Python For loop从云函数在云存储中写入多个文件

我在python中有一个云函数来调用RESTAPI get。函数响应为传递给函数的每个json文件名返回一个json文件。该函数只需一次调用即可完美运行：

def request_func(x):
    url = 'https://use4.dm-us.informaticacloud.com/saas/api/v2/'+ x
    payload_headers = {'content-type': 'application/json','Accept': 'application/json','icSessionId':ricSessionId}
    r = requests.get(url,headers=payload_headers)
    data = r.json()
    return data

def upload_blob(bucket_name, blob_text, destination_blob_name):
    """Uploads a file to the bucket."""
    storage_client = storage.Client()
    bucket = storage_client.get_bucket(bucket_name)
    blob = bucket.blob(destination_blob_name)
    blob.upload_from_string(blob_text)

wf_name = 'JOB_NAME'

data = request_func('workflow/name/'+ wf_name)
rid = data['id']
data = request_func('activity/activityLog?taskId='+rid)

def monitoramento(request):
    BUCKET_NAME = 'bucketname'
    BLOB_NAME = wf_name
    BLOB_STR = json.dumps(data)
    upload_blob(BUCKET_NAME, BLOB_STR, BLOB_NAME)
    return f'{wf_name} forecast json file success uploaded!'

我需要将几个json文件传递给函数，然后我对函数进行了一些调整，以使用FOR循环：

def request_func(x):
    url = 'https://use4.dm-us.informaticacloud.com/saas/api/v2/'+ x
    payload_headers = {'content-type': 'application/json','Accept': 'application/json','icSessionId':ricSessionId}
    r = requests.get(url,headers=payload_headers)
    data = r.json()
    return data

def upload_blob(bucket_name, blob_text, destination_blob_name):
    """Uploads a file to the bucket."""
    storage_client = storage.Client()
    bucket = storage_client.get_bucket(bucket_name)
    blob = bucket.blob(destination_blob_name)
    blob.upload_from_string(blob_text)

list_jobs = ['JOB_NAME1','JOB_NAME2','JOB_NAME3']

for wf_name in list_jobs:

    data = request_func('workflow/name/'+ wf_name)
    rid = data['id']
    data = request_func('activity/activityLog?taskId='+rid)

    def monitoramento(request):
        BUCKET_NAME = 'bucketname'
        BLOB_NAME = wf_name
        BLOB_STR = json.dumps(data)
        upload_blob(BUCKET_NAME, BLOB_STR, BLOB_NAME)

问题是，只有列表的最后一个对象JOB_NAME3 json file被写入bucket！ 我做错了什么？

---

我省略了一些代码。

使用@AlexanderCécile技巧，我可以解决我的问题：

def request_func(x):
    url = 'https://use4.dm-us.informaticacloud.com/saas/api/v2/'+ x
    payload_headers = {'content-type': 'application/json','Accept': 'application/json','icSessionId':ricSessionId}
    r = requests.get(url,headers=payload_headers)
    data = r.json()
    return data

def upload_blob(bucket_name, blob_text, destination_blob_name):
    """Uploads a file to the bucket."""
    storage_client = storage.Client()
    bucket = storage_client.get_bucket(bucket_name)
    blob = bucket.blob(destination_blob_name)
    blob.upload_from_string(blob_text)

list_jobs = ['JOB_NAME1','JOB_NAME2','JOB_NAME3']

for wf_name in list_jobs:
    data = request_func('workflow/name/'+ wf_name)
    rid = data['id']
    data = request_func('activity/activityLog?taskId='+rid)
    BUCKET_NAME = 'monitoramento'
    BLOB_NAME = wf_name
    BLOB_STR = json.dumps(data)
    upload_blob(BUCKET_NAME, BLOB_STR, BLOB_NAME)

---

从哪里调用

monitoramento（）

？提示：不要忽略代码。在请求帮助时创建一个最小且可复制的示例。请不要使用所有大写字母来发布您的邮件。你能把它编辑一下，让它看起来不那么像喊叫吗？我建议你参加Stackoverflow之旅：。创建好的编程问题：为什么要在循环中定义函数？！据我所知，该函数从未运行过。@Robertalmeda它不应该运行，你确定吗？其他函数是否可以创建类似的输出？