来自真实设备的android应用无法连接到mysql

时间:2018-10-15 15:54:04

标签: php android mysql android-studio

嗨,我正在使用真实设备将数据插入db中,当我从浏览器测试本地服务器与我的设备之间的连接时,它可以成功运行,但是在应用程序中,它给了我无效的ip地址toast消息,我的android代码连接到数据库:

        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://192.168.43.210:3306/cfa/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());
    }
}


public void cancel(View view) {

    Intent i = new Intent(this , SignUpAs.class);
    startActivity(i);
    finish();
}
}

和php代码:

<?php
$host='localhost';
$uname='root';
$pwd='';
$db="cfa";

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

$fullName=$_REQUEST['fullName'];
$userName=$_REQUEST['userName'];
$IDNum=$_REQUEST['IDNum'];
$email=$_REQUEST['email'];
$address=$_REQUEST['address'];
$password=$_REQUEST['password='];
 $phone=$_REQUEST['phone'];

$flag['code']=0;

if($r=mysql_query("insert into applicant values(NULL,'$fullName',
'$userName','$IDNum','$phone','$email','$address','$password') ",$con))
{
    $flag['code']=1;
    echo"hi";
}

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

所以我不知道这是什么问题...

0 个答案:

没有答案