ratty-blush

Redirect to same route

When I try to redirect to the same route with a new search object it doesn't keep the search parameter from the redirect. I have tried with my mask and without it.

  beforeLoad: async ({ search }) => {
    console.log({ beforeLoad: search });
    if (search.code && search.state) {
      console.log('redirecting');
      throw redirect({
        to: '.',
        search: { showPriority: true },
      });
    }
  },

  beforeLoad: async ({ search }) => {
    console.log({ beforeLoad: search });
    if (search.code && search.state) {
      console.log('redirecting');
      throw redirect({
        to: '.',
        search: { showPriority: true },
      });
    }
  },

  const search = Route.useSearch();
  console.log({ priority: search });

  const search = Route.useSearch();
  console.log({ priority: search });

Outputs the logs beforeLoad: { code: ..., state: ... } 'redirecting' beforeLoad: { showPriority: true } priority: {}, priority: {},

5 Replies

adverse-sapphire•5mo ago

throw redirect({
  to: '.',
  search: (prev) => ({ ...prev, showPriority: true }),
});

throw redirect({
  to: '.',
  search: (prev) => ({ ...prev, showPriority: true }),
});

ratty-blushOP•5mo ago

Now that the code and state parameters always exist I am getting an endless loop. Because it is always redirecting with those params. I only want showPriority to go through to the component. Am I supposed to be returning something different from beforeLoad when it has the search params I want?

flat-fuchsia•5mo ago

just don't throw a redirect in that case then?

correct-apricot•5mo ago

What's the difference in terms of behavior when you throw a redirect or not?

flat-fuchsia•5mo ago

i meant: just throw the redirect if you want to redirect, here this means: if the search param is already present, then dont throw a redirect

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Programming

Redirect to same route

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