Android 使用HttpUrlConnection将请求发布到服务器
我被夹在中间。我想实现一个POST方法,使用HttpUrlConnection发布电子邮件、名称和密码,以便向服务器注册用户。 这是我的密码:Android 使用HttpUrlConnection将请求发布到服务器,android,post,http-post,httpurlconnection,http-response-codes,Android,Post,Http Post,Httpurlconnection,Http Response Codes,我被夹在中间。我想实现一个POST方法,使用HttpUrlConnection发布电子邮件、名称和密码,以便向服务器注册用户。 这是我的密码: public void createNewProfile(View view){ new Post().execute("http://myurl.com"); } private class Post extends AsyncTask<String, Void, String>{ @Override pro
public void createNewProfile(View view){
new Post().execute("http://myurl.com");
}
private class Post extends AsyncTask<String, Void, String>{
@Override
protected String doInBackground(String... params) {
try {
URL url = new URL("http://myurl.com");
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("POST");
conn.setDoInput(true);
conn.setDoOutput(true);
ContentValues values = new ContentValues();
values.put("email", "abc@xyz.com");
values.put("password", 123);
values.put("name","ABC");
OutputStream os = conn.getOutputStream();
BufferedWriter writer = new BufferedWriter(
new OutputStreamWriter(os, "UTF-8"));
writer.write(getQuery(values));
writer.flush();
writer.close();
os.close();
response = conn.getResponseCode();
conn.connect();
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
Log.i("Result",String.valueOf(e));
}
return null;
}
private String getQuery(ContentValues values) throws UnsupportedEncodingException
{
StringBuilder result = new StringBuilder();
boolean first = true;
for (Map.Entry<String, Object> entry : values.valueSet())
{
if (first)
first = false;
else
result.append("&");
result.append(URLEncoder.encode(entry.getKey(), "UTF-8"));
result.append("=");
result.append(URLEncoder.encode(String.valueOf(entry.getValue()), "UTF-8"));
}
Log.i("Result",result.toString() +" "+ String.valueOf(response));
return result.toString();
}
}
其中,空格后的“0”是http响应代码返回的响应代码。
当我在浏览器中尝试时,我的URL是正确的。
我不知道我在哪里犯了错误;这是我的服务器故障还是我的代码中有错误,因为我认为我的代码没有任何交互处理
我是Android开发的初学者,尝试了很多次,使用了不同的代码,但都出现了错误
请帮忙!
提前谢谢。试试这样的方法:
try {
url = new URL(params[0]);
httpURLConnection = (HttpURLConnection) url.openConnection();
httpURLConnection.setDoInput(true);
httpURLConnection.setDoOutput(true);
httpURLConnection.setRequestMethod("POST");
outputStream = httpURLConnection.getOutputStream();
bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream));
bufferedWriter.write(myValues);
bufferedWriter.flush();
int statusCode = httpURLConnection.getResponseCode();
Log.d(Constants.LOG_TAG, " The status code is " + statusCode);
if (statusCode == 200) {
inputStream = new BufferedInputStream(httpURLConnection.getInputStream());
response = convertInputStreamToString(inputStream);
Log.d(Constants.LOG_TAG, "The response is " + response);
JSONObject jsonObject = new JSONObject(response);
return true;
} else {
return false;
}
} catch (Exception e) {
e.printStackTrace();
} finally {
try {
if (inputStream != null) {
inputStream.close();
}
if (outputStream != null) {
outputStream.close();
}
} catch (Exception e) {
e.printStackTrace();
}
}
在finally语句中获取响应后,尝试关闭编写器将其设置为response=conn.getResponseCode()和conn.connect(),但同样的响应,我也删除了该响应。我已添加代码以供参考。。。检查并让我知道它是否有帮助我现在从这个代码中得到响应代码405。当我为任何响应代码运行inputStream时,再次执行fileNotFoundException。我输入的是我的数据表还是服务器端故障?我应该试试截击吗?你可以试试截击。。但我认为你的url或服务器端代码有问题。你确定代码没有问题吗?我可以使用此POST请求将特定用户注册到API url吗?可以。。从过去6个月以来,我一直在使用上述代码,关于POST请求?只需查询一些我在代码中描述的值,就可以用来注册用户?
try {
url = new URL(params[0]);
httpURLConnection = (HttpURLConnection) url.openConnection();
httpURLConnection.setDoInput(true);
httpURLConnection.setDoOutput(true);
httpURLConnection.setRequestMethod("POST");
outputStream = httpURLConnection.getOutputStream();
bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream));
bufferedWriter.write(myValues);
bufferedWriter.flush();
int statusCode = httpURLConnection.getResponseCode();
Log.d(Constants.LOG_TAG, " The status code is " + statusCode);
if (statusCode == 200) {
inputStream = new BufferedInputStream(httpURLConnection.getInputStream());
response = convertInputStreamToString(inputStream);
Log.d(Constants.LOG_TAG, "The response is " + response);
JSONObject jsonObject = new JSONObject(response);
return true;
} else {
return false;
}
} catch (Exception e) {
e.printStackTrace();
} finally {
try {
if (inputStream != null) {
inputStream.close();
}
if (outputStream != null) {
outputStream.close();
}
} catch (Exception e) {
e.printStackTrace();
}
}