Quick Question: How to display pic from mySQL DB?

sois

I have a picture in a DB. It is stored as a binary longblob. How do I output the picture?

Here is my query:

$query = mysql_query("select PICTURE from table where NAME= 'Aaron'");
$x = mysql_fetch_array($query);

How can I output $x[PICTURE]?

ShawnK

Perhaps you should look up Headers and MIME Types

sois

Thanks Shawn. Is there not a simple way to do this? I will read the article, but if anyone wants to post a quick method, I'd greatly appreciate it.

sois

I don't know if I am attacking the problem correctly. I see you can't send a header again once stuff has been output to the page.

What other method can I try? I had it to where I wrote the file to a directory but I had to chmod 777 to get it to work. Then I kept getting hacked because of that. Any ideas?

kestral321

Does the image need to be stored in the database?

How about storing the images in a directory on your webserver and saving file name within database and printing image on screen?

E.G

$query = mysql_query("select Image_Name from table where NAME= 'Aaron'");
while ($x = mysql_fetch_array($query)){
Print "<img src=\"directory/directory2/\"" . $x['Image_Name'];
}

This would also certainly make a difference to the efficiency of your database if you created an index containing Name,Image_Name

kestral321

Alternatively try this it should work i have used it a while ago in the past, The 2 parts that you need to do this are only really the HEADER line and mysql_fetch_object.

The HEADER line Content-type must be changed to match image ie image/jpg for .jpg image/gif for .gif etc etc etc.

$sql = mysql_query("select PICTURE from table where NAME= 'Aaron'");
$rows = mysql_num_rows($sql);

//check we have a result!
if($rows<1){
// no image matches this query
}
else{
// at least one record has been found
$getimage = mysql_fetch_object($sql);

// and send the correct header to the browser - THIS IS THE IMPORTANT LINE
Header("Content-type: image/jpg");

// now send the image
$image = $getimage->image;
echo $image;
flush();
}

Hope this helps!