Write to Nested Dictionary (Swift 4)












2














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].










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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
















2














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].










share|improve this question
























  • 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














2












2








2







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].










share|improve this question















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






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share|improve this question













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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


















  • 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












2 Answers
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oldest

votes


















2














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.






share|improve this answer





























    1














    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.






    share|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      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.






      share|improve this answer


























        2














        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.






        share|improve this answer
























          2












          2








          2






          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.






          share|improve this answer












          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.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 11 at 23:09









          Gustavo Vollbrecht

          6531515




          6531515

























              1














              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.






              share|improve this answer


























                1














                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.






                share|improve this answer
























                  1












                  1








                  1






                  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.






                  share|improve this answer












                  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.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 11 at 23:26









                  rmaddy

                  238k27309375




                  238k27309375






























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