PHP get image size

Question

I'm using the following code but it gives me an error about an unexpeted " in the statement at this line, but I'm not sure how the syntax should go:

list($width,$height,$type,$attr) = getimagesize("' . $SESSION_["html_folder"] . '/uploadedfiles/' . $row['logo'] . '");

echo "<p>This logo is ".$width; x $.height; echo "pixels in size.</p>";

PHP says the error is on that first line.

The syntax highlighting here should show you what the problem is. You are opening a string with " but not closing it, needlessly as far as I can see. — Pekka
– Pekka, Commented Apr 6, 2011 at 14:58
its storing html_folder in your sessions is what im worried about — RobertPitt
– RobertPitt, Commented Apr 6, 2011 at 15:02

Your Common Sense · Accepted Answer · 2011-04-06 15:19:22Z

2

Try this

list($width,$height,$type,$attr) = getimagesize($SESSION_['html_folder'] . '/uploadedfiles/' . $row['logo']);

The error in your code was that you opened the string "twice", one with " and one with '. If you use a variable as parameter where a string is expected, you do not need to set the variable in quotes.

You should make sure that in $SESSION_['html_folder'] there's no malicious code, e.g. with

if(!is_dir($SESSION_['html_folder']))
   die("ERROR");

edited Apr 6, 2011 at 15:19

Your Common Sense

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answered Apr 6, 2011 at 14:58

strauberry

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4 Comments

RobertPitt Over a year ago

nitpick: the code is not executable, single quotes are such a tiny reduction in performance, and it will also not read escape characters

strauberry Over a year ago

@RobertPitt: the performance aim is not that big, I agree :-) But nevertheless I thought it is worth to say

Your Common Sense Over a year ago

No, it doesn't worth. There is no gain at all but only rumor. Call it your favorite coding style.

strauberry Over a year ago

@Col. Shrapel: thanks for editing my post, I just wanted to remove the unworth words ^^

Galen · Accepted Answer · 2011-04-06 14:59:10Z

0

youre adding all kinds of unnecessary quotes

list($width,$height,$type,$attr) = getimagesize($SESSION_["html_folder"] . '/uploadedfiles/' . $row['logo']);

answered Apr 6, 2011 at 14:59

Galen

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Comments

Mike Lewis · Accepted Answer · 2011-04-06 14:59:42Z

0

You want:

list($width,$height,$type,$attr) = getimagesize($_SESSION["html_folder"] . '/uploadedfiles/' . $row['logo']);

The main issue was that you were using double quotes to open a string, but you didn't close the string with double quotes. The above is a better formated string(and fixed your $_SESSION variable).

answered Apr 6, 2011 at 14:59

Mike Lewis

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