10

For example if I draw a oval and then want to paint part of it into different color, how can I connect edge points of my new path with oval?

paths

  • It looks as though I'm not able to join two open points in a shape either. Have tried a few work arounds, and it's inefficient to build a third shape to cut off the part of the original shape - just because the jin command isn't working. Currently using SKetch3. (Have used illustrator for 15 year....and photoshop, etc) – user46412 Aug 5 '15 at 0:55
  • Sketch is not a drawing tool, so you can't expect the same functionality as Illustrator has. Most of the "obvious" Illustrator solutions are simply not there, and will never be :) – Igor Dec 21 '16 at 15:07
12

What you need to do is this:

  1. Expand the small path, so that its bounds extend over the outline of the oval (you'll need to click 'Close Path' in the inspector to make it a close path):

    Extend Path

  2. Duplicate the big oval:

    Duplicate Oval

  3. Select the newly duplicate oval and the small shape:

    Select Both

  4. Select Layer › Combine › Intersect:

    Intersect

  5. Select Layer › Paths › Flatten:

    Flatten

  6. Ta-da!:

    Done

Hope it helps :)

  • great answer----but with a different kind of shape, what I'd like to do is what illustrator allows me to do-- draw a line from one point and click on an open point of another shape-- to connect two line segments in a situation where where the above solution isn't viable. Can this be done? Thanks! – user46412 Jul 8 '15 at 0:17
0

You could use the big oval as a mask. Then everything you draw on top of it will be clipped by it. This way the shapes are a bit easier to edit than with booleans. Masking in sketch works pretty much the same as using clipping masks in Photoshop.

(The join command only works on two open paths. It will connect them from their end points. You can cut off a section of a closed path with the scissors tool.)

enter image description here

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