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使用PHP for Android Development连接到MySQL数据库

[英]Connecting to MySQL database using PHP for Android Development

I was reading a tutorial about how to connect android to mysql database over the internet using a php interface. 我正在阅读一个关于如何使用php界面通过互联网将android连接到mysql数据库的教程。 I made everything as it said to. 我按照它说的做了一切。

the php code is this: php代码是这样的:

  $q=mysql_query("SELECT * FROM people WHERE birthyear>'".$_REQUEST['year']."'");
  while($e=mysql_fetch_assoc($q))
  $output[]=$e;

  print(json_encode($output));

  mysql_close();
?>

and the java code is this: 而java代码是这样的:

String result = "";
//the year data to send
ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("year","1980"));

//http post
try{
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://example.com/getAllPeopleBornAfter.php");
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
    HttpResponse response = httpclient.execute(httppost);
    HttpEntity entity = response.getEntity();
    InputStream is = entity.getContent();
}catch(Exception e){
    Log.e("log_tag", "Error in http connection "+e.toString());
}
//convert response to string
try{
    BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
    StringBuilder sb = new StringBuilder();
    String line = null;
    while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
    }
    is.close();

    result=sb.toString();
}catch(Exception e){
    Log.e("log_tag", "Error converting result "+e.toString());
}

//parse json data
try{
    JSONArray jArray = new JSONArray(result);
    for(int i=0;i<jArray.length();i++){
            JSONObject json_data = jArray.getJSONObject(i);
            Log.i("log_tag","id: "+json_data.getInt("id")+
                    ", name: "+json_data.getString("name")+
                    ", sex: "+json_data.getInt("sex")+
                    ", birthyear: "+json_data.getInt("birthyear")
            );
    }
}
}catch(JSONException e){
    Log.e("log_tag", "Error parsing data "+e.toString());
}

Now the problem is it gives me an error that the variable 'is' doesn't exist! 现在问题是它给我一个错误,变量'是'不存在! what should I do? 我该怎么办?

The tutorial is from website helloandroid.com : http://www.helloandroid.com/tutorials/connecting-mysql-database 本教程是从网站helloandroid.com: http://www.helloandroid.com/tutorials/connecting-mysql-database

try{
    InputStream is = entity.getContent();
    ...
    ...
} //'is' goes out of scope here! 

The scope is limited to the first try-catch block, and hence, is is unavailable in the second. 的范围不限于第一try-catch块,因此, is在第二不可用。

Do: 做:

InputStream is = entity.getContent(); // 'is' now accessible in both try-catch
HttpEntity entity = response.getEntity(); // 'entity' is now accessible in both try-catch
try{
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://example.com/getAllPeopleBornAfter.php");
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
    HttpResponse response = httpclient.execute(httppost);
    ...
    ...
}

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