Simple PHP Variable Woes

[da_mic_1530]

After taking advice from IIRC DJ Jestar about using the glob function I have ended up with this

		<?php  
  
  foreach glob("'$d_no'.JPG") as $file) {
    
	
    echo '<td>' . "<a href='$file' target=_blank><img  src='$file'  alt='$file'  width='220' height='220'></a>" . '</td>'; 
	}   
echo '</table>'; 

  
?>

Now the above works when I use

(glob("'7.JPG") as $file) {

as there is a file called 7.JPG

However I'm trying to use the variable $d_no, which when echoed on that page displays fine, however if I used the first piece of php code, globing the variable $d_no, I'm met with

"Warning: Invalid argument supplied for foreach() in /home/irgscouk/public_html/userpage.php on line 61"

I'm sure it's just the way I'm declaring my variable

Any ideas?

 

[da_mic_1530]

basically in the same directory as this php script I have a few images

1.JPG
2.JPG etc etc

I want the relevant picture to be displayed upon the choice of the variable

So when id = 1 display 1.JPG

As I said, if I MANUALLY say, display Just 2.JPG, its fine and happy

If I say, dynamically display that variable.JPG, it's not happy

Jestar, I'm using the variable d_no on other things on that page, and they are coming out fine

The variable is definately The number asked

 

[da_mic_1530]

Think you're after:

foreach glob($d_no . ".JPG") as $file) {

and see Jestar's explanation about mixing quotes.

that just gives me

"Parse error: syntax error, unexpected T_STRING, expecting '(' in /home/irgscouk/public_html/userpage.php on line 61
"

 

[da_mic_1530]

Sorted

foreach (glob("$d_no.JPG") as $file) {

Was the end result

 

[da_mic_1530]

no no, I was looking for a file where filename = variable

my point was it wasn't working as I was declaring my variable wrong

The other point was that the script was working due to the fact that if I MANUALLY declared a file, like 2.jpg it worked so I knew it was how I was declaring the variable, hence the thread

Anyhoo it's all sorted now