output from shell_exec

RobbiK

Ehemm, I'm having a small problem . I have a form for some image manipulation, resizing images etc. with ImageMagick. There is a checkbox to convert images to jpeg format, and in my code I want to check first if the image is already in jpeg format...sounds easy??

Here is what I've done:

$output = shell_exec("identify -format %m $image");
if ($output != "JPEG")

If I echo $output I get JPEG but the if sentance doesn't work, i.e. it alway executes the code inside the if clause. Any ideas???

Weedpacket

Well, you kind of stopped just when things were getting interesting with your code, there 🙂, but if you're saying that

$output = shell_exec("identify -format %m $image");
if ($output != "JPEG")
{  echo "AARGH!";
    exit();
}

will output "AARGH!" and stop running (try it as a wee test), then the only answer can be that $output does not equal JPEG.

So change the above test code slightly:

$output = shell_exec("identify -format %m $image");
if ($output != "JPEG")
{  echo "<pre>------\n[$output]\n------</pre>";
    exit();
}

and see just what $output is (the [] and ------ bits are there so that you can see if there are any leading or trailing spaces or newlines or whatnot). What you want to see (i.e., if $output really does equal JPEG) is

[code]

[JPEG]

[/code]

RobbiK

Thank you for your answer, I did find out that strlen($output) was 6 😃

echo "<pre>------\n[$output]\n------</pre>"; is a nice way to see what $output actually "outputs", nice tip Weedpacket

My output was

[JPEG

]

So the $output is actually "JPEG\n\n"...

Just trimmed the output: $output = rtrim($output) (I should have figured this out sooner, guess it is friday morning and I'm excited about the weekend)

It works perfectly now, thanks

Weedpacket

Glad to help 🙂