如何为下面的curl请求编写python代码，以发送post请求并将图像作为二进制文件上载

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编辑：我添加了python示例中的url，因为curl中的url不完整。但是它仍然需要两个值

providerPartition

和

providerPartitionId

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在上，您可以将

curl

转换为python代码。大多数情况下，它是有效的

这里是本页的代码。但我不能测试它

App_Url = "https://p7a0km3l6k.execute-api.us-east-1.amazonaws.com/preprod/v1/images/trademark/metadata/providerPartition/{providerPartition}/providerPartitionId/{providerPartitionId}"
    # f = open('C://Users//UX016491//PycharmProjects//DSSApi//data1.json')
    # requests_json = json.loads(f.read())
    files = {'media' : open('C:\\Users\\UX016491\\Desktop\\images\\image123.jpg','rb') }
    response = requests.request("POST", App_Url, files = files, headers={"content-type": 'application/octet-stream'})
    print(response)

if __name__ == '__main__':
    test_createimage_data()

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您还可以使用url测试这两种方法，它会将从您那里得到的信息发送回来。您可以比较两个请求的结果。我测试了

curl

和

python代码

，我使用了相同的信息，因此它们应该具有相同的效果。

在上，您可以将curl转换为python代码我尝试了代码。。但我正在使用退出代码0maybe完成下面的响应过程，因为您的卷曲不完整。现在我在python代码中看到您使用了不同的url，它有额外数据的位置

{providerPartition}

{providerPartitionId}

App_Url = "https://p7a0km3l6k.execute-api.us-east-1.amazonaws.com/preprod/v1/images/trademark/metadata/providerPartition/{providerPartition}/providerPartitionId/{providerPartitionId}"
    # f = open('C://Users//UX016491//PycharmProjects//DSSApi//data1.json')
    # requests_json = json.loads(f.read())
    files = {'media' : open('C:\\Users\\UX016491\\Desktop\\images\\image123.jpg','rb') }
    response = requests.request("POST", App_Url, files = files, headers={"content-type": 'application/octet-stream'})
    print(response)

if __name__ == '__main__':
    test_createimage_data()

import requests

# incompletet url from curl
#url = 'https://c6y09pww43.execute-api.us-east-1.amazonaws.com/p'

providerPartition = '??'
providerPartitionId = '??'

url = f'https://p7a0km3l6k.execute-api.us-east-1.amazonaws.com/preprod/v1/images/trademark/metadata/providerPartition/{providerPartition}/providerPartitionId/{providerPartitionId}'


headers = {
    'Content-Type': 'application/octet-stream',
}

data = open('C:\\Users\\U6068366\\Downloads\\Koala.jpg', 'rb').read()

response = requests.post(url, headers=headers, data=data)

print(response.text)