How to Build an Express.js Query by Using a URL Parameter?

Question

How to Build an Express.js Query by Using a URL Parameter?

Is there a way to retrieve specific data from my MongoDB database based on the URL? For example, if I navigate to localhost:3000/phones, I want to fetch all entries with the category "phones", and if it's localhost:3000/laptops, I need data related to laptops.

Here is the schema for the data:

    name: {
        required: true,
        type: String
    },
    category: {
        required: true,
        type: String
    },

Currently, my code looks like this:

router.get('/getAll', async (req, res) => {
    try {
        const data = await Model.find();
        res.json(data)
    }
    catch (error) {
        res.status(500).json({ message: error.message })
    }
})

I attempted to use findByID method but encountered some issues.

javascript mongodb express

Answer 1

Answer №1

To easily find the answer, refer to the documentation available at: https://www.mongodb.com/docs/drivers/node/current/quick-reference/

router.get('/phones', async (req, res) => {
    try {
        const data = await Model.find({category: "phones"});
        res.json(data)
    }
    catch (error) {
        res.status(500).json({ message: error.message })
    }
})

If you need to handle both phone and laptop requests using the same query, consider refactoring it to manage in a single endpoint by utilizing query parameters in the request

The URL format should be: localhost:3000/getAll?category=phones

router.get('/getAll', async (req, res) => {
    try {
        const category = req.query.category
        const data = await Model.find({category: category});
        res.json(data)
    }
    catch (error) {
        res.status(500).json({ message: error.message })
    }
})

Answer 2

To easily find the answer, refer to the documentation available at: https://www.mongodb.com/docs/drivers/node/current/quick-reference/

router.get('/phones', async (req, res) => {
    try {
        const data = await Model.find({category: "phones"});
        res.json(data)
    }
    catch (error) {
        res.status(500).json({ message: error.message })
    }
})

If you need to handle both phone and laptop requests using the same query, consider refactoring it to manage in a single endpoint by utilizing query parameters in the request

The URL format should be: localhost:3000/getAll?category=phones

router.get('/getAll', async (req, res) => {
    try {
        const category = req.query.category
        const data = await Model.find({category: category});
        res.json(data)
    }
    catch (error) {
        res.status(500).json({ message: error.message })
    }
})

Answer 3

Answer №2

Ensure to include the query parameter ?category=phones or ?category=laptops in your request

This way, you can retrieve the category using req.query.category.

Here is the code snippet you need:


router.get('/getAll', async (req, res) => {
    try {
        const category = req.query.category;

        const data = await Model.find({
            category,
        });

        res.json(data)
    }
    catch (error) {
        res.status(500).json({ message: error.message })
    }
})

Answer 4

Ensure to include the query parameter ?category=phones or ?category=laptops in your request

This way, you can retrieve the category using req.query.category.

Here is the code snippet you need:


router.get('/getAll', async (req, res) => {
    try {
        const category = req.query.category;

        const data = await Model.find({
            category,
        });

        res.json(data)
    }
    catch (error) {
        res.status(500).json({ message: error.message })
    }
})

How to Build an Express.js Query by Using a URL Parameter?

Answer №1

Answer №2

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