Linq-to-entity subqueries with async - async-await

I have an async method which generates a query where I'm doing something like this:
var query = context.table.AsNoTracking()
.Where(whereQuery)
.GroupJoin(...)
.GroupJoin(... (a, b) => new Lab {
foo = a.bar
}
)
If it's just that, I can do an await query.ToArrayAsync() and all is good. What I'm running into now is that I then need to do a subquery:
var query = context.table.AsNoTracking()
.Where(whereQuery)
.GroupJoin(...)
.GroupJoin(... (a, b) => new Lab {
foo = a.bar,
bar = (
from x in ... select new { ... }
)
}
)
That bar element is multiple items being returned, and so it can't just be done as part of a join. As soon as I do that, I now don't know how I can make that inner query be awaitable. What's the right syntax to use for that?
The main "outer" query is done as method calls because I'm passing an Expression to the method for the whereQuery but sometimes those "inner" queries will be done either way.
So I'm ending up with something like:
[
{
"foo": 17,
"bar": [
{ "a": 12 },
{ "b": 73 }
]
}
]

Related

Null pointer exception while consuming streams

{
"rules": [
{
"rank": 1,
"grades": [
{
"id": 100,
"hierarchyCode": 32
},
{
"id": 200,
"hierarchyCode": 33
}
]
},
{
"rank": 2,
"grades": []
}
]
}
I've a json like above and I'm using streams to return "hierarchyCode" based on some condition. For example if I pass "200" my result should print 33. So far I did something like this:
request.getRules().stream()
.flatMap(ruleDTO -> ruleDTO.getGrades().stream())
.map(gradeDTO -> gradeDTO.getHierarchyCode())
.forEach(hierarchyCode -> {
//I'm doing some business logic here
Optional<SomePojo> dsf = someList.stream()
.filter(pojo -> hierarchyCode.equals(pojo.getId())) // lets say pojo.getId() returns 200
.findFirst();
System.out.println(dsf.get().getCode());
});
So in the first iteration for the expected output it returns 33, but in the second iteration it is failing with Null pointer instead of just skipping the loop since "grades" array is empty this time. How do I handle the null pointer exception here?
You can use the below code snippet using Java 8:
int result;
int valueToFilter = 200;
List<Grade> gradeList = data.getRules().stream().map(Rule::getGrades).filter(x-> x!=null && !x.isEmpty()).flatMap(Collection::stream).collect(Collectors.toList())
Optional<Grade> optional = gradeList.stream().filter(x -> x.getId() == valueToFilter).findFirst();
if(optional.isPresent()){
result = optional.get().getHierarchyCode();
System.out.println(result);
}
I have created POJO's according to my code, you can try this approach with your code structure.
In case you need POJO's as per this code, i will share the same as well.
Thanks,
Girdhar

calling function inside or outside filter

Let's say I have the following 2 functionally equivalent code snippets that return a list of strings that have their reversal also in the list:
var a = Array("abc", "bca", "abc", "aba", "cba")
a.filter(x => a.toSet(x.reverse)).distinct
and
var a = Array("abc", "bca", "abc", "aba", "cba")
var aSet = a.toSet // notice that toSet is called outside filter
a.filter(x => aSet(x.reverse)).distinct
I'm wondering if there is a difference in the time complexity between these snippets, since in the first snippet I call .toSet for every element in a, whereas in the second snippet I call it only at the beginning. However, that being said, something tells me that the compiler might optimize the first call, yielding the 2 snippets equivalent in time complexity.
If the latter is true, could you please refer me to some relevant literature?
Thank you.
Well, lets put it to the test (using Scalameter):
import org.scalameter.{Gen, PerformanceTest}
import org.scalatest._
import scala.collection.mutable
class SOPerformance extends PerformanceTest.Quickbenchmark {
val gen = Gen.unit("unit")
#inline def fn = {
var a = Array("abc", "bca", "abc", "aba", "cba")
a.filter(x => a.toSet(x.reverse)).distinct
}
#inline def fn2 = {
var a = Array("abc", "bca", "abc", "aba", "cba")
var aSet = a.toSet // notice that toSet is called outside filter
a.filter(x => aSet(x.reverse)).distinct
}
performance of "Range" in {
measure method "fn" in {
using(gen) in { gen ⇒
fn
}
}
measure method "fn2" in {
using(gen) in { gen ⇒
fn2
}
}
}
}
Which shows that fn runs on average in 0.005674 millis and fn2 runs on average in 0.003903 millis.
Now lets make this array a bit larger!
import org.scalameter.{Gen, PerformanceTest}
import org.scalatest._
import scala.collection.mutable
class SOPerformance extends PerformanceTest.Quickbenchmark {
var a = (1 to 1000).map(_.toString).toArray
val gen = Gen.unit("unit")
#inline def fn = {
a.filter(x => a.toSet(x.reverse)).distinct
}
#inline def fn2 = {
var aSet = a.toSet // notice that toSet is called outside filter
a.filter(x => aSet(x.reverse)).distinct
}
performance of "Range" in {
measure method "fn" in {
using(gen) in { gen ⇒
fn
}
}
measure method "fn2" in {
using(gen) in { gen ⇒
fn2
}
}
}
}
This shows the true killer. fn takes on average 158.241861 ms, whereas fn2 takes 0.353472 ms! Why? Because creating collections is really expensive! Especially sets, which require making a new HashSet, requires garbage collections and much more.

Is there a better way of coping with Swift's nested "if let" "pyramid of doom?"

Is there a better way of dealing with a chain of optional properties than nested if let statements? I have been advised to use if lets when examining optional properties, which makes sense as it deals with them at compile time rather than run time, but it looks like utter madness! Is there is a better way?
Here is the current "pyramid of doom" I have ended up with, as an example:
( users: [ JSONValue ]? ) in
if let jsonValue: JSONValue = users?[ 0 ]
{
if let json: Dictionary< String, JSONValue > = jsonValue.object
{
if let userIDValue: JSONValue = json[ "id" ]
{
let userID: String = String( Int( userIDValue.double! ) )
println( userID )
}
}
}
Post-script
Airspeed Velocity's answer below is the right answer, but you will need Swift 1.2 to use multiple lets separated by commas as he suggests, which only currently runs in XCode 6.3, which is in beta.
As commenters have said, Swift 1.2 now has multiple-let syntax:
if let jsonValue = users?.first,
json = jsonValue.object,
userIDValue = json[ "id" ],
doubleID = userIDValue.double,
userID = doubleID.map({ String(Int(doubleID))})
{
println( userID )
}
That said, in this instance it looks like you could might be able to do it all via optional chaining in 1.1, depending on what your objects are:
if let userID = users?.first?.object?["id"]?.double.map({String(Int($0))}) {
println(userID)
}
Note, much better to use first (if this is an array) rather than [0], to account for the possibility the array is empty. And map on the double rather than ! (which would blow up if the value is not double-able).
UPDATE for Swift-3 : The syntax has changed :
if let jsonValue = users?.first,
let json = jsonValue.object,
let userIDValue = json[ "id" ],
let doubleID = userIDValue.double,
let userID = doubleID.map({ String(Int(doubleID))})
{
println( userID )
}
In Swift 2, we have the guard statement.
Instead of:
func myFunc(myOptional: Type?) {
if let object = myOptional! {
...
}
}
You can do it like this:
func myFunc(myOptional: Type?) {
guard array.first else { return }
}
Check http://nshipster.com/guard-and-defer/ from NSHipster.

Return an inline-defined enum from a function?

