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java - Android - 通过 POST 方法轻松发送 HTTP 参数

转载 作者:行者123 更新时间:2023-12-02 12:39:01 25 4
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    package demo.com.mysqldbdemo;

import android.content.Context;
import android.os.AsyncTask;
import android.widget.Toast;

import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.net.MalformedURLException;
import java.net.URL;
import java.net.URLEncoder;

import javax.net.ssl.HttpsURLConnection;

/**
* Created by Ahmadzai on 7/10/2017.
*/

public class BackgroundTask extends AsyncTask<String, Void, String> {

Context ctx;
BackgroundTask(Context ctx)
{
this.ctx = ctx;
}

@Override
protected void onPreExecute() {
super.onPreExecute();
}

@Override
protected String doInBackground(String... params) {
String reg_url = "http://192.168.8.100:8081/webapp/register.php";
String login_url = "http://localhost/webapp/login.php";
String method = params[0];
if (method.equals("register"))
{
String name = params[1];
String user_name = params[2];
String user_pass = params[3];

try {
URL url = new URL(reg_url);
HttpsURLConnection httpsURLConnection = (HttpsURLConnection) url.openConnection();
httpsURLConnection.setRequestMethod("POST");
httpsURLConnection.setDoOutput(true);
OutputStream OS = httpsURLConnection.getOutputStream();
BufferedWriter bufferedWriter = new BufferedWriter(new OutputStreamWriter(OS,"UTF-8"));
String data = URLEncoder.encode("user","UTF-8") + "=" +URLEncoder.encode(name,"UTF-8")+"&"+
URLEncoder.encode("user_name","UTF-8") + "=" +URLEncoder.encode(user_name,"UTF-8")+"&"+
URLEncoder.encode("user_pass","UTF-8") + "=" +URLEncoder.encode(user_pass,"UTF-8");
bufferedWriter.write(data);
bufferedWriter.flush();
bufferedWriter.close();
OS.close();

InputStream IS = httpsURLConnection.getInputStream();
IS.close();
return "Registration succeed ....";

} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
return null;
}

@Override
protected void onProgressUpdate(Void... values) {
super.onProgressUpdate(values);
}

@Override
protected void onPostExecute(String result) {
Toast.makeText(ctx,result,Toast.LENGTH_LONG).show();
}
}

我想将数据插入到mysql表中,因为这段代码没有任何错误,但是当我按下界面中的注册按钮时,当我想调试它运行的应用程序时,应用程序停止运行,直到“HttpsURLConnection httpsURLConnection = (HttpsURLConnection ) url.openConnection();"在此之后的代码行将停止调试。任何人都可以在这方面帮助我吗?我一直在使用 stackoverflow 答案,但这是我第一次尝试提问,对于给您带来的不便,我们深表歉意。

最佳答案

开始使用retrofit最适合安卓

关于java - Android - 通过 POST 方法轻松发送 HTTP 参数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45013527/

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