How to display image from mysql BLOB data type using PHP

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maideen
New php-forum User
New php-forum User
Posts: 30
Joined: Mon Mar 07, 2011 11:38 pm

Tue Sep 12, 2017 8:57 pm

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 PHP Insert code

Code: Select all

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");  
    }
}  
It is my insert HTML Code

Code: Select all

                        <div class="col-md-9">
                            <div class="fileinput fileinput-new" data-provides="fileinput">
                                <div class="fileinput-new thumbnail" style="width: 200px; height: 150px;">
                                    <img src="http://www.placehold.it/200x150/EFEFEF/AAAAAA&amp;text=no+image" alt="" /> </div>
                                <div class="fileinput-preview fileinput-exists thumbnail" style="max-width: 200px; max-height: 150px;"> </div>
                                <div>
                                    <span class="btn default btn-file">
                                        <span class="fileinput-new"> Select image </span>
                                        <span class="fileinput-exists"> Change </span>
                                        <input type="file" name="simage"> </span>
                                    <a href="javascript:;" class="btn red fileinput-exists" data-dismiss="fileinput"> Remove </a>
                                </div>
                            </div>
                        </div>
Display PHP Code

Code: Select all

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;

            }
    }
  
?>
Display HTML Code

Code: Select all

                        <div class="col-md-9">
                            <div class="fileinput fileinput-new" data-provides="fileinput">
                                <div class="fileinput-new thumbnail" style="width: 200px; height: 150px;">
                                    <img src="" alt=""  /> 
                                    <?php echo $simage; ?>
                                </div>
                                <div class="fileinput-preview fileinput-exists thumbnail" style="max-width: 200px; max-height: 150px;"></div>
                                <div>
                                    <span class="btn default btn-file">
                                        <span class="fileinput-new"> Select image </span>
                                        <span class="fileinput-exists"> Change </span>
                                        <input type="file" name="simage"> </span>
                                    <a href="javascript:;" class="btn red fileinput-exists" data-dismiss="fileinput"> Remove </a>
                                </div>
                            </div>
                        </div>

chorn
php-forum Active User
php-forum Active User
Posts: 405
Joined: Fri Apr 01, 2016 2:18 am

Tue Sep 12, 2017 10:09 pm

and what is in $_POST['simage']? must be a form field, but no image upload, because that would come from $_FILES.

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