Node.js url.parse result back to string

Question

I am trying to do some simple pagination. To that end, I'm trying to parse the current URL, then produce links to the same query, but with incremented and decremented page parameters.

I've tried doing the following, but it produces the same link, without the new page parameter.

var parts = url.parse(req.url, true);
parts.query['page'] = 25;
console.log("Link: ", url.format(parts));

The documentation for the URL module seems to suggest that format is what I need but I'm doing something wrong.

I know I could iterate and build up the string manually, but I was hoping there's an existing method for this.

Alex Turpin · Accepted Answer · 2011-09-22 15:49:24Z

53

If you look at the latest documentation, you can see that url.format behaves in the following way:

search will be used in place of query

query (object; see querystring) will only be used if search is absent.

And when you modify query, search remains unchanged and it uses it. So to force it to use query, simply remove search from the object:

var url = require("url");
var parts = url.parse("http://test.com?page=25&foo=bar", true);
parts.query.page++;
delete parts.search;
console.log(url.format(parts)); //http://test.com/?page=26&foo=bar

Make sure you're always reading the latest version of the documentation, this will save you a lot of trouble.

answered Sep 22, 2011 at 15:49

Alex Turpin

47.9k23 gold badges118 silver badges146 bronze badges

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

7 Comments

Benjamin Gruenbaum Over a year ago

This is a good answer, I would consider anyone using it to not use the delete operator as it is really slow.

Alex Turpin Over a year ago

Prime example of premature optimization there buddy

Benjamin Gruenbaum Over a year ago

I did not ask you to change the answer, I left a comment. delete is 1000% slower than setting to null or undefined and makes no sense in objects anyway (what does it even mean to remove an object's property, that makes no sense in OO design). In my opinion, setting it to undefined makes much more sense, it still has the property, it is just not defined any more.

Alex Turpin Over a year ago

Sorry, didn't mean to come off as aggressive. You are right, if someone is doing something where speed is critical, they might not want to use delete. Thank you for your comment.

Gyuri Over a year ago

You can always just set it to undefined. Note, that setting to null will throw an error in this case. Also note, that the docs for the url module are hard to read and strangely there are no examples. ;) Note #3: see stackoverflow.com/questions/14967535/… for 'delete' vs 'assign it to undefined' :)

|

Marcel M. · Accepted Answer · 2011-09-22 15:51:58Z

3

Seems to me like it's a bug in node. You might try

// in requires
var url = require('url');
var qs = require('querystring');

// later
var parts = url.parse(req.url, true);
parts.query['page'] = 25;
parts.query = qs.stringify(parts.query);
console.log("Link: ", url.format(parts));

answered Sep 22, 2011 at 15:51

Marcel M.

2,3762 gold badges18 silver badges24 bronze badges

5 Comments

Alex Turpin Over a year ago

It's not a bug, it's clearly documented: nodejs.org/docs/latest/api/url.html

Marcel M. Over a year ago

You're right, but isn't the latest version unstable? I only checked documentation for 0.4.12

Alex Turpin Over a year ago

Indeed, but the documentation has improved, and luckily for us, in this case, is consistent with older versions.

Anirudha Gupta Over a year ago

Is is still querystring or changed to require('qs')

Marcel M. Over a year ago

qs if you want visionmedia's query string parser, querystring for nodejs native parser.

ampersand · Accepted Answer · 2011-09-22 15:50:00Z

2

The other answer is good, but you could also do something like this. The querystring module is used to work with query strings.

var querystring = require('querystring');
var qs = querystring.parse(parts.query);
qs.page = 25;
parts.search = '?' + querystring.stringify(qs);
var newUrl = url.format(parts);

answered Sep 22, 2011 at 15:50

ampersand

4,3162 gold badges26 silver badges33 bronze badges

2 Comments

Marcel M. Over a year ago

parts.query will be an Object if url.parse(..., true) is sent. nodejs.org/docs/v0.4.12/api/url.html#url.parse

ampersand Over a year ago

you're right. This will work if url.parse(req.url) is used, without the last parameter

andrhamm · Accepted Answer · 2019-05-10 20:01:41Z

0

To dry out code and get at URL variables without needing to require('url') I used:

/* 
   Used the url module to parse and place the parameters into req.urlparams.
   Follows the same pattern used for swagger API path variables that load 
   into the req.params scope.

*/
  app.use(function(req, res, next) {
    var url = require('url');
    var queryURL = url.parse(req.url, true);
    req.urlparams = queryURL.query;
    next();
  });

var myID = req.urlparams.myID;

This will parse and move the url variables into the req.urlparams variable. It runs early in the request workflow so is available for all expressjs paths.

edited May 10, 2019 at 20:01

andrhamm

4,4714 gold badges38 silver badges46 bronze badges

answered Jan 22, 2013 at 15:26

d1b1

1441 silver badge5 bronze badges

Collectives™ on Stack Overflow

Node.js url.parse result back to string

4 Answers 4

7 Comments

5 Comments

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

7 Comments

5 Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related