问题描述
我想从URL获取数据。在这种情况下,我已经得到了完整的PHP的数据转换成JSON已经在本地主机上运行(HTTP://localhost/adchara1/index.php/年= 1)
这是PHP脚本
< PHP
的mysql_connect(localhost的根,);
mysql_select_db(测试);
$ Q =请求mysql_query(SELECT * FROM人
哪里
birthyear>'$ _ REQUEST ['年'。');
而($ E = mysql_fetch_assoc($ Q))
$输出[] = $Ë;
打印(json_en code($输出));
则mysql_close(); ?>
这是结果
[{ID:1,名:kongkea,性:1,birthyear:1990},{ID :2,名称:thida,性别:0,birthyear:2000}]≥
我想用按一下按钮,并显示该结果的TextView
公共类MainActivity延伸活动{
AsyncTask的<虚空,虚空,虚空> mTask;
字符串jsonString;
字符串URL = "https://api.twitter.com/1/statuses/user_timeline.json?include_entities=true&include_rts=true&screen_name=50cent&count=2";
按钮B;
@覆盖
公共无效的onCreate(包savedInstanceState){
super.onCreate(savedInstanceState);
的setContentView(R.layout.activity_main);
getActionBar()setDisplayHomeAsUpEnabled(真)。
B =(按钮)findViewById(R.id.btnFetch);
最后的TextView电视=(TextView中)findViewById(R.id.txtView);
mTask =新的AsyncTask<虚空,虚空,虚空> (){
@覆盖
保护无效doInBackground(虚空...... PARAMS){
尝试 {
jsonString = getJsonFromServer(URL);
}赶上(IOException异常E){
// TODO自动生成的catch块
e.printStackTrace();
}
返回null;
}
@覆盖
保护无效onPostExecute(无效的结果){
super.onPostExecute(结果);
tv.setText(jsonString);
}
};
b.setOnClickListener(新OnClickListener(){
公共无效的onClick(视图v){
mTask.execute();
}
});
}
公共静态字符串getJsonFromServer(字符串URL)抛出IOException异常{
BufferedReader中的InputStream = NULL;
URL jsonUrl =新的网址(URL);
的URLConnection直流= jsonUrl.openConnection();
dc.setConnectTimeout(5000);
dc.setReadTimeout(5000);
的InputStream =新的BufferedReader(新的InputStreamReader(
dc.getInputStream()));
//读取JSON结果转换成字符串
串jsonResult = inputStream.readLine();
返回jsonResult;
}
}
获取jsonString从服务器使用这种方法后,您可以分析,并显示在JSON数据。
编辑:你得到的错误,因为你正试图从服务器的JSON出异步task.You需要做的是在后台。您可以使用一个线程或使用AsyncTask的。
I want to get data from url. in this case I've got complete php that data convert to json already and run in localhost (http://localhost/adchara1/index.php/?year=1)
This is the php script
<?php
mysql_connect("localhost","root","");
mysql_select_db("test");
$q=mysql_query("SELECT * FROM people
WHERE
birthyear>'".$_REQUEST['year']."'");
while($e=mysql_fetch_assoc($q))
$output[]=$e;
print(json_encode($output));
mysql_close(); ?>
and this is result
[{"id":"1","name":"kongkea","sex":"1","birthyear":"1990"}, {"id":"2","name":"thida","sex":"0","birthyear":"2000"}]?>
I want to use button click and show this result in textview
public class MainActivity extends Activity {
AsyncTask<Void, Void, Void> mTask;
String jsonString;
String url = "https://api.twitter.com/1/statuses/user_timeline.json?include_entities=true&include_rts=true&screen_name=50cent&count=2";
Button b;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
getActionBar().setDisplayHomeAsUpEnabled(true);
b = (Button) findViewById(R.id.btnFetch);
final TextView tv = (TextView) findViewById(R.id.txtView);
mTask = new AsyncTask<Void, Void, Void> () {
@Override
protected Void doInBackground(Void... params) {
try {
jsonString = getJsonFromServer(url);
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
return null;
}
@Override
protected void onPostExecute(Void result) {
super.onPostExecute(result);
tv.setText(jsonString);
}
};
b.setOnClickListener(new OnClickListener() {
public void onClick(View v) {
mTask.execute();
}
});
}
public static String getJsonFromServer(String url) throws IOException {
BufferedReader inputStream = null;
URL jsonUrl = new URL(url);
URLConnection dc = jsonUrl.openConnection();
dc.setConnectTimeout(5000);
dc.setReadTimeout(5000);
inputStream = new BufferedReader(new InputStreamReader(
dc.getInputStream()));
// read the JSON results into a string
String jsonResult = inputStream.readLine();
return jsonResult;
}
}
After Getting jsonString from server with this method, you can parse and show the data in the Json.
EDIT: You get error because you are trying to get json from server out of async task.You need to do it in the background. You can either use a thread or use AsyncTask.
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