程序奔溃。

请问这是什么原因？怎么解决？

慕莱坞4226677 2018-03-27

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3回答

提问者慕莱坞4226677 2018-03-28 13:31:33

这是解析数据的源码

public class SecondActivity extends AppCompatActivity {
TextView tx1,tx2,tx3;

protected Handler handler=new Handler(){
@Override
public void handleMessage(Message msg) {
super.handleMessage(msg);
Content2 content2=(Content2)msg.obj;
tx1.setText(content2.title);
tx2.setText(content2.author);
tx3.setText(content2.content);

}
};
@Override
protected void onCreate(@Nullable Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_second);
tx1=findViewById(R.id.title);
tx2=findViewById(R.id.author);
tx3=findViewById(R.id.content);

new Thread(){
@Override
public void run() {

try {
URL url = new URL("http://www.imooc.com/api/teacher?type=3&cid=1");

HttpURLConnection coon = (HttpURLConnection) url.openConnection();
coon.setRequestMethod("GET");
coon.setReadTimeout(5000);
if(coon.getResponseCode()==200){
InputStream is=coon.getInputStream();
byte[] b=new byte[1024*512];
int n=0;
ByteArrayOutputStream byteArrayOutputStream=new ByteArrayOutputStream();
while ((n=is.read(b))> -1){

byteArrayOutputStream.write(b,0,n);

}
is.close();
byteArrayOutputStream.close();
String st=byteArrayOutputStream.toString();
Log.e("mseeage",st);
JSONObject jsonObject=new JSONObject(st);
int status=jsonObject.getInt("status");
String msg=jsonObject.getString("msg");
JSONObject data=jsonObject.getJSONObject("data");
String title=data.getString("title");
String author=data.getString("author");
String content=data.getString("content");
Message message=handler.obtainMessage();
Content2 content2=new Content2(title,author,content);
message.obj=content2;
handler.sendMessage(message);

}
}
catch (MalformedURLException e) {
e.printStackTrace();
} catch (ProtocolException e) {
e.printStackTrace();
} catch (JSONException e) {
e.printStackTrace();
}catch (IOException e) {
e.printStackTrace();
}
}

}.start();

}
}

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Tender10 2018-03-28 10:22:26

while ((n=is.read(b,0,n))> -1)这个判断条件修改成 while ((n=is.read(b))> -1)

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提问者慕莱坞4226677 #1

老师那个地方改好了，下面进行数据的解析时出现了下面问题 at com.google.android.apps.gsa.shared.util.concurrent.a.ai.run(SourceFile:6) Caused by: com.google.android.apps.gsa.shared.exception.GsaIOException: Error code: 393238 | Buffer overflow, no available space. at com.google.android.apps.gsa.speech.audio.Tee.f(SourceFile:103) at com.google.android.apps.gsa.speech.audio.au.read(SourceFile:2) at java.io.InputStream.read(InputStream.java:101) at com.google.android.apps.gsa.speech.audio.ao.run(SourceFile:18) at com.google.android.apps.gsa.speech.audio.an.run(SourceFile:2) 请问该如何解决？

2018-03-28 13:30:09
Tender10 回复提问者慕莱坞4226677 #2

这个错误原因是缓存溢出没有可用空间，byte[] b=new byte[1024*512];定义这个的时候，把后面的数再*大一点。

2018-03-28 17:16:56

提问者慕莱坞4226677 2018-03-27 20:33:26

这是第二界面的代码

public class SecondActivity extends AppCompatActivity {
@Override
protected void onCreate(@Nullable Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_second);

new Thread(){
@Override
public void run() {

try {
URL url = new URL("http://www.imooc.com/api/teacher?type=3&cid=1");

HttpURLConnection coon = (HttpURLConnection) url.openConnection();
coon.setRequestMethod("GET");
coon.setReadTimeout(5000);
if(coon.getResponseCode()==200){
InputStream is=coon.getInputStream();
byte[] b=new byte[1024*512];
int n=0;
ByteArrayOutputStream byteArrayOutputStream=new ByteArrayOutputStream();
while ((n=is.read(b,0,n))> -1){

byteArrayOutputStream.write(b,0,n);

}
String st=byteArrayOutputStream.toString();
Log.e("mseeage",st);
is.close();
byteArrayOutputStream.close();

}
}
catch (MalformedURLException e) {
e.printStackTrace();
} catch (ProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}

}.start();

}
}