Trying to get image to display??

karmakiller

Hi,

I am trying to develop sort of like a directory.. but i can get my image to display.. I know it is in my code, but i am really new to php (only 2 weeks) everything else in the database displays but not my image..? I have set my table field that holds the image to longblob and i can upload images to this database.... i just cannot display it...please can you help??

Here is my php code, please be king as i am very new to php... and i know that the code is very simple to you. The name of the image column is 'photo'

<html>
<style type="text/css">

</style>
<body>
<span class="style1"><a href="javascript:history.go(-1);"><font size="2" face="Verdana, Arial, Helvetica, sans-serif">Go
back to Search</font></a>
</span>
<p class="style1"><font size="2" face="Verdana, Arial, Helvetica, sans-serif">Your
Search produced the following results:</font></p>
<p> </p>
<p> <font size="2" face="Verdana, Arial, Helvetica, sans-serif">
<?php

mysql_connect (localhost, root, password);

mysql_select_db (directory);

if ($townorcounty == "")
{$townorcounty = '%';}

if ($county == "")
{$county = '%';}

$result = mysql_query ("SELECT * FROM company
WHERE (town LIKE '$townorcounty%'
OR county LIKE '$townorcounty%')
");

if (!mysql_num_rows($result)) {
print('No search results');
}
else {
print('<table border=0 cellpadding=0 cellspacing=10>');
print('<tr>');
print ("<td><div align=center><strong>Company Name</strong></div></td>");
print ("<td><div align=center><strong>Title</strong></div></td>");
print ("<td><div align=center><strong>Town</strong></div></td>");
print ("<td><div align=center><strong>Tel</strong></div></td>");
print ("<td><div align=center><strong>Photo</strong></div></td>");
print ("</tr>");
while ($row = mysql_fetch_array($result)) {
print('<tr><td>');
print $row["Company"];

print ("</td><td> ");
print $row["title"];
print ("</td><td> ");
print $row["town"];
print ("</td><td> ");
print $row["tel"];
print ("</td></tr>");
print $row["photo"];
print ("</td></tr>");
}
print('</table>');
}
?>
</font> </p>
</body>
</html>

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- 🙂

RyanJones

First let me get some thing clear, is the image in a file or stored in the database with no file required?

If the later then you need to output the image with a mimetype which will cause problems for the rest of your page which will also be treated as an image.

Thats the problem with images stored in a database that way, they normally need to be written to a tempfile before they na be displayed.

If the first then can you point out which part of your code outputes the image, and can you provide an example of your code online?

Also, PLEASE use the PHP UBBC tags 🙂

Cheers,

Ryan Jones

karmakiller

to be honest I dont have a clue about all you have asked.... this is my first php project for myself ( newbie as in baby!!)

the image file is stored in the database... but its the line of code that i need (cos i dont know it) that will show the image.. at the moment all i have is ( print $data["photo"]; ) but i know this isnt the right code to display the image... thats where i am stuck...

here is my code again...

mysql_connect (localhost, root, louvain);

mysql_select_db (directory);

if ($townorcounty == "")
{$townorcounty = '%';}

if ($county == "")
{$county = '%';}

$result = mysql_query ("SELECT * FROM company
WHERE (town LIKE '$townorcounty%'
OR county LIKE '$townorcounty%')
");

print ("</td><td> ");
print $row["title"];
print ("</td><td> ");
print $row["town"];
print ("</td><td> ");
print $row["tel"];
print ("</td></tr>");
print $data["photo"];
print ("</td></tr>");
}
print('</table>');
}
?>
</font> </p>
</body>
</html>

RyanJones

If the file is stored in the database then it will need to be written to a file before you an display it (It can then be displayed by using the HTML IMG tag.) 🙂

Cheers,

Ryan Jones

karmakiller

Ok, thanks for that........ could you point me to how I get this process working... how do i get the image into a temp file.. and then into the page i require?

Is there not a simpler more efficient way to include images... I read that storing images in the database was a bad way of doing things...

If you can point me to any sample code that I can use or play around with, i would be very grateful..

Thanks

RyanJones

Try something like the modifed code in the file attatched.

Cheers,

Ryan Jones

karmakiller

Thanks for that... I have tested it and iut comes back with an error

Warning: imagepng(): supplied argument is not a valid Image resource in d:\inetpub\wwwroot\test44\searchosp2.php on line 70

RyanJones

OOPs, big mistake, its a resource argument and not a string... Sorry about that - I am used to working with files not image strings 🙂

Try the V2 code attatched 🙂

Cheers,

Ryan Jones

karmakiller

I now get the folowing errors....... should i keep my database photo column as 'LONGBLOB' ??

Warning: fopen(w): failed to open stream: No error in d:\inetpub\wwwroot\test44\searchosp2.php on line 70

Warning: fwrite(): supplied argument is not a valid stream resource in d:\inetpub\wwwroot\test44\searchosp2.php on line 71

Warning: fclose(): supplied argument is not a valid stream resource in d:\inetpub\wwwroot\test44\searchosp2.php on line 72

RyanJones

karmakiller wrote:
I now get the folowing errors....... should i keep my database photo column as 'LONGBLOB' ??

Warning: fopen(w): failed to open stream: No error in d:\inetpub\wwwroot\test44\searchosp2.php on line 70

Warning: fwrite(): supplied argument is not a valid stream resource in d:\inetpub\wwwroot\test44\searchosp2.php on line 71

Warning: fclose(): supplied argument is not a valid stream resource in d:\inetpub\wwwroot\test44\searchosp2.php on line 72

I have no idea why those errors are happening.

Would it not be easier to upload the images to a directory and then insert a URL into the database instead?

Also, check that the diretory has the correct permissions to allow files to be written to 🙂

Cheers,

Ryan Jones

karmakiller

I defo have the permissions.

Yeah that would be easier.... you mean have my photo column as a VARCHAR rather than a BLOB? and then have the http:// link to the file?

If this is what you do mean... i still dont know the code to display the image from a http held in the database.....

can you assist in that?

RyanJones

karmakiller wrote:
I defo have the permissions.

Yeah that would be easier.... you mean have my photo column as a VARCHAR rather than a BLOB? and then have the http:// link to the file?

If this is what you do mean... i still dont know the code to display the image from a http held in the database.....

can you assist in that?

Thats it exactly 🙂

If you can give me the column name for the http field, I can modify the code I posted earlier to display the image 🙂

Cheers,

Ryan Jones

karmakiller

The colum is photo 🙂

RyanJones

Same field, OK then 🙂

Try the V3 code attatched.

Cheers,

Ryan Jones

karmakiller

Nearly....

it produces a blank image with the text 'use posted image' but there is no image displayed..

when i right click on the image and view properties it has the link to:

http://localhost/test44 and nothing else?

RyanJones

karmakiller wrote:
Nearly....

it produces a blank image with the text 'use posted image' but there is no image displayed..

when i right click on the image and view properties it has the link to:

http://localhost/test44 and nothing else?

If $data["photo"]; is reading the correct field then the URL to the image is incorrect. Check the link in your database is valid 🙂

Cheers,

Ryan Jones

herve

You can also use this kind of idea :

In the first script, instead of :

print ("<td><div align=center><strong>Photo</strong></div></td>");

you put :

print "<td><div align=center><img src=\"display_img.php?townorcounty=$townorcounty\"></div></td>";

Which mean there's an image, but instead of having a static file
like "image.gif" or "image.jpg" you have a call to another PHP script.

and then the code for the second script :

<?PHP
	$townorcounty = $_REQUEST["townorcounty"];  // Get the id from command line

mysql_connect (localhost, root, louvain);
mysql_select_db (directory);	

$cmd = "SELECT photo FROM company WHERE (town LIKE '$townorcounty%' OR county LIKE '$townorcounty%') "; 
$result = mysql_query($cmd);
$myrow  = mysql_fetch_array($result);

$photo_data = $myrow["photo"];

header ("Content-type: image/jpg"); // Here you may have to change if the image are GIF or PNG

print $photo_data;  // Simply output the raw binary data of the image

?>