correct-apricot

How to make default search params to not be populated when visiting page.

I have a page "/my-files" with a bunch of search params where each parameter have a default value. I have a link to this page looking like so: <Link to="/my-files">. When clicking on this, the URL get's populated with the default search params so it looks like this: /my-files?sort=a-z&page=1&pageSize=10&query=. I would like the URL to be kept as /my-files and only have the URL be populated when a search param is different than the default. Is what I want to do an anti-pattern of sorts? This is my route configuration:

const filesSortSchema = z.enum([
    "newest",
    "oldest",
    "a-z",
    "z-a",
    "smallest",
    "largest",
]);

export type FilesSort = z.infer<typeof filesSortSchema>;

const searchParamsSchema = z.object({
    sort: filesSortSchema.default("a-z"),
    page: z.number().int().min(1).default(1),
    pageSize: z.number().int().min(10).default(10),
    query: z.string().default(""),
});

export type MyFilesSearchParams = z.infer<typeof searchParamsSchema>;

export const Route = createFileRoute("/_authenticated/my-files/")({
    component: RouteComponent,
    validateSearch: zodValidator(searchParamsSchema),
});

const filesSortSchema = z.enum([
    "newest",
    "oldest",
    "a-z",
    "z-a",
    "smallest",
    "largest",
]);

export type FilesSort = z.infer<typeof filesSortSchema>;

const searchParamsSchema = z.object({
    sort: filesSortSchema.default("a-z"),
    page: z.number().int().min(1).default(1),
    pageSize: z.number().int().min(10).default(10),
    query: z.string().default(""),
});

export type MyFilesSearchParams = z.infer<typeof searchParamsSchema>;

export const Route = createFileRoute("/_authenticated/my-files/")({
    component: RouteComponent,
    validateSearch: zodValidator(searchParamsSchema),
});

10 Replies

quickest-silver•12mo ago

we have a concept named "search middlewares" exactly for that purpose

quickest-silver•12mo ago

https://tanstack.com/router/v1/docs/framework/react/guide/search-params#transforming-search-with-search-middlewares

Search Params | TanStack Router React Docs

Similar to how TanStack Query made handling server-state in your React applications a breeze, TanStack Router aims to unlock the power of URL search params in your applications. Why not just use URLSe...

quickest-silver•12mo ago

=> stripSearchParams

correct-apricotOP•12mo ago

oh, damn, I missed that in the bottom, thanks, I'll check it out.

quickest-silver•12mo ago

we have lots of features, so I can absolutely understand if they are bit hard to discover

correct-apricotOP•12mo ago

Cool, it works perfectly:

const filesSortSchema = z.enum([
    "newest",
    "oldest",
    "a-z",
    "z-a",
    "smallest",
    "largest",
]);

export type FilesSort = z.infer<typeof filesSortSchema>;

const defaultSearchParams = {
    sort: "a-z",
    page: 1,
    pageSize: 10,
    query: "",
} as const;

const searchParamsSchema = z.object({
    sort: filesSortSchema.default(defaultSearchParams.sort),
    page: z.number().int().min(1).default(defaultSearchParams.page),
    pageSize: z.number().int().min(10).default(defaultSearchParams.pageSize),
    query: z.string().default(defaultSearchParams.query),
});

export type MyFilesSearchParams = z.infer<typeof searchParamsSchema>;

export const Route = createFileRoute("/_authenticated/my-files/")({
    component: RouteComponent,
    validateSearch: zodValidator(searchParamsSchema),
    search: {
        middlewares: [stripSearchParams(defaultSearchParams)],
    },
});

const filesSortSchema = z.enum([
    "newest",
    "oldest",
    "a-z",
    "z-a",
    "smallest",
    "largest",
]);

export type FilesSort = z.infer<typeof filesSortSchema>;

const defaultSearchParams = {
    sort: "a-z",
    page: 1,
    pageSize: 10,
    query: "",
} as const;

