Enhance your property by adding the isDirty feature

Question

Enhance your property by adding the isDirty feature

Managing changes to properties of classes in TypeScript can be optimized by tracking only the fields that have actually changed. Instead of using an array to keep track of property changes, I am exploring the idea of implementing an isDirty check. By incorporating a simple check like if (property.dirty) then {}, I hope to efficiently determine if and which properties have been modified before updating the database.

I vaguely recall employing a similar mechanism in VB.NET in the past, but the specifics elude me at this moment.

Is the following code implementation feasible?

Current Code

class test{
  private _ID: Guid;
  private _dirty: Array<{}>;

  get ID(): Guid {
    return this._ID;
  }
  set ID(id: Guid) {
    if (this._ID != id) {
        this._ID = id;
        this._dirty.filter(function (f) { return f.Field == "id" }).length > 0 ? this._dirty.filter(function (f) { return f.Field == "id" })[0].Value = id.toString() : this._dirty.push({Field: "id", Value: id});
    }
  }

  get Name(): string {
      return this._Name;
  }
  set Name(name: string) {
      if (this._Name != name) {
          this._Name = name;
          this._DirtyFields.filter(function (f) { return f.Field == "ccseq_name" }).length > 0 ? this._DirtyFields.filter(function (f) { return f.Field == "ccseq_name" })[0].Value = name : this._DirtyFields.push(new EntityField("ccseq_name", name, FieldType.String));
      }
  }
}

Desired Code

class test{
  private _ID: Guid;

  get ID(): Guid {
    return this._ID;
  }
  set ID(id: Guid) {
    if (this._ID != id) {
        this._ID = id;
        this._ID.isDirty = true;
    }
  }

  get Name(): string {
      return this._Name;
  }
  set Name(name: string) {
      if (this._Name != name) {
          this._Name = name;
          this._Name.isDirty = true;
      }
  }
}

javascript typescript properties

Answer 1

Answer №1

When working with JavaScript, it is possible to add a property to an object without any issues:

this._ID.dirty = true;

However, TypeScript presents a challenge as it will throw an error due to the lack of the dirty member in Guid. To address this, you can define it as:

private _ID: Guid & { dirty?: boolean };

Update

In JavaScript, adding properties like so is supported:

obj.dirty = true;

This applies to various types such as booleans, strings, arrays, and even functions. For TypeScript compatibility, you can utilize:

interface Object {
    dirty?: boolean;
}

Just keep in mind that by doing this, you are altering all objects in your codebase. Although it won't impact runtime functionality since you're not modifying the prototype, it will affect typechecking in TypeScript for all instances.

Answer 2

When working with JavaScript, it is possible to add a property to an object without any issues:

this._ID.dirty = true;

However, TypeScript presents a challenge as it will throw an error due to the lack of the dirty member in Guid. To address this, you can define it as:

private _ID: Guid & { dirty?: boolean };

Update

In JavaScript, adding properties like so is supported:

obj.dirty = true;

This applies to various types such as booleans, strings, arrays, and even functions. For TypeScript compatibility, you can utilize:

interface Object {
    dirty?: boolean;
}

Just keep in mind that by doing this, you are altering all objects in your codebase. Although it won't impact runtime functionality since you're not modifying the prototype, it will affect typechecking in TypeScript for all instances.

Answer 3

Answer №2

To tackle this issue, my approach involved creating a specialized Field class that I incorporated as attributes in my Objects.

The Field Class

export class EntityField {
    private _Field: string;
    private _Value: any;
    private _FType: FieldType;
    private _isDirty: boolean;

    constructor(field: string, value: any, fType: FieldType) {
        this._Field = field;
        this._Value = value;
        this._FType = fType;
        this._isDirty = false;
    }

    markClean(): void {
        this._isDirty = false;
    }

    markDirty(): void {
        this._isDirty = true;
    }

    get isDirty(): boolean {
        return this._isDirty;
    }

    get Field(): string {
        return this._Field;
    }
    set Field(field) {
        if (this._Field !== field) {
            this._Field = field;
        }
    }

    get Value(): any {
        return this._Value;
    }
    set Value(value: any) {
        if (this._Value !== value) {
            this._Value = value;
            this._isDirty = true;
        }
    }

    get FType(): FieldType {
        return this._FType;
    }
    set FType(fType: FieldType) {
        if (this._FType != fType) {
            this._FType = fType;
        }
    }
}

Implementation Example

export class Entity{
  public Title: Field
}

Entity sampleEntity = new Entity()
Entity.Title.isDirty() // Returns False
Entity.Title.Value = "Hello";
Entity.Title.isDirty() // Returns True

Answer 4

To tackle this issue, my approach involved creating a specialized Field class that I incorporated as attributes in my Objects.

The Field Class

export class EntityField {
    private _Field: string;
    private _Value: any;
    private _FType: FieldType;
    private _isDirty: boolean;

    constructor(field: string, value: any, fType: FieldType) {
        this._Field = field;
        this._Value = value;
        this._FType = fType;
        this._isDirty = false;
    }

    markClean(): void {
        this._isDirty = false;
    }

    markDirty(): void {
        this._isDirty = true;
    }

    get isDirty(): boolean {
        return this._isDirty;
    }

    get Field(): string {
        return this._Field;
    }
    set Field(field) {
        if (this._Field !== field) {
            this._Field = field;
        }
    }

    get Value(): any {
        return this._Value;
    }
    set Value(value: any) {
        if (this._Value !== value) {
            this._Value = value;
            this._isDirty = true;
        }
    }

    get FType(): FieldType {
        return this._FType;
    }
    set FType(fType: FieldType) {
        if (this._FType != fType) {
            this._FType = fType;
        }
    }
}

Implementation Example

export class Entity{
  public Title: Field
}

Entity sampleEntity = new Entity()
Entity.Title.isDirty() // Returns False
Entity.Title.Value = "Hello";
Entity.Title.isDirty() // Returns True

Enhance your property by adding the isDirty feature

Current Code

Desired Code

Answer №1

Update

Answer №2

The Field Class

Implementation Example

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