database image loader

[Q695]

How do I create an image from a database if there's one avalible, or if there isn't one I use a default photo?

 

Here is what I have so far:

<?php
include "../link/log.php";
include "../link/dead.php";

$sql="SELECT * FROM players WHERE id='$id';";
$result=@mysql_query($sql, $con) or die(death($sql));
$u_row=mysql_fetch_array($result);

if ($u_row[image]){
echo $u_row[image];
} else {
?>
<img src="sword.jpg">
<?php
}
?>

 

note:The two other files are in a different folder at the same level.

[Psycho]

Your code has the right logic, but there are just a few problems.

 

1. Unless you are needing the other fields, just query for the image.

2. You need to enclose the array index name within quotes.

3. You are only checking if the query returns a value for "image". You need to check if the value contains anything. I think that check might work if you are setting the value as NULL, but it won't work if the value is an empty string - which is what most people do

4. Unlsee you are storing the image as the complete html source to display the image (not recommended) you need to add the HTML tags to the path.

 

here is how I would do it:

<?php
include "../link/log.php";
include "../link/dead.php";

$sql="SELECT img FROM players WHERE id='$id';";
$result=@mysql_query($sql, $con) or die(death($sql));
$u_row=mysql_fetch_array($result);

//Determine the image to display
if ($u_row['image'] != '')
{
    $img = $u_row[image]; //Add the appropriate path
} else {
    $img = "sword.jpg"; //Add the appropriate path
}

//Display the image
echo "<img src=\"{$img}\">";

[Q695]

The image is physically being storred in the database as a blob, since it's in it's own folder with a pointer in the image file, kind of like how captchas are done.