Question

我正在尝试更新数据库中的字段，但是从我的php页面获取未定义的变量错误...

我的代码是：

Intent myintent = getIntent();
    //mdate = intent.getStringExtra("mdate");
    //Toast.makeText(getApplicationContext(), mdate.toString(), Toast.LENGTH_LONG).show();
    title=(EditText)findViewById(R.id.editText1);
    Bundle extras = getIntent().getExtras();
    if(extras != null){ 
      mtitle = extras.getString("title");
      stime = extras.getString("mtime");
      sdate = extras.getString("mdate");
      cvenue = extras.getString("venue");
    }

    title.setText(mtitle);
    title.setFocusable(false);
    cdate=(EditText)findViewById(R.id.editText2);
    cdate.setText(sdate);
    cdate.setFocusable(false);
    venue=(EditText)findViewById(R.id.editText4);
    venue.setText(cvenue);
    venue.setFocusable(false);
    ctime=(EditText)findViewById(R.id.editText3);
    ctime.setText(stime);
    ctime.setFocusable(false);
    fdate=cdate.getText().toString();
    ftime=ctime.getText().toString();
    ftitle=title.getText().toString();
    fvenue=venue.getText().toString();
    fattend=uid.toString();

    cnfrm=(Button)findViewById(R.id.button1);       
    cnfrm.setOnClickListener(new View.OnClickListener() 
    {

        @Override
        public void onClick(View v) {
            // TODO Auto-generated method stub
            try
            {
                String str_params=URLEncoder.encode("res="+resy.toString()+
                        "&title="+ftitle+
                        "&venue="+fvenue+
                        "&cdate="+fdate+
                        "&ctime="+ftime+
                        "&attend="+fattend,"UTF-8");                    
                get= new HttpGet("http://10.0.2.2/meetingschedular/confirmmeet.php?"+str_params);                           
                client= new DefaultHttpClient();
                res=client.execute(get);
                in=res.getEntity().getContent();
                StringBuffer str= new StringBuffer();                       
                int ch;
                while((ch=in.read())!=-1)
                    {
                        str.append((char)ch);   
                    }
            String tmp=str.toString();
            tmp=tmp.trim();
            //txt.setText(tmp);
            Toast.makeText(getApplicationContext(), tmp, Toast.LENGTH_LONG).show();
            if(flag.equals(tmp))
                {
                    Toast.makeText(getApplicationContext(), tmp, Toast.LENGTH_LONG).show();
                }   
                else
                {
                    Toast.makeText(getApplicationContext(), "Sorry cannot confirm", Toast.LENGTH_LONG).show();
                }
            }
            catch(Exception e)
                {
                Toast.makeText(getApplicationContext(), e.toString(), Toast.LENGTH_LONG).show();

                }
        }
    });

我的PHP代码是：

 <?php 
// Connects to your Database 
$con=mysql_connect('localhost','root','');
mysql_select_db('meetingschedulardata',$con);
if(!$con)
{
die("can not connect".mysql_error());
}
if (isset($_GET['title']))
{
    $title=$_GET['title'];
}
if(isset($_GET['cdate']))
{
    $cdate=$_GET['cdate'];
}
if(isset($_GET['ctime']))
{
    $ctime=$_GET['ctime'];
}
if(isset($_GET['attend']))
{
    $attend=$_GET['attend'];
}
if(isset($_GET['venue']))
{
    $venue=$_GET['venue'];
}
if(isset($_GET['res']))
{
    $res=$_GET['res'];
}
$sdate = strtotime($cdate);
$new_date = date('Y-m-d', $sdate);
$stime = strtotime($ctime);
$new_time = date('h:i:s', $stime);
//echo "till now it works..";
$order="Update meetingdetails SET status='$res' WHERE status IS NULL AND title='$title' AND mdate='$new_date' AND mtime='$new_time' AND attendees='$attend' AND venue='$venue'";
$result=mysql_query($order,$con);
//echo "It still works";
if($result==1)
{
echo "true";
}
else{
echo "false".mysql_error();
}
mysql_close($con);
?>

我想更新我的数据库中的状态字段但无法执行此操作.. 它显示的错误就像这样

enter image description here

请告诉我一些事情，以便我可以正确运行我的应用程序...... Thnx提前

Answer 1

检查你从android传递的变量以及你在php代码中在服务器端使用的变量。使用print_r（$ _ REQUEST）从php端检查它;出口;在页面的开头。这将帮助您调试，你可以从android传递变量列表。

如果您从中得到所有精细且相同的变量，那么在服务器端您将没有问题。

谢谢:) Kuldip

在从android更新值时，php中未定义的变量错误

1 个答案: