Question

我有这样的卷毛电话：

curl -i -X POST -H "Content-Type: multipart/form-data" -F "file=@data_test/json_test.json" http://domain.com/api/upload_json/

我需要做的就是这个调用的Java实现。我已经制作了这段代码，但服务器上显示的文件似乎是空的。

public static void uploadJson(String url, File jsonFile) {
    try {
        HttpPost request = new HttpPost(url);
        EntityBuilder builder = EntityBuilder
                .create()
                .setFile(jsonFile)
                .setContentType(ContentType
                        .MULTIPART_FORM_DATA)
                .chunked();
        HttpEntity entity = builder.build();
        request.setEntity(entity);
        HttpResponse response = getHttpClient().execute(request);
        logger.info("Response: {}", response.toString());
    } catch (IOException e) {
        logger.error(e.getMessage());
    }
}

构建此请求的正确方法是什么？

Answer 1

CloseableHttpClient httpClient = HttpClientBuilder.create()
        .build();

HttpEntity requestEntity = MultipartEntityBuilder.create()
        .addBinaryBody("file", new File("data_test/json_test.json"))
        .build();
HttpPost post = new HttpPost("http://domain.com/api/upload_json/");
post.setEntity(requestEntity);
try (CloseableHttpResponse response = httpClient.execute(post)) {
    System.out.print(response.getStatusLine());
    EntityUtils.consume(response.getEntity());
}

如何使用Apache HttpPost

1 个答案: