Borislav Hadzhiev
Sun Mar 20 2022·2 min read
Photo by Hiva Sharifi
The error "Property 'files' does not exist on type 'HTMLElement'" occurs when
we try to access the files
property on an element that has a type of
HTMLElement
. To solve the error, use a type assertion to type the element as
HTMLInputElement
before accessing files
.
This is the index.html
file for the examples in this article.
<!DOCTYPE html> <html> <head> <meta charset="UTF-8" /> </head> <body> <label for="avatar">Choose a profile picture:</label> <input type="file" id="avatar" name="avatar" accept="image/png, image/jpeg, image/webp" /> <button id="btn">Log files</button> <script src="./src/index.ts"></script> </body> </html>
And here is an example of how the error occurs in the index.ts
file.
// 👇️ const input: HTMLElement | null const input = document.getElementById('avatar'); const btn = document.getElementById('btn'); btn?.addEventListener('click', () => { // ⛔️ Property 'files' does not exist on type 'HTMLElement'.ts(2339) const files = input?.files; console.log(files); });
The reason we got the error is because the return type of the
document.getElementById
method is HTMLElement | null
and the files
property doesn't exist in the
HTMLElement
type.
To solve the error, use a type assertion to type the element as an
HTMLInputElement
.
const input = document.getElementById('avatar') as HTMLInputElement | null; const btn = document.getElementById('btn'); btn?.addEventListener('click', () => { const files = input?.files; console.log(files); });
Type assertions are used when we have information about the type of a value that TypeScript can't know about.
input
variable stores anHTMLInputElement
or a null
value and not to worry about it.We used a
union type
to specify that the variable could still be null
, because if an HTML element
with the provided id
does not exist in the DOM, the getElementById()
method
returns a null
value.
We used the
optional chaining (?.)
operator to get around the possibly null
value in the example.
const input = document.getElementById('avatar') as HTMLInputElement | null; const btn = document.getElementById('btn'); // 👇️ optional chaining (?.) btn?.addEventListener('click', () => { // 👇️ optional chaining (?.) const files = input?.files; console.log(files); });
The optional chaining operator short-circuits returning undefined
if the
reference is equal to null
or undefined
.
input
variable stores a null
value, we won't attempt to access the files
property on null
and get a runtime error.Alternatively, you can use a simple if
statement that serves as a
type guard.
const input = document.getElementById('avatar') as HTMLInputElement | null; const btn = document.getElementById('btn'); if (btn != null) { btn.addEventListener('click', () => { // 👉️ input has type HTMLInputElement or null here if (input != null) { // 👉️ input has type HTMLInputElement here const files = input.files; console.log(files); } }); }
We explicitly check that the input
variable does not store a null
value.
TypeScript knows that the input
variable has a type of HTMLInputElement
in
the if
block and allows us to directly access the files
property.
Which approach you pick to exclude null
from the type before accessing the
files
property is a matter of personal preference.
However, it's always a best practice to include null
in the type assertion,
because the getElementById
method would return null
if no element with the
provided id
was found.