            logging.info('This msg is my replied msg.')
            return False
        ##0-2 Get channel info
        channel_id = data['channel']
        thread_ts = data['ts']
        global user
        user = data['user']

        ##1 get scenario submsg
        msg = data['text']
        retVal = analysis_msg(msg)
.
.
.
.code continuation 
.
.
.
.mycursor = mydb.cursor()
for value in submsg:
  print(value)
fulfilment=submsg[0]
sku=submsg[1]
environment=submsg[2]
payment=submsg[3]
ts = time.time()
date = datetime.datetime.fromtimestamp(ts).strftime('%Y-%m-%d %H:%M:%S')
orderNumber=res
username=user

print(fulfilment)
print(sku)
print(environment)
print(payment)
print(username)
print(orderNumber)

sqlformula = "INSERT INTO orderDetails (fulfilment,sku,environment,payment,orderNumber,date,user) VALUES (%s,%s,%s,%s,%s,%s,%s)"

#order=("sth",3643387,"sftqa2","AMEX")
#mycursor.execute(sqlformula,order)

mycursor.execute(sqlformula,(fulfilment,sku,environment,payment,orderNumber,date,username))

mydb.commit()
mydb.close()

输出

下面的输出我需要获得松弛的用户名，但我得到了随机字母

spu
3646989
mq1
AMEX
WLCLZ2LDP
"200002086748"

编辑后的代码

@slack.RTMClient.run_on(event='message')
def say_hello(**payload):
    try:
        ##0 get clients and payload
        logging.info('msg received')
        data = payload['data']
        web_client = payload['web_client']
        rtm_client = payload['rtm_client']
        ##0 - 1 Check if it is the first msg, not replied msg by me
        # print(data)
        if data.get('text') == None:
            logging.info('This msg is my replied msg.')
            return False
        ##0-2 Get channel info
        channel_id = data['channel']
        thread_ts = data['ts']

        global user
        user = data['user']
        msg = data['text']
        retVal = analysis_msg(msg)

        response = web_client.users_list()
        assert(response['ok'])
        user_map = {x['id']: x['name'] for x in response['members']}
        global user_name
        user_name = user_map[user] if user in user_map else None
        print(user_name)

情况 1: 编辑代码的输出 - 当其他用户尝试从他们的计算机获取“None”(非预期)时

spu
2469048
sftqa3
AMEX
None
BBY01-200002096736

案例 2:编辑代码的输出:当我从我的机器上运行时，我能够看到我的辅助，即 A6002043(预期)

spu
2469048
sftqa3
VISA
a6002043
BBY01-200002096040

Answer 1

请允许我澄清两个误解:

• 这些不是“随机字母”，而是用户 ID。
• 事件不包含用户名，仅包含用户 ID。

要获取用户名，您需要调用API方法users.list获取用户列表。这样您就可以将 ID 映射到用户名。

以下是如何获取用户 ID 的用户名的示例代码:

slack_token = os.environ['SLACK_TOKEN']
response = client.users_list()
assert(response['ok'])
user_map = {x['id']: x['name'] for x in response['members']}
user_name = user_map[user_id] if user_id in user_map else None
print(user_name)