I'm diving into rust, and I'm trying to do something like this:
match send("select * from User;") {
ConnError => println!("Connection error!"),
DBError(e) => println!("Database error {}", e),
Ok(response) => {
...
}
}
and I'm trying to figure out a compact way of defining the send function. I saw the Result enum, but it only handles one kind of error at a time. I was hoping that I could define my own enum like this:
fn send(query: str) -> enum { Ok(Box<Response>), ConnError, DBError(str) } {
...
}
alas, it is not possible, it's complaining about the unexpected 'enum' keyword. Is there any way to do what I'm trying here, or perhaps make Result handle multiple error types? Thanks!
As you say, you can use Result but you have to define the enum with your error types separately, as you can't define it directly in the return of your function.
Something like this:
use std::rand::distributions::{IndependentSample, Range};
fn main() {
match send("select * from foo") {
Ok(Response) => println!("response"),
Err(e) => match e {
ConnError => println!("connection error"),
DbError(err) => println!("{}", err)
}
}
}
// the enum with your errors
enum DataLayerError {
ConnError,
DbError(String)
}
struct Response; /*...*/
fn send(_query: &str) -> Result<Response, DataLayerError> {
let between = Range::new(0u, 2);
let mut rng = std::rand::task_rng();
// return a random result
match between.ind_sample(&mut rng) {
0 => Ok(Response),
1 => Err(DbError("yikes".to_string())),
2 => Err(ConnError),
_ => unreachable!()
}
}

How to get names of enum entries?

