php代码问题，读取数据库的数据

Question

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

Answer 1

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'];