SectionIndexTitle in TableView
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I am creating list in TableView according to this tutorial. However I would like to have all leters in IndexTitle. It is OK, if I only add all leters to carSectionTitles array? I think it doesn't work correctly in simulator. Could someone help me?
swift tableview
asked Nov 7 at 19:16
Max
12
add a comment |
up vote
-2
down vote
favorite
I am creating list in TableView according to this tutorial. However I would like to have all leters in IndexTitle. It is OK, if I only add all leters to carSectionTitles array? I think it doesn't work correctly in simulator. Could someone help me?
swift tableview
asked Nov 7 at 19:16
Max
12
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I am creating list in TableView according to this tutorial. However I would like to have all leters in IndexTitle. It is OK, if I only add all leters to carSectionTitles array? I think it doesn't work correctly in simulator. Could someone help me?
swift tableview
asked Nov 7 at 19:16
Max
12
I am creating list in TableView according to this tutorial. However I would like to have all leters in IndexTitle. It is OK, if I only add all leters to carSectionTitles array? I think it doesn't work correctly in simulator. Could someone help me?
swift tableview
swift tableview
asked Nov 7 at 19:16
Max
12
asked Nov 7 at 19:16
Max
12
asked Nov 7 at 19:16
Max
12
asked Nov 7 at 19:16
Max
12
asked Nov 7 at 19:16
Max
12
12
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
-1
down vote
This is by no means an elegant solution, but in your view did load, replace:
// 2
carSectionTitles = [String](carsDictionary.keys)
carSectionTitles = carSectionTitles.sorted(by: { $0 < $1 })
with the following:
// 2
carSectionTitles = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "U", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"]
With everything else the same I think this will do what you want.
answered Nov 7 at 20:20
grbcp5
262
How can I do that more elegantly? I would like to learn it. btw, as I mentioned, I tried this, but I think it doesn't work correctly.
– Max
Nov 7 at 20:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
This is by no means an elegant solution, but in your view did load, replace:
// 2
carSectionTitles = [String](carsDictionary.keys)
carSectionTitles = carSectionTitles.sorted(by: { $0 < $1 })
with the following:
// 2
carSectionTitles = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "U", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"]
With everything else the same I think this will do what you want.
answered Nov 7 at 20:20
grbcp5
262
How can I do that more elegantly? I would like to learn it. btw, as I mentioned, I tried this, but I think it doesn't work correctly.
– Max
Nov 7 at 20:34
add a comment |
up vote
-1
down vote
This is by no means an elegant solution, but in your view did load, replace:
// 2
carSectionTitles = [String](carsDictionary.keys)
carSectionTitles = carSectionTitles.sorted(by: { $0 < $1 })
with the following:
// 2
carSectionTitles = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "U", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"]
With everything else the same I think this will do what you want.
answered Nov 7 at 20:20
grbcp5
262
How can I do that more elegantly? I would like to learn it. btw, as I mentioned, I tried this, but I think it doesn't work correctly.
– Max
Nov 7 at 20:34
add a comment |
up vote
-1
down vote
up vote
-1
down vote
This is by no means an elegant solution, but in your view did load, replace:
// 2
carSectionTitles = [String](carsDictionary.keys)
carSectionTitles = carSectionTitles.sorted(by: { $0 < $1 })
with the following:
// 2
carSectionTitles = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "U", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"]
With everything else the same I think this will do what you want.
answered Nov 7 at 20:20
grbcp5
262
This is by no means an elegant solution, but in your view did load, replace:
// 2
carSectionTitles = [String](carsDictionary.keys)
carSectionTitles = carSectionTitles.sorted(by: { $0 < $1 })
with the following:
// 2
carSectionTitles = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "U", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"]
With everything else the same I think this will do what you want.
answered Nov 7 at 20:20
grbcp5
262
answered Nov 7 at 20:20
grbcp5
262
answered Nov 7 at 20:20
grbcp5
262
answered Nov 7 at 20:20
grbcp5
262
262
How can I do that more elegantly? I would like to learn it. btw, as I mentioned, I tried this, but I think it doesn't work correctly.
– Max
Nov 7 at 20:34
add a comment |
How can I do that more elegantly? I would like to learn it. btw, as I mentioned, I tried this, but I think it doesn't work correctly.
– Max
Nov 7 at 20:34
How can I do that more elegantly? I would like to learn it. btw, as I mentioned, I tried this, but I think it doesn't work correctly.
– Max
Nov 7 at 20:34
How can I do that more elegantly? I would like to learn it. btw, as I mentioned, I tried this, but I think it doesn't work correctly.
– Max
Nov 7 at 20:34
add a comment |
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