Querystring.parse returning first value undefined

Question

I am conducting an exercise using nodejs in the REPL, I have been given a URL and am required to use querystring.parse() on it after accessing it via process.argv[] to retrieve the value of a and b and log them to the console. This is the string:

"http://127.0.0.1:8080/test?a=100&b=200"

Here is my code so far

 const qs = require("querystring");   
        function fn() {
            var query = qs.parse(process.argv[2]);
            console.log("query a is " + query["a"]);
            console.log("query b is " + query["b"]);
        }
        module.exports.fn = fn();

The exercise requires the final two console logs to return as follows:

     'query a is 100'
     'query b is 200'

But the output I am getting is:

      query a is undefined
      query b is 200

and when I return the query object itself I get this:

         { 'http://127.0.0.1:8080/test?a': '100', b: '200' }

aldyadk · Accepted Answer · 2019-02-01 09:15:34Z

1

i believe you need the 'url' module to parsed the full url, because the 'querystring' module itself will only correctly parse the query string.

const url = require("url")
const qs = require("querystring");   
    function fn() {
        const parsedUrl = url.parse(process.argv[2])
        const query = qs.parse(parsedUrl.query);
        console.log("query a is " + query["a"]);
        console.log("query b is " + query["b"]);
    }
    module.exports.fn = fn();

answered Feb 1, 2019 at 9:15

aldyadk

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Mebin Joe · Accepted Answer · 2019-02-01 06:54:01Z

1

qs parses query strings rather than URLs.

Try using URL Parser.

(new URL(addr).search.substring(1))

qs.parse("title=querystring&action=edit")

For getting params passed, call

new URL(addr).searchParams

answered Feb 1, 2019 at 6:54

Mebin Joe

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1 Comment

J T Over a year ago

I would but I need to submit something using both the URL and querystring modules nodejs

arizafar · Accepted Answer · 2019-02-01 07:32:26Z

0

Beacuse querystring parses the query only so tis will give you both the queries as parsed by the module

const qs = require("querystring");   
        function fn() {
            var query = qs.parse(process.argv[2]);
            console.log("query a is " + query["http://127.0.0.1:8080/test?a"]);
            console.log("query b is " + query["b"]);
        }
        module.exports.fn = fn();

else you can use url module

var url = require('url');
var url_parts = url.parse('http://127.0.0.1:8080/test?a=100&b=200', true);
var query = url_parts.query;
console.log(query)

answered Feb 1, 2019 at 7:32

arizafar

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