在JAVA中使用JSON字符串的HTTP POST请求
我必须使用已经生成的JSON字符串进行http Post请求。 我尝试了两种不同的方法：
1.HttpURLConnection
2.HttpClient
但是我得到了他们两个相同的“不想要的”结果。 到目前为止我的代码与HttpURLConnection是：
public static void SaveWorkflow() throws IOException {
URL url = null;
url = new URL(myURLgoeshere);
HttpURLConnection urlConn = null;
urlConn = (HttpURLConnection) url.openConnection();
urlConn.setDoInput (true);
urlConn.setDoOutput (true);
urlConn.setRequestMethod("POST");
urlConn.setRequestProperty("Content-Type", "application/json");
urlConn.connect();
DataOutputStream output = null;
DataInputStream input = null;
output = new DataOutputStream(urlConn.getOutputStream());
/*Construct the POST data.*/
String content = generatedJSONString;
/* Send the request data.*/
output.writeBytes(content);
output.flush();
output.close();
/* Get response data.*/
String response = null;
input = new DataInputStream (urlConn.getInputStream());
while (null != ((response = input.readLine()))) {
System.out.println(response);
input.close ();
}
}
到目前为止我的代码与HttpClient是：
public static void SaveWorkflow() {
try {
HttpClient httpClient = new DefaultHttpClient();
HttpPost postRequest = new HttpPost(myUrlgoeshere);
StringEntity input = new StringEntity(generatedJSONString);
input.setContentType("application/json;charset=UTF-8");
postRequest.setEntity(input);
input.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE,"application/json;charset=UTF-8"));
postRequest.setHeader("Accept", "application/json");
postRequest.setEntity(input);
HttpResponse response = httpClient.execute(postRequest);
BufferedReader br = new BufferedReader(
new InputStreamReader((response.getEntity().getContent())));
String output;
while ((output = br.readLine()) != null) {
System.out.println(output);
}
httpClient.getConnectionManager().shutdown();
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
生成JsonString的地方是这样的：
{"description":"prova_Process","modelgroup":"","modified":"false"}
我得到的回应是：
{"response":false,"message":"Error in saving the model. A JSONObject text must begin with '{' at 1 [character 2 line 1]","ids":[]}
请任何想法吗？
最后，我设法找到解决我的问题的办法...
public static void SaveWorkFlow() throws IOException
{
CloseableHttpClient httpClient = HttpClients.createDefault();
HttpPost post = new HttpPost(myURLgoesHERE);
List<NameValuePair> params = new ArrayList<>();
params.add(new BasicNameValuePair("task", "savemodel"));
params.add(new BasicNameValuePair("code", generatedJSONString));
CloseableHttpResponse response = null;
Scanner in = null;
try
{
post.setEntity(new UrlEncodedFormEntity(params));
response = httpClient.execute(post);
// System.out.println(response.getStatusLine());
HttpEntity entity = response.getEntity();
in = new Scanner(entity.getContent());
while (in.hasNext())
{
System.out.println(in.next());
}
EntityUtils.consume(entity);
} finally
{
in.close();
response.close();
}
}
另一种实现方法如下所示：
public static void makePostJsonRequest(String jsonString)
{
HttpClient httpClient = new DefaultHttpClient();
try {
HttpPost postRequest = new HttpPost("Ur_URL");
postRequest.setHeader("Content-type", "application/json");
StringEntity entity = new StringEntity(jsonString);
postRequest.setEntity(entity);
long startTime = System.currentTimeMillis();
HttpResponse response = httpClient.execute(postRequest);
long elapsedTime = System.currentTimeMillis() - startTime;
//System.out.println("Time taken : "+elapsedTime+"ms");
InputStream is = response.getEntity().getContent();
Reader reader = new InputStreamReader(is);
BufferedReader bufferedReader = new BufferedReader(reader);
StringBuilder builder = new StringBuilder();
while (true) {
try {
String line = bufferedReader.readLine();
if (line != null) {
builder.append(line);
} else {
break;
}
} catch (Exception e) {
e.printStackTrace();
}
}
//System.out.println(builder.toString());
//System.out.println("****************");
} catch (Exception ex) {
ex.printStackTrace();
}
}