Android的HttpURLConnection类用JSON字符串 [英] Android HttpURLConnection with json string
本文介绍了Android的HttpURLConnection类用JSON字符串的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我要打电话以下网址：
<$p$p><$c$c>http://192.168.0.196:8080/openapi/localuser/set?{\"syskey\":\"1234\",\"usrname\":\"256\",\"usrpwd\":\"556\"}使用这个地址的新用户添加到数据库中。要做到这一点，我用 HttpURLConnection类在我的的AsyncTask 类
{尝试 网址myUrl =新的URL（PARAMS [0]）;
HttpURLConnection的康恩=（HttpURLConnection类）myUrl.openConnection（）;
conn.setReadTimeout（10000）;
conn.setConnectTimeout（15000）;
conn.setRequestMethod（POST）;
conn.setDoInput（真）;
//开始查询
conn.connect（）;
INT响应= conn.getResponse code（）; Log.d（实验室，响应是：+响应）; statusMap.put（ADDUSER，Integer.toString（响应））; Log.d（实验室，URL：+参数[0]）; }赶上（例外五）{
Log.d（实验室，误差2：+ e.getMessage（））;
}
PARAMS [0] = http://192.168.0.196:8080/openapi/localuser/set?{\"syskey\":\"1234\",\"usrname\":\"256\",\"usrpwd\":\"556\"}
不幸的是，这一呼吁是行不通的。我不明白的错误。
抓返回null
解决方案
尝试这样的方式。你需要在你的code添加几行。
公开的JSONObject makeHtt prequest（字符串requestURL，JSONObject的寄存器）{ 尝试{ URL =新的URL（requestURL）; 连接=（HttpURLConnection类）url.openConnection（）; connection.setReadTimeout（150000）;
connection.setConnectTimeout（150000）; connection.setAllowUserInteraction（真）;
connection.setRequestMethod（POST）;
connection.setRequestProperty（接收字符集，UTF-8）;
connection.setRequestProperty（内容类型，应用程序/ x-WWW的形式urlen codeD）;
connection.setRequestProperty（接受语言，EN-US，连接; Q = 0.5）;
connection.setFixedLengthStreamingMode（register.toString（）的getBytes（）的长度。）;
connection.setDoInput（真）;
connection.setDoOutput（真）;
OutputStreamWriter的OutputStream =新OutputStreamWriter（connection.getOutputStream（））; outputStream.write（register.toString（））; outputStream.flush（）; Log.e（URL，connection.getURL（）的toString（））;
Log.e（JSONObject的，register.toString（））;
}赶上（例外五）{
Log.e（主要例外，e.toString（））;
}
尝试{ INT状态code = connection.getResponse code（）; 如果（状态code == HttpURLConnection.HTTP_OK）{
是= connection.getInputStream（）;
}其他{ }
}赶上（IOException异常五）{
Log.e（IOException异常，e.toString（））;
} 尝试{ RD =新的BufferedReader（新的InputStreamReader（是））; 响应=新的StringBuffer（）; 而（（行= rd.readLine（））！= NULL）{
response.append（线）;
response.append（'\\ n'）;
}
Log.e（回应，response.toString（）+）; rd.close（）;
}赶上（IOException异常五）{
Log.e（BUFFER_READER，e.toString（））;
}赶上（NullPointerException异常五）{
Log.e（NullPointerException异常，e.toString（））;
} {最后
connection.disconnect（）;
} 尝试{
返回新的JSONObject（response.toString（））;
}赶上（JSONException E）{
Log.e（JSONException，e.toString（））; }
返回null;
}
另外你正在使用本地主机必须拥有模拟器可以连接到本地主机。除非它不会去任何设备上运行。
I want to call the following URL:
http://192.168.0.196:8080/openapi/localuser/set?{"syskey":"1234","usrname":"256","usrpwd":"556"}
Use this address to add a new user to the database. To do this I use HttpURLConnection in my AsyncTask class
try {
URL myUrl = new URL(params[0]);
HttpURLConnection conn = (HttpURLConnection) myUrl.openConnection();
conn.setReadTimeout(10000 );
conn.setConnectTimeout(15000);
conn.setRequestMethod("POST");
conn.setDoInput(true);
// Starts the query
conn.connect();
int response = conn.getResponseCode();
Log.d("lab", "The response is: " + response);
statusMap.put("addUser", Integer.toString(response));
Log.d("lab", "URL: " + params[0]);
}catch (Exception e){
Log.d("lab", "Error2: " + e.getMessage());
}
params[0] = http://192.168.0.196:8080/openapi/localuser/set?{"syskey":"1234","usrname":"256","usrpwd":"556"}
Unfortunately, this call is not working. I do not get the error.
catch returns null
解决方案
Try like this way. You need to add few line in your code.
public JSONObject makeHttpRequest(String requestURL, JSONObject register) {
try {
url = new URL(requestURL);
connection = (HttpURLConnection) url.openConnection();
connection.setReadTimeout(150000);
connection.setConnectTimeout(150000);
connection.setAllowUserInteraction(true);
connection.setRequestMethod("POST");
connection.setRequestProperty("Accept-Charset", "UTF-8");
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.setRequestProperty("Accept-Language", "en-US,en;q=0.5");
connection.setFixedLengthStreamingMode(register.toString().getBytes().length);
connection.setDoInput(true);
connection.setDoOutput(true);
OutputStreamWriter outputStream = new OutputStreamWriter(connection.getOutputStream());
outputStream.write(register.toString());
outputStream.flush();
Log.e("URL", connection.getURL().toString());
Log.e("JSONObject", register.toString());
} catch (Exception e) {
Log.e("MAIN Exception", e.toString());
}
try {
int statuscode = connection.getResponseCode();
if (statuscode == HttpURLConnection.HTTP_OK) {
is = connection.getInputStream();
} else {
}
} catch (IOException e) {
Log.e("IOException", e.toString());
}
try {
rd = new BufferedReader(new InputStreamReader(is));
response = new StringBuffer();
while ((line = rd.readLine()) != null) {
response.append(line);
response.append('\n');
}
Log.e("Response", response.toString() + " ");
rd.close();
} catch (IOException e) {
Log.e("BUFFER_READER", e.toString());
} catch (NullPointerException e) {
Log.e("NullPointerException", e.toString());
} finally {
connection.disconnect();
}
try {
return new JSONObject(response.toString());
} catch (JSONException e) {
Log.e("JSONException", e.toString());
}
return null;
}
Also You are using localHost you must have emulator which can connect to localhost. Unless it will not going to work on any device.
这篇关于Android的HttpURLConnection类用JSON字符串的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
查看全文
