Question #272088 on Sikuli changed: https://answers.launchpad.net/sikuli/+question/272088
Status: Open => Answered RaiMan proposed the following answer: Yes, one event for one image (currently ;-) This is the appropriate solution for version 1.1.0+: r = Region(Region(2516,495,694,273)) leftSide = [("left1.png"), ("left2.png"), ("left3.png"), ("left4.png"), ("left5.png"), ("left6.png"), ("left7.png"), ("left8.png"), ("left9.png"), ("left10.png"), ("left11.png")] rightSide = [("right1.png"), ("right2.png"), ("right3.png"), ("right4.png"), ("right5.png"), ("right6.png"), ("right7.png"), ("right8.png"), ("right9.png"), ("right10.png"), ("right11.png"), ("right12.png")] neitherSide = [("center1.png"), ("center2.png"), ("center3.png"), ("center4.png")] def moveLeft(event): type(Key.LEFT) event.repeat(0.2) # continue observation after 0.2 seconds def moveRight(event): type(Key.RIGHT) event.repeat(0.2) # continue observation after 0.2 seconds # register the events with region r # one event == one image for ls in leftSide: r.onAppear(ls, moveRight) for rs in rightSide: r.onAppear(rs, moveLeft) nor rs in neitherSide: r.onAppear(ns, moveRight) # run the observation r.observe(FOREVER) -- You received this question notification because your team Sikuli Drivers is an answer contact for Sikuli. _______________________________________________ Mailing list: https://launchpad.net/~sikuli-driver Post to : sikuli-driver@lists.launchpad.net Unsubscribe : https://launchpad.net/~sikuli-driver More help : https://help.launchpad.net/ListHelp