Seeing if a variable contains something?

DoD

Is there a way that, i could get a small script that would check if a varibale (such as an image name) had .jpg within it? I so far have not found anything on this...

juschillinnow

if you just want to check to see if the "image name " have the ".jpg" in it.....

use preg_match or a related check. There's another one for doing it but it doesn't come to mind right off hand.

Psyche

$myImage = "blah.jpg";
$ImageExtension = explode(".", $myImage);
if ($ImageExtension[1] == "jpg")
{
  echo "Hurray! It's a jpg! :)";
}
else
{
  echo "Uhuhu...";
}

DoD

I was reasing preg_match_all on php.net and it gave me that idea as it exploded them... you just gave me the code... thanks!

I have the file uploading from a form... From their machine. Is the variable $POST_["Image"] going to be the name of this image? or the actual image? As i want to see if its jpg or gif before its saved...

devinemke

problems would arise in Psyche's code if the file name was something like "this.is.my.file.jpg" or if the file name was "myfile.JPG".

i suggest this instead:

$file = 'filename.jpg';
$file_array = explode ('.', strtolower ($file));
$count = count ($file_array) - 1;
if ($file_array[$count] == 'jpg') {echo "<b>$file</b> is a JPG image";}
else {echo "<b>$file</b> is not a JPG image";}

philipolson

See also pathinfo. Here's a simple example:

<?php
$dir = '.';
$fd = opendir($dir);

$types = array('jpg', 'jpeg', 'gif', 'png', 'bmp');

$html = '';
while ($file = readdir($fd)) {
    $path = pathinfo($file);
    if (is_readable("$dir/$file") && in_array(strtolower($path['extension']), $types)) {
        $html .= "<p><img src='$dir/$file' alt='$file'></p>";
    }
}
?>

DoD

Path info did it!

	$imgs=$HTTP_POST_FILES['simage']['name'];

	$path_parts=pathinfo ("$imgs");
	$dot=".";
	$ext=$path_parts["extension"];
	$imgname="$id$dot$ext";

su1d

Guys, are you intentionally choose the longer, the difficulter, and ...the worse way? =)

if('jpg' === strtolower(substr($filename, 1+strrpos($filename, '.'))))
     echo ".JPG file";

devinemke

...and what if the file is called "image.jpeg"?
sometimes a longer, difficulter, and "worse" way is required.

DoD

I think this is much easier:

        $path_parts=pathinfo ("$imgs");
       echo $path_parts["extension"];

su1d

Originally posted by devinemke
...and what if the file is called "image.jpeg"?
sometimes a longer, difficulter, and "worse" way is required.

pathinfo() returns 'jpg' even when the extension is 'jpeg'? =)
there's no need to call a parser function (and, btw, waste an array variable) just to get an extension of a file.
when you need to kill a fly, you don't use an AK-47, do you? =)

DoD

Yours IS longer and more complicated. I can understand mine but not yours.

Also.

jpeg jpg... whats the difference? one letter! no file type, no nothing! so i can save as .jpg if i want to or whatever it spurts out.

philipolson

Using pathinfo() is perfectly fine here, it's not too much overhead. There are a million ways to get a files extension, what I like about pathinfo is it makes it very clear what is happening in the code. It's called personal preference.

And no, pathinfo does not return jpg if the extension is jpeg, that is bogus.