Reputation: 4057
I have a component that needs to copy all of the query parameters from the URL. However I do not want to copy dynamic route parameters. So if it is used in a component like: /pages/model/{modelId}.tsx
and the url is /model/123456?page=2&sort=column&column=value
, I would like to retrieve the following object:
{
page: 2,
sort: "column",
column: "value"
}
I would use useRouter().query
but this would include modelId
. And the router object doesn't seem to have any other list of parameters.
So how would I distinguish regular URL parameters from dynamic route parameters?
Upvotes: 4
Views: 5186
Reputation: 26384
Why not use the built-in URL API instead:
const url = new URL(window.location.href);
url.searchParams.get("page"); // a string or null if not present
Going a little further, you could also wrap this in a hook for future use.
If you want the search parameters as an object, you could wrap it in Object.fromEntries
too:
const params = Object.fromEntries(new URL(window.location.href).searchParams);
params.page; // string or undefined now
Upvotes: 3
Reputation: 1738
In my case I stripped the unwanted key from the query object. In this case the dynamic url param.
I used lodash for this, but this could be perfectly done with Javascript only.
export const ignoreQueryParam = (obj, key) => _.omit(obj, key);
Applied to your case:
const router = useRouter();
const searchQuery = ignoreQueryParam(router.query, 'modelId');
I hope it helps.
Upvotes: 0