HttpServlet接收客户端json数据的问题

qq_18454837 2014-09-09 02:10:49

客户端通过HttpPost提交的json数据在服务器上要怎么解析？？代码如下：

服务器：
public class testGET implements ServletRequestAware,ServletResponseAware {

HttpServletRequest request;
HttpServletResponse response;

@Override
public void setServletRequest(HttpServletRequest request) {
// TODO Auto-generated method stub
this.request = request;
}

@Override
public void setServletResponse(HttpServletResponse response) {
// TODO Auto-generated method stub
this.response = response;
}

public void login(){
response.setContentType("text/html;charset=utf-8");
response.setCharacterEncoding("UTF-8");
String username = request.getParameter("userName");
String password = request.getParameter("passWord");
System.out.print("userName=" + username);
System.out.println("passWord=" + password);

try {
getConn getconn = new getConn();
Statement st = getconn.getConnection().createStatement();
String sql = "insert into aaa values (333 , 333)";
int b = st.executeUpdate(sql);
if(b>0){
System.out.println("执行条数为："+ b);
//json();
}else{
System.out.print("失败");
}

} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();

}
}
}

客户端：
public class webPOST extends Activity{

@Override
protected void onCreate(Bundle savedInstanceState) {
// TODO Auto-generated method stub
super.onCreate(savedInstanceState);
setContentView(R.layout.test);
Button bt = (Button) findViewById(R.id.web1);
bt.setOnClickListener(new MyOnClickListener());
}

class MyOnClickListener implements OnClickListener{

@Override
public void onClick(View v) {
Runnable runnable = new Runnable() {

@Override
public void run() {
try {
JSONObject json = new JSONObject();
json.put("userName", "lzw");
json.put("passWord", "love");
String sjson = json.toString();
Log.v("sjson", sjson);

HttpClient httpClient = new DefaultHttpClient();
String url = "http://172.16.0.244:8080/abc/login";
HttpPost httpPost = new HttpPost(url);
//设置httpPost为json、utf-8格式
httpPost.addHeader("Content-Type", "application/json;charset=utf-8");
//封装json
StringEntity se = new StringEntity(sjson, "UTF-8");
httpPost.setEntity(se);
HttpResponse httpResponse = httpClient.execute(httpPost);
int code = httpResponse.getStatusLine().getStatusCode();
if(code == 200){
Log.v("连接结果", "连接成功");

}else{
Log.v("连接结果", "连接失败");
System.out.println(code);
}

} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (ClientProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
};

Thread th = new Thread(runnable);
th.start();
}

}
}

如果客户端用HttpGet传输数据的话可以接收到，如：http://172.16.0.244:8080/abc/login?userName=333&passWord=333
用HttpPost不知道在服务器端怎么处理

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liangfeiyue 2015-08-07

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服务端顶你刻画吨

红袖满楼招 2014-09-09

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一个读一个写，

Kenney_Qin 2014-09-09

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可以用JSONArray，JSONObject（如果你没有的话就去下一个jar包）轻松解析json，对于你的程序，可以在服务器端这样解析：

JSONArray jArray = new JSONArray(str);//假设str是你在服务端接收到的json字符串
			JSONObject json_data = jArray.getJSONObject(0);
			String username=json_data.getString("userName");//userName对应你当时put时的名字
			String password=json_data.getString("passWord");//passWord对应你当时put时的名字