How to display image from mysql BLOB data type using PHP

Hi
I am new. I am using Bootstrap css.
I have an issue to display image with data type BLOB, Really I don’t know how to do this.
Pls advise me how to do this.
Thanks
maideen

My Insert PHP Code
[php]
if(isset($_POST[‘add’]))
{
if($_SERVER[“REQUEST_METHOD”] == “POST”)
{

  $sid =$_POST['sid'];
  $sname =$_POST['sname'];
  $nric=$_POST['nric'];
  $gender=$_POST['gender'];
  $dob=$_POST['dob'];
  $courseid=$_POST['courseid'];
  $pname=$_POST['pname'];
  $emailid=$_POST['emailid'];
  $hpno=$_POST['hpno'];
  $homeno=$_POST['homeno'];
  $offno=$_POST['offno'];
  $emergencyno=$_POST['emergencyno'];
  $address=$_POST['address'];
  $city=$_POST['city'];
  $state=$_POST['state'];
  $country=$_POST['country'];
  $hname=$_POST['hname'];
  $alergies=$_POST['alergies'];
  $milkpowder=$_POST['milkpowder'];
  $specialfoods=$_POST['specialfoods'];
  $specactivy1=$_POST['specactivy1'];
  $specactivy2=$_POST['specactivy2'];
  $specactivy3=$_POST['specactivy3'];
  $specactivy4=$_POST['specactivy4'];
  $simage=$_POST['simage'];
  $createdby =$_POST['username'];
  $time = strftime("%X");
  $date = strftime("%B %d,%Y");
  //$createdon= $date;
  $createdon = date("Y-m-d H:i:s");
  $bool = true;
  $sql="insert into tbl_student_master(sid,sname,nric,gender,dob,courseid,pname,emailid,hpno,homeno,offno,emergencyno,address,city,state,country,hname,alergies,milkpowder,
        specialfoods,specactivy1,specactivy2,specactivy3,specactivy4,simage,createdby,createdon) 
        values ('$sid','$sname','$nric','$gender','$dob','$courseid','$pname','$emailid','$hpno','$homeno','$offno','$emergencyno','$address','$city','$state','$country',
                '$hname','$alergies','$milkpowder','$specialfoods','$specactivy1','$specactivy2','$specactivy3','$specactivy4','$simage','$createdby','$createdon')";
  $stmt=$pdo->prepare($sql);
  $stmt->execute();
  $pdo = null;
  print '<script>alert("Saved");</script>';
  header("location:../student/student_add.php");  
}

}
[/php]

My HTML code

[php]








Select image
Change

Remove



[/php]

My display php code

[php]

<?php if($_SERVER["REQUEST_METHOD"] == "GET") { $id=$_GET['id']; $sql ="Select * from tbl_student_master where id ='$id'"; $stmt = $pdo->prepare($sql); $stmt->execute(); while ($row = $stmt->fetch()) { $id =$row['id']; $sid =$row['sid']; $sname =$row['sname']; $nric=$row['nric']; $gender=$row['gender']; $dob=$row['dob']; $courseid=$row['courseid']; $pname=$row['pname']; $emailid=$row['emailid']; $hpno=$row['hpno']; $homeno=$row['homeno']; $offno=$row['offno']; $emergencyno=$row['emergencyno']; $address=$row['address']; $city=$row['city']; $state=$row['state']; $country=$row['country']; $hname=$row['hname']; $alergies=$row['alergies']; $milkpowder=$row['milkpowder']; $specialfoods=$row['specialfoods']; $specactivy1=$row['specactivy1']; $specactivy2=$row['specactivy2']; $specactivy3=$row['specactivy3']; $specactivy4=$row['specactivy4']; $simage=$row['simage']; //header("Content-Type: image/jpeg"); //echo $simage; } } ?>

[/php]

display HTML Code

[php]





<?php echo $simage; ?>




Select image
Change

Remove



[/php]

You have way more code then you need to have. Do not create variables for nothing. You also need to use prepared statements. You never ever put variables in your query. It also appears that you have a problem with your database design. When you start adding numbered column names that is a red flag.

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