為什么我收到HTTP 400錯誤請求

Question

我正在使用HTTP客戶端（從http://www.mkyong.com/java/apache-httpclient-examples/復制的代碼）發送帖子請求。 我一直在嘗試將它與http://postcodes.io一起使用，以查找大量郵政編碼，但失敗了。 根據http://postcodes.io，我應該以以下JSON格式向http://api.postcodes.io/postcodes發送發布請求： {"postcodes" : ["OX49 5NU", "M32 0JG", "NE30 1DP"]}但我總是得到HTTP響應代碼400。我在下面包括了我的代碼。 請告訴我我做錯了什么？ 謝謝

private void sendPost() throws Exception {

    String url = "http://api.postcodes.io/postcodes";
    HttpClient client = HttpClientBuilder.create().build();
    HttpPost post = new HttpPost(url);

    List<NameValuePair> urlParameters = new ArrayList<NameValuePair>();
    urlParameters.add(new BasicNameValuePair("postcodes", "[\"OX49 5NU\", \"M32 0JG\", \"NE30 1DP\"]"));
    post.setEntity(new UrlEncodedFormEntity(urlParameters));
    HttpResponse response = client.execute(post);

    System.out.println("Response Code : " 
                + response.getStatusLine().getStatusCode());
    System.out.println("Reason : " 
            + response.getStatusLine().getReasonPhrase());
    BufferedReader br = new BufferedReader(
            new InputStreamReader(response.getEntity().getContent()));
    StringBuffer result = new StringBuffer();
    String line = "";
    while ((line = br.readLine()) != null) {
        result.append(line);
    }
    br.close();
    System.out.println(result.toString());
}

Answer 1

這有效，不建議使用HTTP.UTF_8：

String url = "http://api.postcodes.io/postcodes";
HttpClient client = HttpClientBuilder.create().build();
HttpPost post = new HttpPost(url);

StringEntity params =new StringEntity("{\"postcodes\" : [\"OX49 5NU\", \"M32 0JG\", \"NE30 1DP\"]}");
post.addHeader("Content-Type", "application/json");
post.setEntity(params);

Answer 2

Jon Skeet是正確的（通常，我可能會補充說），您基本上是在發送表單，並且默認情況下為form-url-encoding。 您可以嘗試這樣的方法：

String jsonString = "{\"postcodes\" : [\"OX49 5NU\", \"M32 0JG\", \"NE30 1DP\"]}";
StringEntity entity = new StringEntity(jsonObj.toString(), HTTP.UTF_8);
entity.setContentType("application/json");
post.setEntity(entity);