const searchParamsSchema = z.object({
    sort: filesSortSchema.default(defaultSearchParams.sort),
    page: z.number().int().min(1).default(defaultSearchParams.page),
    pageSize: z.number().int().min(10).default(defaultSearchParams.pageSize),
    query: z.string().default(defaultSearchParams.query),
});

export type MyFilesSearchParams = z.infer<typeof searchParamsSchema>;

export const Route = createFileRoute("/_authenticated/my-files/")({
    component: RouteComponent,
    validateSearch: zodValidator(searchParamsSchema),
    search: {
        middlewares: [stripSearchParams(defaultSearchParams)],
    },
});

quickest-silver•12mo ago

great!

correct-apricot•5mo ago

import {
  createFileRoute,
  Link,
  stripSearchParams,
} from "@tanstack/react-router";
import { zodValidator } from "@tanstack/zod-adapter";
import { z } from "zod/v4";

const defaultSearch = {
  name: "world",
};

export const Route = createFileRoute("/hello")({
  component: RouteComponent,
  validateSearch: zodValidator(
    z.object({
      name: z.string().default(defaultSearch.name),
    })
  ),
  search: {
    middlewares: [stripSearchParams(defaultSearch)],
  },
});

function RouteComponent() {
  return (
    <div className="flex flex-col gap-4">
      <Link from="/hello" search={{ name: "world" }} to=".">
        {({ isActive }) => (
          // URL will be "/hello"
          // Will be active even if URL is "/hello?name=world2"
          <div>
            <h1>Go to "/hello?name=world"</h1>
            <p>Is active: {isActive ? "true" : "false"}</p>
          </div>
        )}
      </Link>
      <Link from="/hello" search={{ name: "world2" }} to=".">
        {({ isActive }) => (
          // URL will be "/hello?name=world2"
          // Will be active only if URL is "/hello?name=world2"
          <div>
            <h1>Go to "/hello?name=world2"</h1>
            <p>Is active: {isActive ? "true" : "false"}</p>
          </div>
        )}
      </Link>
    </div>
  );
}

import {
  createFileRoute,
  Link,
  stripSearchParams,
} from "@tanstack/react-router";
import { zodValidator } from "@tanstack/zod-adapter";
import { z } from "zod/v4";

const defaultSearch = {
  name: "world",
};

export const Route = createFileRoute("/hello")({
  component: RouteComponent,
  validateSearch: zodValidator(
    z.object({
      name: z.string().default(defaultSearch.name),
    })
  ),
  search: {
    middlewares: [stripSearchParams(defaultSearch)],
  },
});

function RouteComponent() {
  return (
    <div className="flex flex-col gap-4">
      <Link from="/hello" search={{ name: "world" }} to=".">
        {({ isActive }) => (
          // URL will be "/hello"
          // Will be active even if URL is "/hello?name=world2"
          <div>
            <h1>Go to "/hello?name=world"</h1>
            <p>Is active: {isActive ? "true" : "false"}</p>
          </div>
        )}
      </Link>
      <Link from="/hello" search={{ name: "world2" }} to=".">
        {({ isActive }) => (
          // URL will be "/hello?name=world2"
          // Will be active only if URL is "/hello?name=world2"
          <div>
            <h1>Go to "/hello?name=world2"</h1>
            <p>Is active: {isActive ? "true" : "false"}</p>
          </div>
        )}
      </Link>
    </div>
  );
}

Just discovering stripSearchParams and it seems exactly what I need! However, I'm finding that <Link> always is active even if the search param is different from the default. I can get around this by using exact: true but then that would break certain use cases. Not sure if this is expected or not, or if there is a different solution that would work better here.

rival-black•5mo ago

I think you could also do default in typescript vs in zod const { sort = "a-z". page = 1 ... } = Route.useSearch();

flat-fuchsia•5mo ago

Wanted to achieve something similar without luck. So finally I removed the "defaults" params

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How to make default search params to not be populated when visiting page.

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