[SOLVED] Disrorted image

[oceans]

Dear Friends,

 

I src a jpg file, I noted it gets distored, there are not many paramerters to play.

So I conclude, if the image width and height does not equal to src width and height, the image will be bad. Am I right?

 

If so how can i get arround, because it is sick to walk at pixel level. :'(

[AndyB]

If you distort an image (by setting width or height to incorrect values or an incorrect ratio) then yes it gets distorted. What else would you expect??? That's it. There's no way around it.

 

Quite what 'walk at the pixel level' means beats me.

 

I wonder if getimagesize() has anything to do with a solution - if that's what you need.

[oceans]

ha ha ha ha, you are funny, really I am laughing    , I was typing fast earlier, also had a bit more serious thing in my mind while typing this issue.

 

Later I noted, if the proportion ratio is maintained thing are slightly ok, but i am doing an image upload from clients, they have all sort of sizes, yes i can mention i only accept this size, ya by the way can i get the width and height of an image, that way i can force the client to crop to my size, may be your getimagesize() will help me

[Dragen]

getimagesize() wil give you the width and height of any image.

You could use it in whatever upload form you have for clients to check that the image is a ertain size..

or you could use it wherever you display the image and set the images width and height to whatever it actually is, instead of having to guess

[oceans]

Dear Dragen,

 

I tried it is not giving me a value,

 

I use this to validate a potential upload file size $_FILES["file"]["size"],

 

I want to validate width & height how can I,

 

I tried getimagesize(["file"]), it is not working, I am unsure how can a function retuen 2 values?

 

error I got

"

Warning: getimagesize(Array) [function.getimagesize]: failed to open stream: No such file or directory

"

[Dragen]

use it like this:

$image = 'http://www.mydomain.com/myimage.gif'
list($width, $height) = getimagesize($image)

echo '<img src="' . $image . '" width="' . $width . '" height="' . $height . '" border="0" alt="" />