Question

我正在尝试将数据从android推送到mysql。它说无效的IP地址。为什么会这样？问题出在PHP代码或Java的MainActivity上吗？

JAVA CODE：

`public class MainActivity extends Activity {

String name;
String id;
InputStream is=null;
String result=null;
String line=null;
int code;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    final EditText e_id=(EditText) findViewById(R.id.Text1);
    final EditText e_name=(EditText) findViewById(R.id.Text2);
    Button insert=(Button) findViewById(R.id.but1);

    insert.setOnClickListener(new View.OnClickListener() {

    @Override
    public void onClick(View v) {
        // TODO Auto-generated method stub

        id = e_id.getText().toString();
        name = e_name.getText().toString();

        insert();
    }
});
}

public void insert()
{
    ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();

nameValuePairs.add(new BasicNameValuePair("id",id));
nameValuePairs.add(new BasicNameValuePair("name",name));

    try
    {
    HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://www.shoaib.byethost9.com/insert.php");
        httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpclient.execute(httppost); 
        HttpEntity entity = response.getEntity();
        is = entity.getContent();
        Log.e("pass 1", "connection success ");
}
    catch(Exception e)
{
        Log.e("Fail 1", e.toString());
        Toast.makeText(getApplicationContext(), "Invalid IP Address",
        Toast.LENGTH_LONG).show();
}     

    try
    {
        BufferedReader reader = new BufferedReader
        (new InputStreamReader(is,"iso-8859-1"),8);
        StringBuilder sb = new StringBuilder();
        while ((line = reader.readLine()) != null)
    {
            sb.append(line + "\n");
        }
        is.close();
        result = sb.toString();
    Log.e("pass 2", "connection success ");
}
    catch(Exception e)
{
        Log.e("Fail 2", e.toString());
}     

try
{
        JSONObject json_data = new JSONObject(result);
        code=(json_data.getInt("code"));

        if(code==1)
        {
    Toast.makeText(getBaseContext(), "Inserted Successfully",
        Toast.LENGTH_SHORT).show();
        }
        else
        {
     Toast.makeText(getBaseContext(), "Sorry, Try Again",
        Toast.LENGTH_LONG).show();
        }
}
catch(Exception e)
{
        Log.e("Fail 3", e.toString());
}
}

@Override
public boolean onCreateOptionsMenu(Menu menu) {
    getMenuInflater().inflate(R.menu.main,menu);
    return true;
}

}

PHP代码：

enter code here<?php
$host='sql100.byethost9.com';
$uname='b9_14989725';
$pwd='';
$db='b9_14989725_android';


$con = mysql_connect($host,$uname,$pwd) or die("connection failed");
mysql_select_db($db,$con) or die("db selection failed");

$id=$_REQUEST['id'];
$name=$_REQUEST['name'];

$flag['code']=0;

if($r=mysql_query("insert into android (id,name) values('$id','$name') ",$con))
{
    $flag['code']=1;
    echo"hi";
}

print(json_encode($flag));
mysql_close($con);

＆GT;

有什么问题？为什么IP地址无效？

Answer 1

@SuppressLint("NewApi")
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
    StrictMode.setThreadPolicy(policy);
    setContentView(R.layout.xyz);

无效的IP地址

1 个答案: