Take the 2-minute tour ×
Graphic Design Stack Exchange is a question and answer site for professional graphic designers and non-designers trying to do their own graphic design. It's 100% free, no registration required.

I am designing a website with Adobe Fireworks and I want to have a design where I have divs with rounded corners. Inside those divs I want to have an image that gets cut (rounded) by the containing div (rectangle in Fireworks) with rounded corners.

I found this tutorial, though it cuts out everything inside the the vector image. With a bit of tricking (ctrl-x) the marquee and delete the old image underneath I was able to do this.

I can't imagine there isn't a better way, but how?

BTW: this is not in HTML or CSS but in Adobe Fireworks.

share|improve this question
1  
Set the image as a background image for the div in the CSS. Otherwise, the image itself must be saved with rounded corners. This is a question better suited for stackoverflow.com if you need CSS advice. –  Scott Apr 17 '13 at 14:02
    
I meant in fireworks :) –  Bart Burg Apr 17 '13 at 14:44
add comment

2 Answers

up vote 1 down vote accepted

Sounds like you're describing Fireworks masks (the equivalent of a clipping mask in Illustrator).

In Fireworks:

  • Put the vector you want to crop the image over the image where you want it cropped.
  • Cut it (it remembers where it was)
  • Select the image to be cropped
  • Edit > Paste as Mask.

This crops the image to that shape. Unlike chopping it up manually with a marquee, it keeps the hidden parts of the image intact behind the scenes. If you grab the anchor points of the vector and move them around, the mask moves revealing and hiding different areas of the original image.

share|improve this answer
    
Thanks, that's what I was looking for :) –  Bart Burg Apr 18 '13 at 7:52
add comment

You can also try http://johndunning.com/fireworks/about/SmartPunch very very useful, even allows vectors to punch through bitmaps.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.