I would like to iterate a TypeScript enum object and get each enumerated symbol name, for example:
enum myEnum { entry1, entry2 }
for (var entry in myEnum) {
// use entry's name here, e.g., "entry1"
}
Though the answer is already provided, Almost no one pointed to the docs
Here's a snippet
enum Enum {
A
}
let nameOfA = Enum[Enum.A]; // "A"
Keep in mind that string enum members do not get a reverse mapping generated at all.
The code you posted will work; it will print out all the members of the enum, including the values of the enum members. For example, the following code:
enum myEnum { bar, foo }
for (var enumMember in myEnum) {
console.log("enum member: ", enumMember);
}
Will print the following:
Enum member: 0
Enum member: 1
Enum member: bar
Enum member: foo
If you instead want only the member names, and not the values, you could do something like this:
for (var enumMember in myEnum) {
var isValueProperty = Number(enumMember) >= 0
if (isValueProperty) {
console.log("enum member: ", myEnum[enumMember]);
}
}
That will print out just the names:
Enum member: bar
Enum member: foo
Caveat: this slightly relies on an implementation detail: TypeScript compiles enums to a JS object with the enum values being members of the object. If TS decided to implement them different in the future, the above technique could break.
For me an easier, practical and direct way to understand what is going on, is that the following enumeration:
enum colors { red, green, blue };
Will be converted essentially to this:
var colors = { red: 0, green: 1, blue: 2,
[0]: "red", [1]: "green", [2]: "blue" }
Because of this, the following will be true:
colors.red === 0
colors[colors.red] === "red"
colors["red"] === 0
This creates a easy way to get the name of an enumerated as follows:
var color: colors = colors.red;
console.log("The color selected is " + colors[color]);
It also creates a nice way to convert a string to an enumerated value.
var colorName: string = "green";
var color: colors = colors.red;
if (colorName in colors) color = colors[colorName];
The two situations above are far more common situation, because usually you are far more interested in the name of a specific value and serializing values in a generic way.
If you only search for the names and iterate later use:
Object.keys(myEnum).map(key => myEnum[key]).filter(value => typeof value === 'string') as string[];
Assuming you stick to the rules and only produce enums with numeric values, you can use this code. This correctly handles the case where you have a name that is coincidentally a valid number
enum Color {
Red,
Green,
Blue,
"10" // wat
}
var names: string[] = [];
for(var n in Color) {
if(typeof Color[n] === 'number') names.push(n);
}
console.log(names); // ['Red', 'Green', 'Blue', '10']
It seems that none of the answers here will work with string-enums in strict-mode.
Consider enum as:
enum AnimalEnum {
dog = "dog", cat = "cat", mouse = "mouse"
}
Accessing this with AnimalEnum["dog"] may result in an error like:
Element implicitly has an 'any' type because expression of type 'any' can't be used to index type 'typeof AnimalEnum'.ts(7053).
Proper solution for that case, write it as:
AnimalEnum["dog" as keyof typeof AnimalEnum]
With current TypeScript Version 1.8.9 I use typed Enums:
export enum Option {
OPTION1 = <any>'this is option 1',
OPTION2 = <any>'this is option 2'
}
with results in this Javascript object:
Option = {
"OPTION1": "this is option 1",
"OPTION2": "this is option 2",
"this is option 1": "OPTION1",
"this is option 2": "OPTION2"
}
so I have to query through keys and values and only return values:
let optionNames: Array<any> = [];
for (let enumValue in Option) {
let optionNameLength = optionNames.length;
if (optionNameLength === 0) {
this.optionNames.push([enumValue, Option[enumValue]]);
} else {
if (this.optionNames[optionNameLength - 1][1] !== enumValue) {
this.optionNames.push([enumValue, Option[enumValue]]);
}
}
}
And I receive the option keys in an Array:
optionNames = [ "OPTION1", "OPTION2" ];
As of TypeScript 2.4, enums can contain string intializers https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-4.html
This allows you to write:
enum Order {
ONE = "First",
TWO = "Second"
}
console.log(`One is ${Order.ONE.toString()}`);
and get this output:
One is First
