php代码问题，读取数据库的数据

do{
$d=mysqli_query($con,"SELECT AID FROM id WHERE BID='$g'");
mysqli_query($con,"DELETE * FROM id WHERE BID='$g'");
$g=$d;
}
while($d==0);
这段php代码中$d=mysqli_query($con,"SELECT AID FROM id WHERE BID='$g'");出问题了
Catchable fatal error: Object of class mysqli_result could not be converted to string in C:\wamp\www\
该怎么解决

意思是$d是结果集，不能赋值给字符串$g，要想赋值则需要把值从结果集中取出，如果只有一个值，可以这样
$row=mysqli_fetch_assoc($d);
$g=$row['AID'];

 $sql="select * from qa limit 0,10";
 $result=$mysqli->query($sql);
 while ($row = mysqli_fetch_assoc($result))
  {
    echo "question_title : {$row['question_title']} <br>";
    echo "answer_content1: {$row['answer_content1']} <br>";
  }
 /* free result set */
    $result->close();