select BLOB类型图片，插入到另一个表中

Question

I have two table one is storing basic info with image type is blob and the other table is for attendance table, Now my problem is when i select all the table with basic info with image and insert it to another table the image column is only blank the rest has a value, Hope you can help fix it thanks.

here is my php code:

<?php 

include_once('connection.php');
if(isset($_POST['submit0'])){
$rfid = $_POST['rfid'];

$sql = mysqli_query($conn,"select * from stud where rfid_num ='$rfid'");
$count = mysqli_num_rows($sql);

if ($count == 0) {
header("location:notexist.php");

    } else{

while( $row = mysqli_fetch_array($sql)) {
                  $rfid=$row['rfid_num'];
                  $id=$row['id'];
                   $name0 = $row['name'];
                   $course0 = $row['course'];
                  $image = $row['image'];

     $InsertSql = "INSERT INTO student_att(rfid_num,id,name,course,image) VALUES ('$rfid','$id','$name0','$course0','$image')";
        $res = mysqli_query($conn, $InsertSql); 
}
}
}
?> 

Answer 1

Instead of executing two separate queries I will suggest you to have a single query which will select from one table and insert into another table.

Example query:

INSERT INTO table2 (st_id,uid,changed,status,assign_status)
SELECT st_id,from_uid,now(),'Pending','Assigned'
FROM table1

NOTE: It's not good to save images in database. We usually keep them in folders and save the path of it in database.

Answer 2

The simplest way to solve your error is by doing like this

$image=mysqli_real_escape_string($mysqli,$row['image'];

$mysqli is connection variable. and then in your query use $image to insert image hope it will helps.