This solution work too.
enum ScreenType {
Edit = 1,
New = 2,
View = 4
}
var type: ScreenType = ScreenType.Edit;
console.log(ScreenType[type]); //Edit
In a nutshell
if your enums is as below:
export enum Colors1 {
Red = 1,
Green = 2,
Blue = 3
}
to get specific text and value:
console.log(Colors1.Red); // 1
console.log(Colors1[Colors1.Red]); // Red
to get list of value and text:
public getTextAndValues(e: { [s: number]: string }) {
for (const enumMember in e) {
if (parseInt(enumMember, 10) >= 0) {
console.log(e[enumMember]) // Value, such as 1,2,3
console.log(parseInt(enumMember, 10)) // Text, such as Red,Green,Blue
}
}
}
this.getTextAndValues(Colors1)
if your enums is as below:
export enum Colors2 {
Red = "Red",
Green = "Green",
Blue = "Blue"
}
to get specific text and value:
console.log(Colors2.Red); // Red
console.log(Colors2["Red"]); // Red
to get list of value and text:
public getTextAndValues(e: { [s: string]: string }) {
for (const enumMember in e) {
console.log(e[enumMember]);// Value, such as Red,Green,Blue
console.log(enumMember); // Text, such as Red,Green,Blue
}
}
this.getTextAndValues(Colors2)
Another interesting solution found here is using ES6 Map:
export enum Type {
low,
mid,
high
}
export const TypeLabel = new Map<number, string>([
[Type.low, 'Low Season'],
[Type.mid, 'Mid Season'],
[Type.high, 'High Season']
]);
USE
console.log(TypeLabel.get(Type.low)); // Low Season
TypeLabel.forEach((label, value) => {
console.log(label, value);
});
// Low Season 0
// Mid Season 1
// High Season 2
Let ts-enum-util (github, npm) do the work for you and provide a lot of additional type-safe utilities. Works with both string and numeric enums, properly ignoring the numeric index reverse lookup entries for numeric enums:
String enum:
import {$enum} from "ts-enum-util";
enum Option {
OPTION1 = 'this is option 1',
OPTION2 = 'this is option 2'
}
// type: ("OPTION1" | "OPTION2")[]
// value: ["OPTION1", "OPTION2"]
const keys= $enum(Option).getKeys();
// type: Option[]
// value: ["this is option 1", "this is option 2"]
const values = $enum(Option).getValues();
Numeric enum:
enum Option {
OPTION1,
OPTION2
}
// type: ("OPTION1" | "OPTION2")[]
// value: ["OPTION1", "OPTION2"]
const keys= $enum(Option).getKeys();
// type: Option[]
// value: [0, 1]
const values = $enum(Option).getValues();
In TypeScript, an enum is compiled to a map (to get the value from the key) in javascript:
enum MyEnum {
entry0,
entry1,
}
console.log(MyEnum['entry0']); // 0
console.log(MyEnum['entry1']); // 1
It also creates a reversed map (to get the key from the value):
console.log(MyEnum[0]); // 'entry0'
console.log(MyEnum[0]); // 'entry1'
So you can access the name of an entry by doing:
console.log(MyEnum[MyEnum.entry0]); // 'entry0'
console.log(MyEnum[MyEnum.entry1]); // 'entry1'
However, string enum has no reverse map by design (see comment and pull request) because this could lead to conflict between keys and values in the map object.
enum MyEnum {
entry0 = 'value0',
entry1 = 'value1',
}
console.log(MyEnum['value0']); // undefined
console.log(MyEnum['value1']); // undefined
If you want to force your string enum to compile a reversed map (you then have to make sure all the keys and values are different), you can use this trick:
enum MyEnum {
entry0 = <any>'value0',
entry1 = <any>'value1',
}
console.log(MyEnum['value0']); // 'entry0'
console.log(MyEnum['value1']); // 'entry1'
console.log(MyEnum[MyEnum.entry0]); // 'entry0'
console.log(MyEnum[MyEnum.entry1]); // 'entry1'
I got tired looking through incorrect answers, and did it myself.
THIS ONE HAS TESTS.
Works with all types of enumerations.
Correctly typed.
type EnumKeys<Enum> = Exclude<keyof Enum, number>
const enumObject = <Enum extends Record<string, number | string>>(e: Enum) => {
const copy = {...e} as { [K in EnumKeys<Enum>]: Enum[K] };
Object.values(e).forEach(value => typeof value === 'number' && delete copy[value]);
return copy;
};
const enumKeys = <Enum extends Record<string, number | string>>(e: Enum) => {
return Object.keys(enumObject(e)) as EnumKeys<Enum>[];
};
const enumValues = <Enum extends Record<string, number | string>>(e: Enum) => {
return [...new Set(Object.values(enumObject(e)))] as Enum[EnumKeys<Enum>][];
};
enum Test1 { A = "C", B = "D"}
enum Test2 { A, B }
enum Test3 { A = 0, B = "C" }
