如何从 curl 生成正确的 RestTemplate 用法？

Question

我有一个可用的 curl 命令，如下所示：

curl -v  -H "Content-Type:application/octet-stream" \
         -H "x-amz-server-side-encryption:aws:kms" \
         -H "x-amz-server-side-encryption-aws-kms-key-id:abcdef81-abcd-4c85-b1d8-ee540d0a5f5d" \
         --upload-file /Users/fd/Downloads/video.mp4 \
         'https://video-uploads-prod.s3-accelerate.amazonaws.com/ABCDEAQGZHEhM55fvvA/ads-aws_userUploadedVideo?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20200106T165718Z&X-Amz-SignedHeaders=content-type%3Bhost%3Bx-amz-server-side-encryption%3Bx-amz-server-side-encryption-aws-kms-key-id&X-Amz-Expires=86400&X-Amz-Credential=ABCDEFHLWTCWZ2MUPPBQ%2F20200106%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=037949abcd1234b063c75d3d505dd9120dd3fa9250c1ababa152e91fee123ca0'

卷曲工作正常：

* We are completely uploaded and fine
< HTTP/1.1 200 OK

但是，当我尝试使用 RestTemplate（我使用的是 spring boot 1.5.6）时，我无法让它工作。 我使用的代码是：

byte[] media = //video in mp4//;
String uploadUrl = "https://video-uploads-prod.s3-accelerate.amazonaws.com/ABCDEAQGZHEhM55fvvA/ads-aws_userUploadedVideo?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20200106T165718Z&X-Amz-SignedHeaders=content-type%3Bhost%3Bx-amz-server-side-encryption%3Bx-amz-server-side-encryption-aws-kms-key-id&X-Amz-Expires=86400&X-Amz-Credential=ABCDEFHLWTCWZ2MUPPBQ%2F20200106%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=037949abcd1234b063c75d3d505dd9120dd3fa9250c1ababa152e91fee123ca0";

RestTemplate restTemplate = new RestTemplate();

HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_OCTET_STREAM);
headers.set("x-amz-server-side-encryption", encryption);
headers.set("x-amz-server-side-encryption-aws-kms-key-id", awsKmsKeyId);

HttpEntity entity = new HttpEntity<>(media, headers);

ResponseEntity<String> respEntity = restTemplate.exchange(uploadUrl, HttpMethod.POST, entity, String.class);

我从 AWS 得到的错误是：

<Error>
   <Code>AuthorizationQueryParametersError</Code>
   <Message>Error parsing the X-Amz-Credential parameter; the Credential is mal-formed; expecting "&lt;YOUR-AKID&gt;/YYYYMMDD/REGION/SERVICE/aws4_request".</Message>
   <RequestId>51FF099744C43804</RequestId>
   <HostId>FOlLws+txYMP0hKEg7aDjQeeARdn7bJN+lw7q/aGA48hRnr1YEsJrVmRi6oEz+mkpHlTIax5MkI=</HostId>
</Error>

我怀疑 RestTemplate 正在更改 URL 的编码。 无论如何，是否可以使用 RestTemplate 复制与 curl 完全相同的内容？

Answer 1

我还没有测试以下方法，但是如果您将扩展的 url 传递给 restTemplate 实例怎么办？

String uploadUrl = "...{your_params_in_placeholders}";
URI expanded = new UriTemplate(url).expand(uploadUrl, <param_values>);
url = URLDecoder.decode(expanded.toString(), "UTF-8");
restTemplate.exchange(url, HttpMethod.POST, entity, String.class);

如果它不起作用，您可以尝试注意 URL 中的编码值（我的意思是“%2F”等）

Answer 2

在为此苦苦挣扎之后，我可以让它发挥作用。 该代码非常丑陋，但使技巧起作用。 发布给其他人以备不时之需。 无论如何，这并没有回答我的问题，我想从 curl 命令中获取确切的 RestTemplate 配置。

我必须使用URLDecoder.decode手动解码每个查询参数，然后使用UriComponentsBuilder.fromHttpUrl构建 URI 并添加先前解码的参数

byte[] media = //video in mp4//;
String uploadUrl = "https://video-uploads-prod.s3-accelerate.amazonaws.com/ABCDEAQGZHEhM55fvvA/ads-aws_userUploadedVideo?X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20200106T165718Z&X-Amz-SignedHeaders=content-type%3Bhost%3Bx-amz-server-side-encryption%3Bx-amz-server-side-encryption-aws-kms-key-id&X-Amz-Expires=86400&X-Amz-Credential=ABCDEFHLWTCWZ2MUPPBQ%2F20200106%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=037949abcd1234b063c75d3d505dd9120dd3fa9250c1ababa152e91fee123ca0";

RestTemplate restTemplate = new RestTemplate();

HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_OCTET_STREAM);
headers.set("x-amz-server-side-encryption", encryption);
headers.set("x-amz-server-side-encryption-aws-kms-key-id", awsKmsKeyId);

HttpEntity entity = new HttpEntity<>(media, headers);

String url = uploadUrl.split("\\?")[0];
String[] urlParams = uploadUrl.split("\\?")[1].split("&");

MultiValueMap<String, String> params = new LinkedMultiValueMap<String, String>();

for (String p : urlParams) {
    String key = p.split("=")[0];
    String value = URLDecoder.decode(p.split("=")[1], StandardCharsets.UTF_8.toString());
    params.put(key, Collections.singletonList(value));
}

String uri = UriComponentsBuilder.fromHttpUrl(url)
        .queryParams(params)
        .build()
        .toUriString();

ResponseEntity<String> respEntity = restTemplate.exchange(uri, HttpMethod.PUT, entity, String.class);

Answer 3

您可以在RestTemplate传递URI而不是String ，这应该可以消除您的错误。 在这里发生了同样的事情。

URI uri = URI.create(urlAsString);
Map response = new RestTemplate().exchange(uri, GET, new HttpEntity(headers), Map.class).getBody();

使用 String 时似乎斜线被十六进制转义，因此%2F变为%252F 。 直接传入 URI 可以避免这种情况。