facultyname
的回显显示值,但fid
没有得到值。
我想将fid
的值分配给变量。
$facultyname=$_POST['facultyname'];
echo $facultyname;
$query="SELECT fid FROM faculty WHERE fname='$facultyname'";
$fid_result=mysqli_query ($link,$query);
$finfo = mysqli_fetch_field($fid_result);
printf("FacultyID :- %d ",$finfo->fid);
最佳答案
将字段更改为assoc
$finfo = mysqli_fetch_field($fid_result);
至
$finfo = mysqli_fetch_assoc($fid_result);
阅读-供参考
Fetch field
Fetch Assoc
关于php - mysqli_result值转换为php中的变量,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/30886352/