I have a straight forward react-comp. which I want to test the styling depending on the react state. The comp looks like the following:
const Backdrop = ({ showBackdrop }) => {
const backdropRef = useRef();
function getBackdropHeight() {
if (typeof window !== 'undefined') {
return `calc(${document.body.clientHeight}px -
${backdropRef.current?.offsetTop || 0}px)`;
}
return 0;
}
return (
<div
data-testid="backdrop"
className={classNames(styles.backdrop, showBackdrop ? styles.show : styles.hide)}
ref={backdropRef}
style={{ height: getBackdropHeight() }}
/>
);
};
.backdrop {
width: 100%;
position: absolute;
left: 0;
right: 0;
top: 156px;
background-color: rgba(0, 0, 0, 0.7);
z-index: 3;
...
}
.show {
opacity: 0;
visibility: hidden;
transition: visibility 0.25s, opacity 0.25s ease-out;
}
.hide {
opacity: 1;
visibility: hidden;
transition: opacity 0.25s ease-in;
}
And the error that I always get from the test is, that the element is visible:
Received element is visible:
<div class="backdrop hide" data-testid="backdrop" style="height: calc(0px -
0px);" />
21 | const { getByTestId } = renderWithIntl(<Backdrop showBackdrop={false} />);
22 |
> 23 | expect(getByTestId('backdrop')).not.toBeVisible();
| ^
24 | });
25 | });
26 |
it("should not render visible backdrop on falsy state", () => {
const { getByTestId } = render(<Backdrop showBackdrop={false} />);
expect(getByTestId('backdrop')).not.toBeVisible();
});
Any way on how to get the element as not visible without using react inline styling!?
You can use toBeVisible()
function from RTL.
Here you have docs:
https://github.com/testing-library/jest-dom#tobevisible
Example:
// element should not be visible on screen
expect(screen.queryByText('1')).not.toBeVisible();
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