enum Test4 { A = "0", B = "C" }
enum Test5 { undefined = "A" }
enum Test6 { A = "undefined" }
enum Test7 { A, B = "A" }
enum Test8 { A = "A", B = "A" }
enum Test9 { A = "B", B = "A" }
console.log(enumObject(Test1)); // {A: "C", B: "D"}
console.log(enumObject(Test2)); // {A: 0, B: 1}
console.log(enumObject(Test3)); // {A: 0, B: "C"}
console.log(enumObject(Test4)); // {A: "0", B: "C"}
console.log(enumObject(Test5)); // {undefined: "A"}
console.log(enumObject(Test6)); // {A: "undefined"}
console.log(enumObject(Test7)); // {A: 0,B: "A"}
console.log(enumObject(Test8)); // {A: "A", B: "A"}
console.log(enumObject(Test9)); // {A: "B", B: "A"}
console.log(enumKeys(Test1)); // ["A", "B"]
console.log(enumKeys(Test2)); // ["A", "B"]
console.log(enumKeys(Test3)); // ["A", "B"]
console.log(enumKeys(Test4)); // ["A", "B"]
console.log(enumKeys(Test5)); // ["undefined"]
console.log(enumKeys(Test6)); // ["A"]
console.log(enumKeys(Test7)); // ["A", "B"]
console.log(enumKeys(Test8)); // ["A", "B"]
console.log(enumKeys(Test9)); // ["A", "B"]
console.log(enumValues(Test1)); // ["C", "D"]
console.log(enumValues(Test2)); // [0, 1]
console.log(enumValues(Test3)); // [0, "C"]
console.log(enumValues(Test4)); // ["0", "C"]
console.log(enumValues(Test5)); // ["A"]
console.log(enumValues(Test6)); // ["undefined"]
console.log(enumValues(Test7)); // [0, "A"]
console.log(enumValues(Test8)); // ["A"]
console.log(enumValues(Test9)); // ["B", "A"]
Online version.
Assume you have an enum
export enum SCROLL_LABEL_OFFSET {
SMALL = 48,
REGULAR = 60,
LARGE = 112
}
and you want to create a type based on enum but not just copy and paste.
You could use an enum to create your type like this:
export type ScrollLabelOffset = keyof typeof SCROLL_LABEL_OFFSET;
In result you will receive a type with possible values as 'SMALL' | 'REGULAR' | 'LARGE'
Starting from TypeScript 2.4, the enum would not contain the key as a member anymore. source from TypeScript readme
The caveat is that string-initialized enums can't be reverse-mapped to get the original enum member name. In other words, you can't write Colors["RED"] to get the string "Red".
My solution:
export const getColourKey = (value: string ) => {
let colourKey = '';
for (const key in ColourEnum) {
if (value === ColourEnum[key]) {
colourKey = key;
break;
}
}
return colourKey;
};
Based on some answers above I came up with this type-safe function signature:
export function getStringValuesFromEnum<T>(myEnum: T): (keyof T)[] {
return Object.keys(myEnum).filter(k => typeof (myEnum as any)[k] === 'number') as any;
}
Usage:
enum myEnum { entry1, entry2 };
const stringVals = getStringValuesFromEnum(myEnum);
the type of stringVals is 'entry1' | 'entry2'
See it in action
According to TypeScript documentation, we can do this via Enum with static functions.
Get Enum Name with static functions
enum myEnum {
entry1,
entry2
}
namespace myEnum {
export function GetmyEnumName(m: myEnum) {
return myEnum[m];
}
}
now we can call it like below
myEnum.GetmyEnumName(myEnum.entry1);
// result entry1
for reading more about Enum with static function follow the below link
https://basarat.gitbooks.io/typescript/docs/enums.html
I wrote an EnumUtil class which is making a type check by the enum value:
export class EnumUtils {
/**
* Returns the enum keys
* #param enumObj enum object
* #param enumType the enum type
*/
static getEnumKeys(enumObj: any, enumType: EnumType): any[] {
return EnumUtils.getEnumValues(enumObj, enumType).map(value => enumObj[value]);
}
/**
* Returns the enum values
* #param enumObj enum object
* #param enumType the enum type
*/
static getEnumValues(enumObj: any, enumType: EnumType): any[] {
return Object.keys(enumObj).filter(key => typeof enumObj[key] === enumType);
}
}
export enum EnumType {
Number = 'number',
String = 'string'
}
How to use it:
enum NumberValueEnum{
A= 0,
B= 1
}
enum StringValueEnum{
A= 'A',
B= 'B'
}
EnumUtils.getEnumKeys(NumberValueEnum, EnumType.Number);
EnumUtils.getEnumValues(NumberValueEnum, EnumType.Number);
EnumUtils.getEnumKeys(StringValueEnum, EnumType.String);
EnumUtils.getEnumValues(StringValueEnum, EnumType.String);
Result for NumberValueEnum keys: ["A", "B"]
Result for NumberValueEnum values: [0, 1]
Result for StringValueEnumkeys: ["A", "B"]
Result for StringValueEnumvalues: ["A", "B"]
