donvo24600 2017-12-03 18:34
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将对象上传到AWS S3,而无需使用aws-sdk-go创建文件

I am trying to upload an object to AWS S3 using golang sdk without needing to create a file in my system (trying to upload only the string). But I am having difficulties to accomplish that. Can anybody give me an example of how can I upload to AWS S3 without needing to create a file?

AWS Example of how to upload a file:

// Creates a S3 Bucket in the region configured in the shared config
// or AWS_REGION environment variable.
//
// Usage:
//    go run s3_upload_object.go BUCKET_NAME FILENAME
func main() {
    if len(os.Args) != 3 {
        exitErrorf("bucket and file name required
Usage: %s bucket_name filename",
            os.Args[0])
    }

    bucket := os.Args[1]
    filename := os.Args[2]

    file, err := os.Open(filename)
    if err != nil {
        exitErrorf("Unable to open file %q, %v", err)
    }

    defer file.Close()

    // Initialize a session in us-west-2 that the SDK will use to load
    // credentials from the shared credentials file ~/.aws/credentials.
    sess, err := session.NewSession(&aws.Config{
        Region: aws.String("us-west-2")},
    )

    // Setup the S3 Upload Manager. Also see the SDK doc for the Upload Manager
    // for more information on configuring part size, and concurrency.
    //
    // http://docs.aws.amazon.com/sdk-for-go/api/service/s3/s3manager/#NewUploader
    uploader := s3manager.NewUploader(sess)

    // Upload the file's body to S3 bucket as an object with the key being the
    // same as the filename.
    _, err = uploader.Upload(&s3manager.UploadInput{
        Bucket: aws.String(bucket),

        // Can also use the `filepath` standard library package to modify the
        // filename as need for an S3 object key. Such as turning absolute path
        // to a relative path.
        Key: aws.String(filename),

        // The file to be uploaded. io.ReadSeeker is preferred as the Uploader
        // will be able to optimize memory when uploading large content. io.Reader
        // is supported, but will require buffering of the reader's bytes for
        // each part.
        Body: file,
    })
    if err != nil {
        // Print the error and exit.
        exitErrorf("Unable to upload %q to %q, %v", filename, bucket, err)
    }

    fmt.Printf("Successfully uploaded %q to %q
", filename, bucket)
}

I already tried to create the file programmatically but it is creating the file on my system and then uploading it to S3.

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3条回答 默认 最新

  • doumei1203 2017-12-03 18:59
    关注

    The Body field of the UploadInput struct is just an io.Reader. So pass any io.Reader you want--it doesn't need to be a file.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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