Displaying images from mysql database

How to display images already stored in mysql database with other row data

[okonjiaustin]

I have just been able to successfully insert images into mysql database.
The code is below:

<?php
$errmsg = "";
if (! @mysql_connect("localhost","root","")) {
$errmsg = "Cannot connect to database";
}
@mysql_select_db("adim");

if(isset($_REQUEST['submit']))
{

$imgtype=$_FILES['uploadfile']['type'];
$name=$_REQUEST['name'];
$address=$_REQUEST['address'];
$dateofbirth=$_REQUEST['dateofbirth'];

if($imgtype=="image/jpeg" ¦¦ $imgtype=="image/jpg" ¦¦ $imgtype=="image/pjpeg" ¦¦ $imgtype=="image/gif" ¦¦ $imgtype=="image/x-png" ¦¦ $imgtype=="image/bmp")
{

$image=$_FILES['uploadfile']['tmp_name'];
$fp = fopen($image, 'r');
$content = fread($fp, filesize($image));
$content = addslashes($content);
fclose($fp);
$sql="insert into img_tab1 (name,image,address,dateofbirth) values ('$name','$content','$address','$dateofbirth')";
$res=mysql_query($sql) or die (mysql_error());
}
}
?>
Now I want to retrieve the row data into an html page including the image.

Please let someone help.

Austin

[Habtom]

$query = "SELECT image, image_type from img_tab1 where id=$id";
$result = @MYSQL_QUERY($query);

$data = @MYSQL_RESULT($result,0,"image");
$type = @MYSQL_RESULT($result,0,"image_type");

header( "Content-type: $type");
echo $data;

Two points to remember:

The data type of the image folder needs to be set to the correct type. You might also need to save the image type in a separate column as shown in the example above.

Saving image to a database is not the ideal way of storing images for websites, unless you have a very specific reason for it. Images can be just uploaded to a certain file directory, and the path could be saved in the database, that makes it much more effecient and less maintenance hassles.

[okonjiaustin]

Thanks for your reply Habtom,

While I am trying out the solution you preferred for now, can you kindly give me a code for the separate storage in another directory and save the path in the database as you proposed.

Thanks, I will appreciate.

Austin

[okonjiaustin]

Hello Habtom,

I have quickly tried the code you gave but what I get is an empty page.

The code is below:

<?php
$errmsg = "";
if (! @mysql_connect("localhost","root","")) {
$errmsg = "Cannot connect to database";
}
@mysql_select_db("adim");

$query = "SELECT image, image_type from img_tab1 where id=$id";
$result = @MYSQL_QUERY($query);

$data = @MYSQL_RESULT($result,0,"image");
$type = @MYSQL_RESULT($result,0,"image_type");

echo $data;

?>

Please you can go through and see where the error is.

Thanks.

Austin

[Habtom]

You will need the header:

header( "Content-type: $type");
echo $data;

[okonjiaustin]

Hi Habtom,

I have tried including headers as instructed but I get the error below:

Warning: Cannot modify header information - headers already sent by (output started at D:\wamp\www\imagesfolder\image_display.php:2) in D:\wamp\www\imagesfolder\image_display.php on line 16

And this is the code being reffered to in line 16:

<?php
$errmsg = "";
if (! @mysql_connect("localhost","root","")) {
$errmsg = "Cannot connect to database";
}
@mysql_select_db("adim");

$query = "SELECT image, image_type from img_tab1 where id=$id";
$result = @MYSQL_QUERY($query);

$data = @MYSQL_RESULT($result,0,"image");
$type = @MYSQL_RESULT($result,0,"image_type");

header( "Content-type: $type");
echo $data;

?>
Please go through and let me know where I went wrong.

Thanks

Austin

[wheelie34]

Austin, as Habtom says storing the images in the DB isn't a good idea, if you don't really need to store them in the DB have a look here

[webmasterworld.com ]

I have just finished an upload script that adds the data also to a mysql DB, it is very flexible and easy to work with it does what I need and it looks like it will do what you need too.

The only thing missing on that page is the finished image saving script that now can resize the saved images I finished that yesterday, if you want it just ask and I will post it at the bottom of that thread.

[edited by: jatar_k at 4:49 pm (utc) on Feb. 14, 2008]
[edit reason] fixed link [/edit]

[okonjiaustin]

Hi wheelie34,

I sure need the script, please do post it.

Thanks.

Austin.

[okonjiaustin]

Hi wheelie34,

Please when you get the display code let me know also.

Thanks.

[okonjiaustin]

Please try and help me with an image download script.

I have successfully uploaded image to mysql database using a form: This is the code:

<?php
$errmsg = "";
if (! @mysql_connect("localhost","root","")) {
$errmsg = "Cannot connect to database";
}
@mysql_select_db("adim");

if(isset($_REQUEST['submit']))
{

$imgtype=$_FILES['uploadfile']['type'];
$name=$_REQUEST['name'];
$address=$_REQUEST['address'];
$dateofbirth=$_REQUEST['dateofbirth'];

if($imgtype=="image/jpeg" ¦¦ $imgtype=="image/jpg" ¦¦ $imgtype=="image/pjpeg" ¦¦ $imgtype=="image/gif" ¦¦ $imgtype=="image/x-png" ¦¦ $imgtype=="image/bmp")
{

$image=$_FILES['uploadfile']['tmp_name'];
$fp = fopen($image, 'r');
$content = fread($fp, filesize($image));
$content = addslashes($content);
fclose($fp);
$sql="insert into img_tab1 (name,image,address,dateofbirth) values ('$name','$content','$address','$dateofbirth')";
$res=mysql_query($sql) or die (mysql_error());
}
}
?>

Now I need to display the row data including the image, can you please offer a code that will solve the problem.

Thanks

Austin

[edited by: jatar_k at 11:22 pm (utc) on Feb. 15, 2008]

[wheelie34]

As requested here is the final script that saves the image and data.

$uploaddir = "../apartments/images/";
$target = "350";
foreach ($_FILES as $key=>$value){
if ($value[size] > '0'){

$add = "../apartments/images/".$propname."+".$key.".jpg";
$sql = "insert into images values('', '$propname-$key.jpg', '$propname', '$key')";
if (mysql_query($sql, $dblink)){
echo "IMAGE added!<br><br>";
}else{
echo "Something was wrong<br><br>";
}
copy($_FILES[$key][tmp_name], $add);
chmod("$add",0644);

$new_pic = "$propname+$key.jpg";
$new_image = "$uploaddir$new_pic";

$sourcefile = "$uploaddir$new_pic";
$picsize = getimagesize("$sourcefile");
$source_x = $picsize[0];
$source_y = $picsize[1];

if ($source_x > $source_y)
{
$percentage = ($target / $source_x);
} else {
$percentage = ($target / $source_y);
}
$dest_x = round($source_x * $percentage);
$dest_y = round($source_y * $percentage);

$targetfile = "$uploaddir".$propname."-".$key.".jpg";
$jpegqual = 90;
$source_id = imagecreatefromjpeg("$sourcefile");
$target_id = imagecreatetruecolor($dest_x, $dest_y);
$target_pic = imagecopyresized($target_id,$source_id,0,0,0,0,$dest_x,$dest_y,$source_x,$source_y);
imagejpeg($target_id,"$targetfile",$jpegqual);
unlink("$new_image");
}
}

Hope it helps.