They have provided a concept called 'reverse-mapping' in their official documentation. It helped me:
https://www.typescriptlang.org/docs/handbook/enums.html#reverse-mappings
The solution is quite straight forward:
enum Enum {
A,
}
let a = Enum.A;
let nameOfA = Enum[a]; // "A"
Having numeric enum:
enum MyNumericEnum {
First = 1,
Second = 2
}
You need to convert it to array first:
const values = Object.values(MyNumericEnum);
// ['First', 'Second', 1, 2]
As you see, it contains both keys and values. Keys go first.
After that, you can retrieve its keys:
values.slice(0, values.length / 2);
// ['First', 'Second']
And values:
values.slice(values.length / 2);
// [1, 2]
For string enums, you can use Object.keys(MyStringEnum) in order to get keys and Object.values(MyStringEnum) in order to get values respectively.
Though it's somewhat challenging to extract keys and values of mixed enum.
There are already a lot of answers here but I figure I'll throw my solution onto the stack anyway.
TypeScript Playground
enum AccountType {
Google = 'goo',
Facebook = 'boo',
Twitter = 'wit',
}
type Key = keyof typeof AccountType // "Google" | "Facebook" | "Twitter"
// this creates a POJO of the enum "reversed" using TypeScript's Record utility
const reversed = (Object.keys(AccountType) as Key[]).reduce((acc, key) => {
acc[AccountType[key]] = key
return acc
}, {} as Record<AccountType, string>)
For Clarity:
/*
* reversed == {
* "goo": "Google",
* "boo": "Facebook",
* "wit": "Twitter",
* }
* reversed[AccountType.Google] === "Google" 👍
*/
Reference for TypeScript Record
To iterate:
for (const [key, value] of Object.entries(reversed)) {
console.log(`${key}: ${value}`);
}
A nice helper function:
const getAccountTypeName = (type: AccountType) => {
return reversed[type]
};
// getAccountTypeName(AccountType.Twitter) === 'Twitter'
The only solution that works for me in all cases (even if values are strings) is the following :
var enumToString = function(enumType, enumValue) {
for (var enumMember in enumType) {
if (enumType[enumMember]==enumValue) return enumMember
}
}
You can use the enum-values package I wrote when I had the same problem:
Git: enum-values
var names = EnumValues.getNames(myEnum);
I found this question by searching "TypeScript iterate over enum keys". So I just want to post solution which works for me in my case. Maybe it'll help to someone too.
My case is the following: I want to iterate over each enum key, then filter some keys, then access some object which has keys as computed values from enum. So this is how I do it without having any TS error.
enum MyEnum = { ONE = 'ONE', TWO = 'TWO' }
const LABELS = {
[MyEnum.ONE]: 'Label one',
[MyEnum.TWO]: 'Label two'
}
// to declare type is important - otherwise TS complains on LABELS[type]
// also, if replace Object.values with Object.keys -
// - TS blames wrong types here: "string[] is not assignable to MyEnum[]"
const allKeys: Array<MyEnum> = Object.values(MyEnum)
const allowedKeys = allKeys.filter(
(type) => type !== MyEnum.ONE
)
const allowedLabels = allowedKeys.map((type) => ({
label: LABELS[type]
}))
You can get an array of names from Enum in this way:
const enumNames: string[] = Object.keys(YourEnum).filter(key => isNaN(Number(key)));
Old question, but, why do not use a const object map?
Instead of doing this:
enum Foo {
BAR = 60,
EVERYTHING_IS_TERRIBLE = 80
}
console.log(Object.keys(Foo))
// -> ["60", "80", "BAR", "EVERYTHING_IS_TERRIBLE"]
console.log(Object.values(Foo))
// -> ["BAR", "EVERYTHING_IS_TERRIBLE", 60, 80]
Do this (pay attention to the as const cast):
const Foo = {
BAR: 60,
EVERYTHING_IS_TERRIBLE: 80
} as const
console.log(Object.keys(Foo))
// -> ["BAR", "EVERYTHING_IS_TERRIBLE"]
console.log(Object.values(Foo))
// -> [60, 80]
If you have enum
enum Diet {
KETO = "Ketogenic",
ATKINS = "Atkins",
PALEO = "Paleo",
DGAF = "Whatever"
}
Then you can get key and values like:
Object.keys(Diet).forEach((d: Diet) => {
console.log(d); // KETO
console.log(Diet[d]) // Ketogenic
});
I find that solution more elegant:
for (let val in myEnum ) {
if ( isNaN( parseInt( val )) )
console.log( val );
}
It displays:
bar
foo
To get the list of the enum values you have to use:
enum AnimalEnum {
DOG = "dog",
CAT = "cat",
MOUSE = "mouse"
}
Object.values(AnimalEnum);

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