Error TS2339: The type 'Element' does not have a property named 'style'

Question

Error TS2339: The type 'Element' does not have a property named 'style'

Below is the code snippet to change the style of elements:

const test = Array.from(document.getElementsByClassName('mat-form-field-infix'));
test.forEach((element) => {
    element.outerHTML = '<div class="good-day-today" style="width: 0px;"></div>'; // This line is functioning correctly!
    element.style.padding = '10px';
    element.style.borderTop = '0';
});

The error message I encounter during compilation states:

ERROR in src/app//.component.ts(101,21): error TS2339: Property 'style' does not exist on type 'Element'. src/app//.component.ts(102,21): error TS2339: Property 'style' does not exist on type 'Element'.

Any suggestions on how to resolve this issue?

I have attempted various approaches such as removing the Array.from... section, using for of and for in, utilizing as any, but none seem to be working except for the current method.

javascript typescript

Answer 1

Answer №1

To properly access the style property, you'll need to use a typecast:

Array.from(document.getElementsByClassName('mat-form-field-infix') as HTMLCollectionOf<HTMLElement>)

The reason for this is that when using getElementsByClassName, it only returns an HTMLCollection<Element>, which does not have a style property. On the other hand, HTMLElement implements this property through its ElementCSSInlineStyle extended interface.

It's important to note that by applying this typecast, you ensure type safety, considering that every Element is either an HTMLElement or an SVGElement. Hopefully, your SVG Elements do not contain any classes conflicting with this logic.

Answer 2

To properly access the style property, you'll need to use a typecast:

Array.from(document.getElementsByClassName('mat-form-field-infix') as HTMLCollectionOf<HTMLElement>)

The reason for this is that when using getElementsByClassName, it only returns an HTMLCollection<Element>, which does not have a style property. On the other hand, HTMLElement implements this property through its ElementCSSInlineStyle extended interface.

It's important to note that by applying this typecast, you ensure type safety, considering that every Element is either an HTMLElement or an SVGElement. Hopefully, your SVG Elements do not contain any classes conflicting with this logic.

Answer 3

Answer №2

One alternative method is using querySelectorAll along with a type parameter. Unlike getElementsByClassName, which is not generic, querySelectorAll allows you to specify the type of elements to be selected easily:

const test = document.querySelectorAll<HTMLElement>('.mat-form-field-infix');

This approach eliminates the need for type casting and enables immediate usage of forEach without converting it to an array first. (While getElementsByClassName returns an HTMLCollection without a forEach method, querySelectorAll returns a NodeList which does support forEach on modern browsers. For older browsers, a polyfill or array conversion may be necessary.)

If only one element is needed, querySelector can be used as well:

const elm = document.querySelector<HTMLElement>('.foo')!;
elm.style.padding = '10px';

An additional advantage of using querySelectorAll (and

querySelector</code) is their acceptance of CSS selector strings, offering more flexibility and precision. For instance, the selector string</p>
<pre><code>.container > input:checked

will target checked <input> elements that are children of

<div class="container">

.

Answer 4

One alternative method is using querySelectorAll along with a type parameter. Unlike getElementsByClassName, which is not generic, querySelectorAll allows you to specify the type of elements to be selected easily:

const test = document.querySelectorAll<HTMLElement>('.mat-form-field-infix');

This approach eliminates the need for type casting and enables immediate usage of forEach without converting it to an array first. (While getElementsByClassName returns an HTMLCollection without a forEach method, querySelectorAll returns a NodeList which does support forEach on modern browsers. For older browsers, a polyfill or array conversion may be necessary.)

If only one element is needed, querySelector can be used as well:

const elm = document.querySelector<HTMLElement>('.foo')!;
elm.style.padding = '10px';

An additional advantage of using querySelectorAll (and

querySelector</code) is their acceptance of CSS selector strings, offering more flexibility and precision. For instance, the selector string</p>
<pre><code>.container > input:checked

will target checked <input> elements that are children of

<div class="container">

.

