Making a variable an image

7724

I have this code which extracts data from mysql database:

<?

# select all records ordered by date
$SQL = "SELECT * FROM `pics`" ;

# make the query (assume connection is there)
$result = mysql_query($SQL) OR die(mysql_error());


# iterate the result set
while ($row = mysql_fetch_assoc($result)){
   # build display rows by concatenation

$output .= "<table width='38%' border='0' cellpadding='1' cellspacing='1' bgcolor='#CCCCCC' align='center' vspace='6' hspace='6'>";
$output .= "<tr bgcolor='#FFFFFF'>\n";
$output .= "<td><img src=../uploader/news/<?php echo $name; ?>.jpg></td>\n";
$output .= "</tr>\n";
$output .= "<tr bgcolor='#EEEEEE'>\n";
$output .= "<td>".$row['caption']."</td>\n";
$output .= "</tr>\n";
$output .= "<br>\n";
}
# check if anything is in output
if (!$output){
   $output = "<tr><td>No pics found</td></tr>\n";
}
# print the table
echo "<table width='38%' border='0' cellpadding='1' cellspacing='1' bgcolor='#CCCCCC' align='center' vspace='6' hspace='6'>\n".$output."</table>";
# done
?>

Now this line:

$output .= "<td><img src=../uploader/news/<?php echo $name; ?>.jpg></td>\n";

I want the code to output as an image rather than name.jpg, how would I modify the code to do this?

Thanks

Chris

justsomeone

I'm not sure I understand your question.

The code snippet you've shown doesn't set any value for $name. Is this correct? has the value been set outside of the snippet?

Or should you be using $row['name'] ? Is it a field returned from the query?

The following line confuses me further:

$output .= "<td><img src=../uploader/news/<?php echo $name; ?>.jpg></td>\n";

Is this line part of your main PHP code (It looks like it)? If so, then you shouldn't have the extra "<?php ... ?>" tags in there.

Assuming that $name is actually populated with something, say a value of "news01", then

$output .= "<td><img src=../uploader/news/$name.jpg></td>\n";

followed by an echo or print of $output will send a command to the browser to download and display the image found at ../uploader/news/news01.jpg on your site.

Does that help clarify at all?

7724

Thanks:

1) Yep the value $name has been set outside the code snippett. I just did not want to paste the whole thing.

2) This code:

$output .= "<td><img src=../uploader/news/<?php echo $name; ?>.jpg></td>\n";

I have added thanks, but it just outputs as: uploader/news/.jpg, ignoring the variable.

Now I know you say about echoing, but surely where I echo the $output in the table in the third but last line should make this variable appear?

Thanks

justsomeone

I take it that this line outputs what you expect:
$output .= "<td>".$row['caption']."</td>\n";

and that this one doesn't:
$output .= "<td><img src=../uploader/news/<?php echo $name; ?>.jpg></td>\n";

It looks to me like the nested "<?php ... ?>" tags are the problem.

You only need these when you're dropping some php into html. but in this case you're already in php, so you don't need 'em!

Have you tried:
$output .= "<td><img src=../uploader/news/".$name.".jpg></td>\n";