Simple PHP Variable Woes

da_mic_1530 · 19 Mar 2009 at 11:39

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

PHP:

		<?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

PHP:

(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?

Dj_Jestar · 19 Mar 2009 at 11:42

Be sure to know exactly what is being sent to glob..

PHP:

$d_no = 'foo';
echo "'$d_no'.JPG";

this will output:

Code:

'foo'.JPG

I'm not sure you want that?

inph · 19 Mar 2009 at 11:45

Think you're after:

Code:

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

and see Jestar's explanation about mixing quotes.

da_mic_1530 · 19 Mar 2009 at 11:51

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 · 19 Mar 2009 at 11:52

inph said:
Think you're after:

Code:

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 · 19 Mar 2009 at 11:57

Sorted

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

Was the end result

Inquisitor · 19 Mar 2009 at 12:34

If you only want a single file, whose name you already know, then why are you looping over glob? :confused:

Surely this does exactly the same thing:

PHP:

<?php

$file = $d_no . '.JPG';
echo '<td><a href="' . $file . '" target="_blank"><img  src="' . $file . '"  alt="' . $file . '"  width="220" height="220" /></a></td>'; 
echo '</table>';

?>

Also, just to be pedantic: TD elements should be contained within TR elements in HTML tables.

da_mic_1530 · 19 Mar 2009 at 13:00

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