Get name of variable in typescript

Question

How can I get a variables name in typescript? I want something like this:

var name = "Foo";
alert(getVariableName(name)); //Prints "name"

Answer 1

Expanding on basarat's answer, you need to create function that takes as a parameter a function that will contain the access to your variable. Because in JavaScript you can access the code of any function it then becomes a simple matter of using a regex to extract the variable name.

var varExtractor = new RegExp("return (.*);");
export function getVariableName<TResult>(name: () => TResult) {
    var m = varExtractor.exec(name + "");
    if (m == null) throw new Error("The function does not contain a statement matching 'return variableName;'");
    return m[1];
}

var foo = "";
console.log(getVariableName(() => foo));

Answer 2

One approach is to store such values in an object:

var o = {
    firstName: "Homer",
    lastName: "Simpson"
};

We can't get the name of o, but we can get the names (or "keys") of its two properties:

var k = Object.keys(o);
console.log(k[0]); // prints "firstName"

Answer 3

Expanding Cernicova-Dragomir's answer:

Expanding on basarat's answer, you need to create function that takes as a parameter a function that will contain the access to your variable. Because in JavaScript you can access the code of any function it then becomes a simple matter of using a regex to extract the variable name.

to also support a field of an object:

var varExtractor = new RegExp("(.*)");

export function getVariableName<TResult>(getVar: () => TResult) {
    var m = varExtractor.exec(getVar + "");

    if (m == null)
        throw new Error("The function does not contain a statement matching 'return variableName;'");

    var fullMemberName = m[1];
    var memberParts = fullMemberName.split('.');

    return memberParts[memberParts.length-1];
}

var foo = { bar: "" };
var varName = getVariableName(() => foo.bar ); //prints "bar"

Notice that I've deleted the "return" because it doesn't work on Node.js when targeting ES6.

Answer 4

TypeScript is JavaScript at runtime. So the same limitations as there apply : Get the 'name' of a variable in Javascript

However you can do stuff like

alert(getVariableName(()=>name)) 

Here you would parse the body of the function passed into getVariableName and get that as a string.

Answer 5

//typescript
function getVarName(obj: object): string {
  const keys = Object.keys(obj);
  console.assert(keys.length == 1, 'key count must be 1');
  return keys[0];
}
const simpleName = 123;
console.log(getVarName({ simpleName }), simpleName);

or

//javascript
"use strict";
function getVarName(obj) {
    const keys = Object.keys(obj);
    console.assert(keys.length == 1, 'key count must be 1');
    return keys[0];
}
const simpleName = 123;
console.log(getVarName({ simpleName }), simpleName);

it will output

[LOG]: simpleName, 123

but i think it should be finished in compile duration will be better in instead of runtime

Answer 6

If the variable is part of a class/interface, there is a really simple solution:

function getFieldName<T, K extends keyof T = keyof T>(name: K): K {
    return name;
}

I.e. const name = getFieldName<SomeInterface>('some_field') provides a compile-time guarantee that name contains 'some_field', and some_field is a valid field of SomeInterface.

Answer 7

Look at my answer for JavaScript migrated to TypeScript:

const nameOf = (f: () => void) => (f).toString().replace(/[ |()=>]/g, '');

Examples:

nameOf(() => myVariable)             // myVariable
nameOf(() => myVariable.name)        // myVariable.name
nameOf(() => myVariable.name.length) // myVariable.name.length
nameOf(() => myVariable.name[10])    // myVariable.name[10]
nameOf(() => MySuperClass)           // MySuperClass