或者我可以在将项目添加到 ScrollView
后访问它吗?小部件?
例子:
local scrollView = widget.newScrollView {...}
scrollView:insert(display.newImage("img1.png", 0, 0))
scrollView:insert(display.newImage("img2.png", 100, 0))
接下来我想从
scrollView
中删除第一张图片:scrollView:remove(1) -- has no effect
更新 :我的解决方案:
local scrollView = widget.newScrollView {...}
scrollView.content = {}
scrollView.content[#scrollView.content+1]= display.newImage("img1.png", 0, 0)
scrollView:insert(scrollView.content[#scrollView.content])
scrollView.content[#scrollView.content+1]= display.newImage("img2.png", 0, 0)
scrollView:insert(scrollView.content[#scrollView.content])
...
-- at some point I want to delete some item
scrollView.content[n]:removeSelf()
table.remove(scrollView.content, n)
最佳答案
你可以这样做:
local scrollView = widget.newScrollView {...}
local img_1 = display.newImage("img1.png", 0, 0)
local img_2 = display.newImage("img2.png", 100, 0)
scrollView:insert(img_1)
scrollView:insert(img_2)
然后:
img_1:removeSelf()
-- or
img_2:removeSelf()
继续编码……………… :)
关于lua - 有没有办法从 ScrollView 小部件中删除项目?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/19659310/