What is the proper way to create a child JSON object in Angular using TypeScript?

Question

What is the proper way to create a child JSON object in Angular using TypeScript?

I have a model.ts file.

export class MainClass {
    someVar: string;
    var2: string;
    subClass: SubClass;
    contact: ContactList;
}

export class SubClass {
  subItem: {
    item1: string;
    item2: string;
    item3: string;
  }
  constructor() {}
}


export class ContactList {
  itemList: ListItems[];
  constructor() {}
}


export class ListItems {
  list1: string;
  list2: string;
  list3: string;  
}

In my coding journey, I encountered a service.ts file where critical setter methods reside to update the model.

constructor(private model: SomeModel) {}

//some code

updateList(param1, param2, param3) {
    this.model.subClass.subItem.item1 = param1;//doesn't work
    this.model.contact.item[0].list1 = param2;//doesn't work
    this.model.someVar = param3;//works
}

While trying to enhance the functionality, I faced the obstacle of "Can't read property subItem of undefined" error message when attempting to update subClass and contact. It's clear that initializing the subItem and itemList within the model's constructor is necessary, but grasping the correct syntax seems to be challenging. How can I effectively initialize and modify the model structure?

angular typescript angular-services

Answer 1

Answer №1

In order to properly set up the constructor within the model, I followed this structure:

export class MainClass {
  someVar: string;
  var2: string;
  subClass: SubClass;
  contact: ContactList;
  constructor() {
    this.subClass = new SubClass();
    this.contact = new ContactList();
  }
}

export class SubClass {
  subItemVar: SubItem;
  constructor() {
    this.subItemVar = new SubItem();
  }
}

export class SubItem {
  item1: string;
  item2: string;
  item3: string;
}

export class ContactList {
  itemList: ListItems[];
  constructor() {
    this.itemList = new Array<ListItems>();
  }
}

export class ListItems {
  list1: string;
  list2: string;
  list3: string;
}

Answer 2

In order to properly set up the constructor within the model, I followed this structure:

export class MainClass {
  someVar: string;
  var2: string;
  subClass: SubClass;
  contact: ContactList;
  constructor() {
    this.subClass = new SubClass();
    this.contact = new ContactList();
  }
}

export class SubClass {
  subItemVar: SubItem;
  constructor() {
    this.subItemVar = new SubItem();
  }
}

export class SubItem {
  item1: string;
  item2: string;
  item3: string;
}

export class ContactList {
  itemList: ListItems[];
  constructor() {
    this.itemList = new Array<ListItems>();
  }
}

export class ListItems {
  list1: string;
  list2: string;
  list3: string;
}

Answer 3

Answer №2

The reason why subClass and subItem are showing as undefined is because they have not been initialized yet. To resolve this, you need to create a new instance for them before assigning values:

Update SomeModel to MainClass or utilize the .map function for mapping purposes.

this.model.subClass = new SubClass();
this.model.subClass.subItem =  {item1:'123', item2: '222', item3: '333'};

Answer 4

The reason why subClass and subItem are showing as undefined is because they have not been initialized yet. To resolve this, you need to create a new instance for them before assigning values:

Update SomeModel to MainClass or utilize the .map function for mapping purposes.

this.model.subClass = new SubClass();
this.model.subClass.subItem =  {item1:'123', item2: '222', item3: '333'};

Answer 5

Answer №3

Make sure to call the constructor of any complex types you want to initialize in your MainClass.

export class MainClass {
    someVar: string;
    var2: string;
    subClass: SubClass();
    contact: ContactList();
}

Pay attention to the parenthesis. If you have more complicated initialization steps, there may be other considerations to take into account. For example, in your SubClass, ensure that subItem has a constructor if you want it to be initialized when you create an instance of SubClass.

An alternative approach for initializing subItem in your parent class would be to include something like this in the constructor:

constructor() {
        //... some init code
        this.subItem = new Array<SubItemModel>();
        this.subItem.push(new SubItemModel());
    }

I hope this explanation is helpful,

Answer 6

Make sure to call the constructor of any complex types you want to initialize in your MainClass.

export class MainClass {
    someVar: string;
    var2: string;
    subClass: SubClass();
    contact: ContactList();
}

Pay attention to the parenthesis. If you have more complicated initialization steps, there may be other considerations to take into account. For example, in your SubClass, ensure that subItem has a constructor if you want it to be initialized when you create an instance of SubClass.

An alternative approach for initializing subItem in your parent class would be to include something like this in the constructor:

constructor() {
        //... some init code
        this.subItem = new Array<SubItemModel>();
        this.subItem.push(new SubItemModel());
    }

I hope this explanation is helpful,

Answer 7

Answer №4

The reason behind encountering this error is that the variables subClass, subItem, and contact have not been properly defined. In order to resolve this issue, it is recommended to make adjustments to the models as indicated below:

export class MainClass {
    someVar: string;
    var2: string;
    subClass: SubClass;
    contact: ContactList;
    constructor() {
      if(!this.subClass)
      this.subClass =new SubClass();
      if(!this.contact)
      this.contact =new ContactList();
    }
}

export class SubClass {
  subItem: SubItem;
  constructor() {
    if(!this.subItem)
     this.subItem =new SubItem();
  }
}
export class SubItem {

    item1: string;
    item2: string;
    item3: string;

  constructor() {}
}


export class ContactList {
  itemList: ListItems[]=[];
  constructor() {}
}
export class ListItems {
  list1: string;
  list2: string;
  list3: string;  
}

Implementation in service:

constructor(private model: SomeModel) {}

//Some code

updateList(param1, param2, param3) {
    this.model.subClass.subItem.item1 = param1;

 this.model.contact.itemList = [];
 let listItem: ListItems =new ListItems();
 listItem.list1 =param2
 this.model.contact.itemList.push(listItem)
    this.model.someVar = param3;
}

This guidance aims to assist you in recognizing the errors made and adapting the code according to your needs. We hope this information proves useful.

Answer 8

The reason behind encountering this error is that the variables subClass, subItem, and contact have not been properly defined. In order to resolve this issue, it is recommended to make adjustments to the models as indicated below:

export class MainClass {
    someVar: string;
    var2: string;
    subClass: SubClass;
    contact: ContactList;
    constructor() {
      if(!this.subClass)
      this.subClass =new SubClass();
      if(!this.contact)
      this.contact =new ContactList();
    }
}

export class SubClass {
  subItem: SubItem;
  constructor() {
    if(!this.subItem)
     this.subItem =new SubItem();
  }
}
export class SubItem {

    item1: string;
    item2: string;
    item3: string;

  constructor() {}
}


export class ContactList {
  itemList: ListItems[]=[];
  constructor() {}
}
export class ListItems {
  list1: string;
  list2: string;
  list3: string;  
}

Implementation in service:

constructor(private model: SomeModel) {}

//Some code

updateList(param1, param2, param3) {
    this.model.subClass.subItem.item1 = param1;

 this.model.contact.itemList = [];
 let listItem: ListItems =new ListItems();
 listItem.list1 =param2
 this.model.contact.itemList.push(listItem)
    this.model.someVar = param3;
}

This guidance aims to assist you in recognizing the errors made and adapting the code according to your needs. We hope this information proves useful.

What is the proper way to create a child JSON object in Angular using TypeScript?

Answer №1

Answer №2

Answer №3

Answer №4

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