Answer 5

Answer №3

If you're looking for a quick fix, you can try the following approach:

element["style"].padding = '10px';
element["style"].borderTop = '0';

It might not be the most optimal solution, but in my experience, it has proven to be effective on multiple occasions :)

Answer 6

If you're looking for a quick fix, you can try the following approach:

element["style"].padding = '10px';
element["style"].borderTop = '0';

It might not be the most optimal solution, but in my experience, it has proven to be effective on multiple occasions :)

Answer 7

Answer №4

Here is a potential solution:

const element = document.querySelector('.className');

Answer 8

Here is a potential solution:

const element = document.querySelector('.className');

Answer 9

Answer №5

Although I believe type casting won't create any issues in this scenario, it's always best to steer clear of it whenever feasible. To avoid type casting here, you can utilize instanceof narrowing:

const test = Array.from(
  document.getElementsByClassName('mat-form-field-infix')
)
test.forEach((element) => {
  if (!(element instanceof HTMLElement)) {
    throw new TypeError(`Expected an object of Type HTMLElement`)
  }

  element.outerHTML =
    '<div class="good-day-today" style="width: 0px;"></div>' // Just a note that this line works perfectly!
  element.style.padding = '10px'
  element.style.borderTop = '0'
})

Answer 10

Although I believe type casting won't create any issues in this scenario, it's always best to steer clear of it whenever feasible. To avoid type casting here, you can utilize instanceof narrowing:

const test = Array.from(
  document.getElementsByClassName('mat-form-field-infix')
)
test.forEach((element) => {
  if (!(element instanceof HTMLElement)) {
    throw new TypeError(`Expected an object of Type HTMLElement`)
  }

  element.outerHTML =
    '<div class="good-day-today" style="width: 0px;"></div>' // Just a note that this line works perfectly!
  element.style.padding = '10px'
  element.style.borderTop = '0'
})

Answer 11

Answer №6

Modifying the outerHTML of an element results in the destruction of its original form, causing styling issues.

If you opt to manipulate the innerHTML instead, you'll find that your desired styling changes take effect successfully.

While this approach may not achieve identical outcomes, it should serve as a helpful guide for your progress.

const test = Array.from(document.getElementsByClassName('mat-form-field-infix'));
test.forEach((element) => {
    element.innerHTML = '<div class="good-day-today" style="width: 0px;"></div>'; // It's worth noting that this line executes as intended!
    element.style.padding = '10px';
    element.style.borderTop = '0';
});

Answer 12

Modifying the outerHTML of an element results in the destruction of its original form, causing styling issues.

If you opt to manipulate the innerHTML instead, you'll find that your desired styling changes take effect successfully.

While this approach may not achieve identical outcomes, it should serve as a helpful guide for your progress.

const test = Array.from(document.getElementsByClassName('mat-form-field-infix'));
test.forEach((element) => {
    element.innerHTML = '<div class="good-day-today" style="width: 0px;"></div>'; // It's worth noting that this line executes as intended!
    element.style.padding = '10px';
    element.style.borderTop = '0';
});

Answer 13

Answer №7

While encountering a similar issue during my project, I found a workaround by adding an extra class instead of using the style property.

document.querySelectorAll(".<className>");

For instance, in my case:

//css file
    .<classname> {
        display: none;
      }
    .<classname>.show {
        display: flex;
      }

//ts file
elements.forEach((ele, index) => {
const errors = something.length;
if (index < errors) {
  ele.classList.add("show");
} else {
  ele.classList.remove("show");
}

}); };

Answer 14

While encountering a similar issue during my project, I found a workaround by adding an extra class instead of using the style property.

document.querySelectorAll(".<className>");

For instance, in my case:

//css file
    .<classname> {
        display: none;
      }
    .<classname>.show {
        display: flex;
      }

//ts file
elements.forEach((ele, index) => {
const errors = something.length;
if (index < errors) {
  ele.classList.add("show");
} else {
  ele.classList.remove("show");
}

}); };

Answer 15

Answer №8

Discovering a simpler approach:

To streamline the process, you can simply generate an index.d.ts file and include the following code:

interface Element {
    style: CSSStyleDeclaration
}

Answer 16

Discovering a simpler approach:

To streamline the process, you can simply generate an index.d.ts file and include the following code:

interface Element {
    style: CSSStyleDeclaration
}

Error TS2339: The type 'Element' does not have a property named 'style'

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

Answer №8

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