Typescript - How do you create a GET endpoint with params?

Question

I am using express and trying to create an API endpoint to accept a GET request like GET /dogs?breed=GOLDEN&name=rex.

What is the best approach to pass parameters into a url? While testing, I created two endpoints as shown below. Neither return a success

router.get('/dogs?breed=GOLDEN&name=rex', breedAndNameController.fetch);

router.get('/dogs?breed=:breed&name=:name', breedAndNameController.fetch);

export const fetch = async (req: any, res: any) => {
    try {
        console.log("success")
        return res.status(200).send("success");
    } catch (err) {
        console.log(err)
    }
}

Are you using express?

r0-
– r0-

2022-06-21 18:28:23 +00:00
Commented Jun 21, 2022 at 18:28 — r0-
– r0-, Commented Jun 21, 2022 at 18:28

r0- · Accepted Answer · 2022-06-21 18:42:18Z

2

It looks like you're using express.js (just a guess).

Query-params are not a part of your route.

You have two options now:

Use query strings

Using query-string arguments your url looks like this:

router.get('/dogs', breedAndNameController.fetch);

And your route:

 export const fetch = async (req: any, res: any) => {
        try {

  console.log(req. query);
            return res.status(200).send("success");
        } catch (err) {
            console.log(err)
        }
    }

The call looks like /dogs?breed=asdf&name=test

Use params

Using params your url looks like this:

router.get('/dogs/:breed/:name', breedAndNameController.fetch);

which gives you access like this:

export const fetch = async (req: any, res: any) => {
    try {

        console.log(req.params)
        return res.status(200).send("success");
    } catch (err) {
        console.log(err)
    }
}

And your url looks like /dogs/tset/asdf.

edited Jun 21, 2022 at 18:42

answered Jun 21, 2022 at 18:37

r0-

2,5181 gold badge25 silver badges29 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Sarun · Accepted Answer · 2022-06-21 18:40:54Z

0

You should keep your route as /dogs and supply the query params when you actually call the query

router.get('/dogs', breedAndNameController.fetch);

and then you make a request to as /dogs?breed=GOLDEN&name=rex

answered Jun 21, 2022 at 18:40

Sarun

594 bronze badges

Collectives™ on Stack Overflow

Typescript - How do you create a GET endpoint with params?

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related