php - Unable to retrieve image data from table and display image -

i'm trying upload image file database php , mysqli, i'm coming across confusing piece of error. table has 2 column, 'title' , 'image'. 'title' file's name , 'image' image data. both tables not allowed accept nulls. when upload file, data stored table columns. 'title' contains right values, 'image' column contains '<binary data>'.

since table column not accept null values assumed file's data, when try retrieve , display image data in showimage.php tells me image data null.

i using blob datatype store image data in table. far concerned based off online resources , examples should work. thanks.

code:

php:

uploads.php

if (isset($_post['submit'])) {     $title = $_files['image']['name'];     $data = $_files['image']['tmp_name'];     $content = file_get_contents($data);     $query = "insert images (title, image) values (?, ?)";     $statement = $databaseconnection->prepare($query);     $statement->bind_param('sb', $title, $content);     $statement->execute();     $statement->store_result();     $creationwassuccessful = $statement->affected_rows == 1 ? true : false;     if ($creationwassuccessful)     {         echo "works!";     } else {         echo 'failed';     } } 

showimage.php

if (isset($_get['id'])) {      $id = $_get['id'];     $query = "select * images id = ?";     $statement = $databaseconnection->prepare($query);     $statement->bind_param('i', $id);      $statement->execute();     $statement->store_result();      if ($statement->num_rows >= 1)     {         $statement->bind_result($imageid, $title, $image)         while ($statement->fetch()) {             if ($image == null) {                 echo "image data not exist!";             } else {                 header("content-type: image/jpeg");                 echo $image;             }          }             } } 

html

<form action="uploads.php" method="post" enctype="multipart/form-data">     <input type="file" name="image">     <input type="submit" name="submit"> </form> 

first need save image output file_get_contents database. , put imagecreatefromstring , show image.

here's simple example. maybe out :)

$data = file_get_contents("acl.jpg"); $img = imagecreatefromstring($data); header("content-type: image/jpeg"); imagejpeg($img); 

edit :

you need put code :

$statement->bind_result($imageid, $title, $image) while ($statement->fetch()) {     if ($image == null) {         echo "image data not exist!";     } else {         $img = imagecreatefromstring($image);         header("content-type: image/jpeg");         imagejpeg($img);     }  } 

edit fix :

uploads.php

in file need change $statement->bind_param('sb', $title, $content); become $statement->bind_param('ss', $title, $content);

if (isset($_post['submit'])) {     $title = $_files['image']['name'];     $data = $_files['image']['tmp_name'];     $content = file_get_contents($data);     $query = "insert images (title, image) values (?, ?)";     $statement = $databaseconnection->prepare($query);     $statement->bind_param('ss', $title, $content);     $statement->execute();     $statement->store_result();     $creationwassuccessful = $statement->affected_rows == 1 ? true : false;     if ($creationwassuccessful)     {         echo "works!";     } else {         echo 'failed';     } } 

showimage.php , in show using :

$img = imagecreatefromstring($image); header("content-type: image/jpeg"); imagejpeg($img); in last statement

if (isset($_get['id'])) {      $id = $_get['id'];     $query = "select id,title,image images id = ?";     $statement = $databaseconnection->prepare($query);     $statement->bind_param('i', $id);      $statement->execute();     $statement->store_result();      if ($statement->num_rows >= 1)     {         $statement->bind_result($imageid, $title, $image)         while ($statement->fetch()) {             if ($image == null) {                 echo "image data not exist!";             } else {                 $img = imagecreatefromstring($image);                 header("content-type: image/jpeg");                 imagejpeg($img);             }          }             } } 

hope work fine too, i've tested , run well... :)