java - HTTPClient Apache 模拟 cURL

Question

我正在尝试使用他们的 api 将数据上传到 CrowdFlower，为此我正在编写 JAVA 包装器。我正在使用 HttpClient Apache。

CrowdFlower 给出的 cURL 示例如下:curl -T 'sampledata.xlsx' -H 'Content-Type: application/vnd.ms-excel' https://api.crowdflower.com/v1/jobs/upload.json?key={api_key}

这是我的代码:

public InputStream HTTPmethodPostUpload (String authKey, File file) throws ClientProtocolException, IOException{

        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost("https://api.crowdflower.com/v1/jobs/upload.json?key="+authKey);

        MultipartEntity mpEntity = new MultipartEntity();
        ContentBody cbody = new FileBody( file,"application/vnd.ms-excel");
        mpEntity.addPart("sampledata.xlsx", cbody );
        httpPost.setEntity(mpEntity);
        HttpResponse response = httpclient.execute(httpPost);
        HttpEntity entityResponse = response.getEntity();
        return  entityResponse.getContent(); }

但是它返回一个错误并显示以下消息:

{ Not Acceptable 格式，内容类型必须是“格式”中列出的格式之一，但您发送了“multipart/form-data;boundary=yTuwTm4hWmnasxIMB9dC-sxdELIGoNJVudjJdCz\””，“格式” :[“application/vnd.oasis.opendocument.spreadsheet”，“application/vnd.openxmlformats-officedocument.spreadsheetml.sheet”，“application/vnd.ms-excel”，“text/csv”，“text/plain”] }}

我不太了解 Apache HttpClient，所以我不明白我的代码中的问题出在哪里。

Answer 1

你在java中的哪里设置了 header ？

-H 'Content-Type: application/vnd.ms-excel'

尝试以下方法:

mpEntity.setContentType("application/vnd.ms-excel");