Java android-如何获取web源代码?(空异常)
我想获取网页源代码(由我用php编写),然后在textview中显示。但是,它总是返回null 我使用permisson(互联网),但它不起作用 当我运行此应用程序时,TextView显示:“Kaynak Kod:null” 这是我的活动代码:Java android-如何获取web源代码?(空异常),java,android,httpwebrequest,httpwebresponse,Java,Android,Httpwebrequest,Httpwebresponse,我想获取网页源代码(由我用php编写),然后在textview中显示。但是,它总是返回null 我使用permisson(互联网),但它不起作用 当我运行此应用程序时,TextView显示:“Kaynak Kod:null” 这是我的活动代码: protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.activi
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
tv = (TextView)findViewById(R.id.textView1);
Button button = (Button)findViewById(R.id.button1);
button.setOnClickListener(new OnClickListener(){
@Override
public void onClick(View v) {
try {
String source = getData("http://www.oeaslan.com/excel/index_.php?gun=1");
tv.setText("Kaynak Kod: "+source);
} catch (Exception e) {
tv.setText("Hata: "+e.getMessage());
}
}
});
}
private String getData(String url){
try{
HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet(url);
HttpResponse response = client.execute(request);
String html = "";
InputStream in = response.getEntity().getContent();
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
StringBuilder str = new StringBuilder();
String line = null;
while((line = reader.readLine()) != null)
{
str.append(line);
}
in.close();
html = str.toString();
return html;
}catch(Exception e){
return e.getMessage();
}
}
舱单:
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.omer.text"
android:versionCode="1"
android:versionName="1.0" >
<uses-sdk
android:minSdkVersion="8"
android:targetSdkVersion="21" />
<uses-permission android:name="android.permission.INTERNET" />
<application
android:allowBackup="true"
android:icon="@drawable/ic_launcher"
android:label="@string/app_name"
android:theme="@style/AppTheme" >
<activity
android:name=".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>
问候。你可以试试这样的东西:
HttpGet request = new HttpPost(url);
HttpResponse response = httpclient.execute(request);
String responseBody = EntityUtils.toString(response.getEntity());
您有一个
networkMainThreadException
在LogCat
中查看它。您必须将代码放入异步任务
或线程
您所要做的就是在代码中包含此方法:
public static String getSourceCodeOfWebsite(String urlget){
String fetched_data ="";
try {
URL url = null;
url = new URL(urlget);
HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.connect();
StringBuilder response = new StringBuilder(50000);
InputStream inputStream = connection.getInputStream();
BufferedReader rd = new BufferedReader(new InputStreamReader(inputStream));
int i = 0;
while ((i = rd.read()) > 0) {
response.append((char)i);
}
fetched_data = response.toString();
} catch (Exception e) {
e.printStackTrace();
}
return fetched_data;
}
将URL作为参数传入,然后返回代码如何放置此代码?你能举个例子吗?这个网站上有很多例子。谷歌搜索“使用asynctask上传”或dowmload。没有什么区别。谢谢,但它与查看网页源代码相关吗?它与从internet下载数据相关。然后你对下载的数据做什么是无关紧要的。我看到很多android的web服务应用程序。没有任何应用程序使用asynctask或线程。但是,我无法获取源代码。。