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php – How to retrieve images from MySQL database and display in an html tag

Posted by: admin April 23, 2020 Leave a comment

Questions:

I created a MySQL database with a table using phpmyadmin. I created this table with a BLOB column to hold a jpeg file.

I have issues with regards to the php variable $result here.

My code so far: (catalog.php):

<body>
<?php
  $link = mysql_connect("localhost", "root", "");
  mysql_select_db("dvddb");
  $sql = "SELECT dvdimage FROM dvd WHERE id=1";
  $result = mysql_query("$sql");
  mysql_close($link);

?>
<img src="" width="175" height="200" />
</body>

How can I get the variable $result from PHP into the HTML so I can display it in the <img> tag?

How to&Answers:

You can’t. You need to create another php script to return the image data, e.g. getImage.php. Change catalog.php to:

<body>
<img src="getImage.php?id=1" width="175" height="200" />
</body>

Then getImage.php is

<?php

  $id = $_GET['id'];
  // do some validation here to ensure id is safe

  $link = mysql_connect("localhost", "root", "");
  mysql_select_db("dvddb");
  $sql = "SELECT dvdimage FROM dvd WHERE id=$id";
  $result = mysql_query("$sql");
  $row = mysql_fetch_assoc($result);
  mysql_close($link);

  header("Content-type: image/jpeg");
  echo $row['dvdimage'];
?>

Answer:

Technically, you can too put image data in an img tag, using data URIs.

<img src="data:image/jpeg;base64,<?php echo base64_encode( $image_data ); ?>" />

There are some special circumstances where this could even be useful, although in most cases you’re better off serving the image through a separate script like daiscog suggests.

Answer:

You need to retrieve and disect the information into what you need.

while($row = mysql_fetch_array($result)) {
 echo "img src='",$row['filename'],"' width='175' height='200' />";
}

Answer:

First off you need to fetch the resulting row from the resultset of the query. For that you can use mysql_fetch_row. Now that you have the fetched row you can access the retrieved value and echo it into the src.

For example:

$sql = "SELECT dvdimage FROM dvd WHERE id=1";
$result = mysql_query($sql);
$row = mysql_fetch_row($result);
?>
<img src="<?=$row[0]?>" width="175" height="200" />
<?

Answer:

add $row = mysql_fetch_object($result); after your mysql_query();

your html <img src="<?php echo $row->dvdimage; ?>" width="175" height="200" />

Answer:

I have added slashes before inserting into database so on the time of fetching i removed slashes again stripslashes() and it works for me. I am sharing the code which works for me.

How i inserted into mysql db (blob type)

$db = mysqli_connect("localhost","root","","dName"); 
$image = addslashes(file_get_contents($_FILES['images']['tmp_name']));
$query = "INSERT INTO student_img (id,image) VALUES('','$image')";  
$query = mysqli_query($db, $query);

Now to access the image

$sqlQuery = "SELECT * FROM student_img WHERE id = $stid";
$rs = $db->query($sqlQuery);
$result=mysqli_fetch_array($rs);
echo '<img src="data:image/jpeg;base64,'.base64_encode( stripslashes($result['image']) ).'"/>';

Hope it will help someone

Thanks.