Write to Nested Dictionary (Swift 4)
I have declared a dictionary in Swift as so: var dict = [String: [String: [String]]]()
.
What I am trying to do now is to write to the nested dictionary. I have used both codes below, however, none of them work as the initial key does not exist:
dict["Test"]?["One"] = ["Failed"]
dict["Test"]!["One"] = ["Failed"]
What I am trying to do is to create a key for ["One"]
, much like how you can create a key for a normal dictionary using dict[key]
.
swift dictionary collections
add a comment |
I have declared a dictionary in Swift as so: var dict = [String: [String: [String]]]()
.
What I am trying to do now is to write to the nested dictionary. I have used both codes below, however, none of them work as the initial key does not exist:
dict["Test"]?["One"] = ["Failed"]
dict["Test"]!["One"] = ["Failed"]
What I am trying to do is to create a key for ["One"]
, much like how you can create a key for a normal dictionary using dict[key]
.
swift dictionary collections
did you solve it?
– Gustavo Vollbrecht
Nov 13 at 16:09
yes i did! thank you, but I don't know why I can't accept your answer. I don't see the green checkmark.
– M.I
Nov 13 at 20:45
add a comment |
I have declared a dictionary in Swift as so: var dict = [String: [String: [String]]]()
.
What I am trying to do now is to write to the nested dictionary. I have used both codes below, however, none of them work as the initial key does not exist:
dict["Test"]?["One"] = ["Failed"]
dict["Test"]!["One"] = ["Failed"]
What I am trying to do is to create a key for ["One"]
, much like how you can create a key for a normal dictionary using dict[key]
.
swift dictionary collections
I have declared a dictionary in Swift as so: var dict = [String: [String: [String]]]()
.
What I am trying to do now is to write to the nested dictionary. I have used both codes below, however, none of them work as the initial key does not exist:
dict["Test"]?["One"] = ["Failed"]
dict["Test"]!["One"] = ["Failed"]
What I am trying to do is to create a key for ["One"]
, much like how you can create a key for a normal dictionary using dict[key]
.
swift dictionary collections
swift dictionary collections
edited Nov 11 at 22:58
Hamish
45k7103158
45k7103158
asked Nov 11 at 22:55
M.I
13512
13512
did you solve it?
– Gustavo Vollbrecht
Nov 13 at 16:09
yes i did! thank you, but I don't know why I can't accept your answer. I don't see the green checkmark.
– M.I
Nov 13 at 20:45
add a comment |
did you solve it?
– Gustavo Vollbrecht
Nov 13 at 16:09
yes i did! thank you, but I don't know why I can't accept your answer. I don't see the green checkmark.
– M.I
Nov 13 at 20:45
did you solve it?
– Gustavo Vollbrecht
Nov 13 at 16:09
did you solve it?
– Gustavo Vollbrecht
Nov 13 at 16:09
yes i did! thank you, but I don't know why I can't accept your answer. I don't see the green checkmark.
– M.I
Nov 13 at 20:45
yes i did! thank you, but I don't know why I can't accept your answer. I don't see the green checkmark.
– M.I
Nov 13 at 20:45
add a comment |
2 Answers
2
active
oldest
votes
You need to instantiate every inner dictionary.
var dict = [String : [String : [String]]]()
dict["Test"] = [String : [String]]()
dict["Test"]?["One"] = ["Worked"]
print(dict)
Make sure to avoid force unwrapping.
add a comment |
dict
is empty. There is no value for the "Test"
key.
One option is to provide a default:
dict["Test", default: [:]]["One"] = ["A", "B"]
You can take this one step further:
dict["Test2", default: [:]]["Two", default: ].append("Hello")
That last line will work for any combination of the keys "Test2"
and "Two"
existing or not before that is used.
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You need to instantiate every inner dictionary.
var dict = [String : [String : [String]]]()
dict["Test"] = [String : [String]]()
dict["Test"]?["One"] = ["Worked"]
print(dict)
Make sure to avoid force unwrapping.
add a comment |
You need to instantiate every inner dictionary.
var dict = [String : [String : [String]]]()
dict["Test"] = [String : [String]]()
dict["Test"]?["One"] = ["Worked"]
print(dict)
Make sure to avoid force unwrapping.
add a comment |
You need to instantiate every inner dictionary.
var dict = [String : [String : [String]]]()
dict["Test"] = [String : [String]]()
dict["Test"]?["One"] = ["Worked"]
print(dict)
Make sure to avoid force unwrapping.
You need to instantiate every inner dictionary.
var dict = [String : [String : [String]]]()
dict["Test"] = [String : [String]]()
dict["Test"]?["One"] = ["Worked"]
print(dict)
Make sure to avoid force unwrapping.
answered Nov 11 at 23:09
Gustavo Vollbrecht
6531515
6531515
add a comment |
add a comment |
dict
is empty. There is no value for the "Test"
key.
One option is to provide a default:
dict["Test", default: [:]]["One"] = ["A", "B"]
You can take this one step further:
dict["Test2", default: [:]]["Two", default: ].append("Hello")
That last line will work for any combination of the keys "Test2"
and "Two"
existing or not before that is used.
add a comment |
dict
is empty. There is no value for the "Test"
key.
One option is to provide a default:
dict["Test", default: [:]]["One"] = ["A", "B"]
You can take this one step further:
dict["Test2", default: [:]]["Two", default: ].append("Hello")
That last line will work for any combination of the keys "Test2"
and "Two"
existing or not before that is used.
add a comment |
dict
is empty. There is no value for the "Test"
key.
One option is to provide a default:
dict["Test", default: [:]]["One"] = ["A", "B"]
You can take this one step further:
dict["Test2", default: [:]]["Two", default: ].append("Hello")
That last line will work for any combination of the keys "Test2"
and "Two"
existing or not before that is used.
dict
is empty. There is no value for the "Test"
key.
One option is to provide a default:
dict["Test", default: [:]]["One"] = ["A", "B"]
You can take this one step further:
dict["Test2", default: [:]]["Two", default: ].append("Hello")
That last line will work for any combination of the keys "Test2"
and "Two"
existing or not before that is used.
answered Nov 11 at 23:26
rmaddy
238k27309375
238k27309375
add a comment |
add a comment |
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did you solve it?
– Gustavo Vollbrecht
Nov 13 at 16:09
yes i did! thank you, but I don't know why I can't accept your answer. I don't see the green checkmark.
– M.I
Nov 13 at 20:45