PHP getimagesize with variable

Question

I'm trying to use the getimagesize function to get the height and with of an image. I'm pulling the image URL from a database. (The field ProjectURL contains a line such as xxx.jpg). However I'm getting an error.

Code:

$testing = "projects/'.$row['ProjectURL'].'";
    list($width, $height, $type, $attr) = getimagesize($testing);
    echo "Image width " .$width;
echo "<br />";
echo "Image height " .$height;

Error:

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING

Your first ' should be a ". Use a decent IDE or editor with syntax highlighting, it will make finding such errors much easier — Pekka
– Pekka, Commented Jan 19, 2011 at 23:49

Paul · Accepted Answer · 2011-01-19 23:49:31Z

6

it's because you are mixing single and double quotes...

this should be ok:

$testing = "projects/" . $row['ProjectURL'];
list($width, $height, $type, $attr) = getimagesize($testing);
echo "Image width " . $width;
echo "Image height " . $height;

You might also have noticed that I removed the echo "";... this one was useless :)

answered Jan 19, 2011 at 23:49

Paul

3,0202 gold badges23 silver badges16 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Nikola Sivkov Over a year ago

echo "<br />"; is not useless , otherwise it echoes them on the same line :) the better approach would be echo "Image width " . $width. "\n";

Paul Over a year ago

Last time I saw the article, there was no "<br />" but a \n... :)

Ian Over a year ago

Thanks very much Paul, it was the mixture of quotes. Working now, cheers :D

Collectives™ on Stack Overflow

PHP getimagesize with variable

1 Answer 1

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related