使用boto3将文件上传到S3。我不知道用什么替换<FMI>

Question

import boto3
S3API = boto3.client("s3", region_name="us-east-1")
S3API.**<FMI>**(Bucket="tiwaladebucket")

import boto3
S3API = boto3.client("s3", region_name="us-east-1") 
bucket_name = "tiwaladebucketcatfound"

filename = "../cat.jpg"
S3API.**<FMI>**(filename, tiwaladebucketcatfound, "cat.jpg", ExtraArgs={'**<FMI>**': "image/jpg", "CacheControl": "max-age=0"})

filename = "../index.html"
S3API.**<FMI>**(filename, tiwaladebucketcatfound, "**<FMI>**", ExtraArgs={'ContentType': "text/html", "CacheControl": "max-age=0"})

Answer 1

您是s3客户端，需要使用put_object接口上传文件到S3

S3API.put_object(Body=binary_data, Bucket='my_bucket_name', Key='s3folder/path/filename.txt')

or 

S3API.put_object(Body=file_object, Bucket='my_bucket_name', Key='s3folder/path/filename.txt')

参考：https://boto3.amazonaws.com/v1/documentation/api/latest/reference/services/s3.html#S3.Client.put_object

或者，您也可以使用S3资源而不是S3客户端来执行相同操作

s3_resource = boto3.resource('s3')    
s3_resource.Bucket('bucketname').upload_file('/local/file/a.txt','folder/path/to/s3key')

Answer 2

有多个FMI，正如我所看到的，它们不能被相同的命令替换。更多关于代码尝试的信息会更好，但我猜你想要：

创建存储桶"iwaladebucket"

• upload两个文件到所述存储桶“../.html”和"../cat.jpg"

使用上面的代码，下面的代码将实现这一点：

import boto3
s3client = boto3.client("s3", region_name="us-east-1")
s3client.create_bucket(Bucket="tiwaladebucket")

bucket_name = "tiwaladebucket"

filename = "../cat.jpg"
s3client.put_object(filename, bucket_name, "cat.jpg", ExtraArgs={'ContentType': "image/jpg", "CacheControl": "max-age=0"})

filename = "../index.html"
s3client.put_object(filename, bucket_name, "index.html", ExtraArgs={'ContentType': "text/html", "CacheControl": "max-age=0"})