我得到了我的代码连接到我的计算机php文件并在java程序中输出正确的文本 . 当我试图将它添加到我的android项目以显示高分时它总是抛出IOException而我无法弄清楚原因 . 这是我的代码 . 任何帮助,将不胜感激 .
package com.enophz.spacetrash;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.MalformedURLException;
import java.net.URL;
import java.net.URLConnection;
import android.app.Activity;
import android.content.Intent;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.TextView;
public class Scores extends Activity {
//private TextView HscoreText;
/** Called when the activity is first created. */
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.scores);
Button next = (Button) findViewById(R.id.Button01);
next.setOnClickListener(new View.OnClickListener() {
public void onClick(View view) {
Intent myIntent = new Intent(view.getContext(), Menu.class);
startActivity(myIntent);
}
});
TextView HscoreText = (TextView) findViewById(R.id.text);
try
{
URL page = new URL("http://192.168.1.108/score.php");
URLConnection pageconnection = page.openConnection();
BufferedReader in = new BufferedReader(
new InputStreamReader(
pageconnection.getInputStream()));
in.close();
HscoreText.setText("It Worked!");
}
catch(MalformedURLException e)
{
HscoreText.setText("MalformedURL");
}
catch (IOException e)
{
HscoreText.setText("IOException");
}
}
}
1 回答
我不知道你在php页面中期待什么类型的内容..但是以下内容可以帮助你从web服务器获取内容:
现在JSON将是单行,因此您可以使用:
否则将整个内容作为字符串获取: