[SOLVED] Show images using mysql data.

[marksie1988]

ok i have a script that can do ip2country and i need to show the flags on it but i dont know how to call the data from the database and then add the .png image extention on the end of it anyone know?

 

cheers

Steve

[MadTechie]

why not just use

<?php
header("Content-Type: image/png");
imagepng($DBimg);
?>

 

of course $DBimg is the DB image

[marksie1988]

would this work as the database doesnt know they are images the database has the country code for example united kingdom is UK and the image name is uk.png but the database doesnt have a specific image entry.

[MadTechie]

but the database doesnt have a specific image entry.

 

what do you mean ?

[marksie1988]

ok i have the following database (also included one line of data for you)

 

CREATE TABLE `IPCountry` (
  `IP_START` varchar(50) default NULL,
  `IP_END` varchar(50) default NULL,
  `IP_FROM` double default NULL,
  `IP_TO` double default NULL,
  `COUNTRY_CADE2` char(2) default NULL,
  `COUNTRY_NAME` varchar(50) default NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

-- 
-- Dumping data for table `IPCountry`
-- 

INSERT INTO `IPCountry` VALUES ('2.6.190.56', '2.6.190.63', 33996344, 33996351, 'GB', 'United Kingdom');

 

And here is the php all is does is uses data in the database to find out what country an ip address is from what i want to be able to do is when the country is shown a small image of that countries flag is shown.

 

but the database doesnt have the image locations in it and with over 60000 rows in the database its not possible to do so.

 

<?php
$DatabaseServer = "dbserver"; 
$Username = "dbuser"; 
$Password = "dbpass"; 
$DatabaseName = "dbname"; 
$link = mysql_connect($DatabaseServer, $Username, $Password) or die('Could not connect: ' . mysql_error()); 
mysql_select_db($DatabaseName) or die('Could not select database'); 
$IP = $_SERVER["REMOTE_ADDR"]; //Get the IP address 
$res = mysql_query("SELECT country_cade2,country_name FROM IPCountry WHERE IP_FROM<=inet_aton('$IP') AND IP_TO>=inet_aton('$IP')");//look up IP address 
$Codes = mysql_fetch_array($res); //get result 
$CountryCode = $Codes['country_cade2']; //two-letter country code 
$CountryName = $Codes['country_name']; //full country name 
echo "$CountryCode - $CountryName"; //print it out, just as an example 
mysql_close($link); //clean up
?>

[MadTechie]

ahh so

echo "<img src='image/$CountryCode.png' alt='$CountryName' />";

 

i assume that you have an image in an image folder called GB.png etc

 

 

EDIT Changed alt to $CountryName

[marksie1988]

ok i tried that and it works but i had to change the image name to uppercase as the GB is shown in uppercase is it supposed to be case specific ?

[penguin0]

yes images are case specific.  GB is not going to work with gb.

[MadTechie]

Make all image names uppercase and use this

 

$CountryCode = strtoupper($CountryCode);
echo "<img src='image/$CountryCode.PNG' alt='$CountryName' />";

 

EDIT:

images are case specific ON *nix servers not windows, the above code will sort this

[marksie1988]

ok then i guess at least it works now will just need to get round to renaming all 502 images to uppercase :S not fun

 

thanx for the help

[MadTechie]

If they are lower case use this

$CountryCode = strtolower($CountryCode);
echo "<img src='image/$CountryCode.png' alt='$CountryName' />";

[marksie1988]

aww well good it works now you guys are great help here and very speedy reply:)