Question

我试图允许用户登录自己的帐户细节存储在MySQL数据库中，并且有一个php文件可以帮助指导和验证他们的输入。

但是，当我启动代码时，没有任何事情发生我不确定我犯了哪些错误，我已经编辑并重新编写了三天的代码而没有任何结果。

我将不胜感激。

更新：看起来错误可能是String链接，因为catch正在打印网站地址

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.net.HttpURLConnection;
import java.net.URI;
import java.net.URL;
import java.net.URLConnection;
import java.net.URLEncoder;



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

public class SigninActivity  extends AsyncTask<String,Void,String>{
    private TextView statusField,roleField;
    private Context context;


    //flag 0 means get and 1 means post.(By default it is get.)
    public SigninActivity(Context context,TextView statusField,TextView roleField) {
        this.context = context;
        this.statusField = statusField;
        this.roleField = roleField;
    }

    protected void onPreExecute(){

    }

    @Override
    protected String doInBackground(String... arg0) {

            try{
                String username = (String)arg0[0];
                String password = (String)arg0[1];

                String link = "https://*/*/LoginFile.php";
                String data  = URLEncoder.encode("username", "UTF-8") + "=" + URLEncoder.encode(username, "UTF-8");
                data += "&" + URLEncoder.encode("password", "UTF-8") + "=" + URLEncoder.encode(password, "UTF-8");

                URL url = new URL(link);
                URLConnection conn = url.openConnection();

                conn.setDoOutput(true);
                OutputStreamWriter wr = new OutputStreamWriter(conn.getOutputStream());

                wr.write( data );
                wr.flush();

                BufferedReader reader = new BufferedReader(new InputStreamReader(conn.getInputStream()));

                StringBuilder sb = new StringBuilder();
                String line = null;

                // Read Server Response
                while((line = reader.readLine()) != null)
                {
                    sb.append(line);
                    break;
                }
                return sb.toString();
            }
            catch(IOException e){
                return new String("Exception: " + e.getMessage());
            }

    }

    @Override
    protected void onPostExecute(String result){
        this.statusField.setText("Login Successful");
        this.roleField.setText(result);
    }
}

然后通过MainActivity

提交代码

public void loginPost(View view){
        String username = usernameField.getText().toString();
        String password = passwordField.getText().toString();
        method.setText("Post Method");
        new SigninActivity(this,status,role).execute(username,password);
        method.setText(username);

    }

清单

  <?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.example.*.loginsystem" >

    <uses-permission android:name="android.permission.INTERNET"/>
    <uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />

    <application
        android:allowBackup="true"
        android:label="loginsystem"
        android:theme="@style/AppTheme" >

        <activity
            android:name="com.example.*.loginsystem.MainActivity"
            android:label="@string/app_name" >

            <intent-filter>
                <action android:name="android.intent.action.MAIN" />
                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>

        </activity>

    </application>

</manifest>

Java - 无法连接到MySQL，但是没有显示错误消息

0 个答案: