Regex to match the url pattern in nodejs

Question

I am working on a node application, i need a regex to match the url pattern and get information out of the url, suggest the possible solutions.

This are the url patterns:
1) www.mysite.com/Paper/cat_CG10
2) www.mysite.com/White-Copy-Printer-Paper/cat_DP5027
3) www.mysite.com/pen/directory_pen?
4) www.mysite.com/Paper-Mate-Profile-Retractable-Ballpoint-Pens-Bold-Point-Black-Dozen/product_612884
5) www.mysite.com/22222/directory_22222?categoryId=12328

These is what is want from the above url:
1) name= "cat" value="CG10"
2) name= "cat" value="DP5027"
3) name= "directory" value ="pen"
4) name="product" value ="612884"
5) name="directory" value="22222" params = {categoryId : 12328}

I want a regex which can match the url pattern and get the values like name, value and params out of the urls.

colbydauph · Accepted Answer · 2014-01-23 04:47:14Z

1

This function does the trick for the urls and desired matches you've provided. It will also parse out an infinite number of query parameters.

Fiddle: http://jsfiddle.net/8a9nK/

function parseUrl(url)
{
    var split = /^.*\/(cat|directory|product)_([^?]*)\??(.*)$/gi.exec(url);
    var final_params = {};
    split[3].split('&').forEach(function(pair){
       var ps = pair.split('=');
       final_params[ps[0]] = ps[1];
    });
    return {
        name: split[1], 
        value: split[2], 
        params: final_params
    };
}

Explanation

^ Start from the beginning of the string
.* Match any number of anything (The beginning of the url we don't care about)
\/ Match a single backslash (The last one before the things we care about)
(cat|directory|product) Match and capture the word cat OR directory OR product (This is our name)
_ Match an underscore (The character separating our name and value)
([^?]*) Match and capture any number of anything EXCEPT a question mark (This is our value)
\?? Match a question mark if it exists, otherwise don't worry about it (The start of a potential query string)
(.*) Match and capture any number of anything (This is the entire query string that we will split into param later)
$ Match the end of the string

edited Jan 23, 2014 at 4:47

answered Jan 22, 2014 at 14:22

colbydauph

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6 Comments

Manu Over a year ago

can you explain this part "/^.*\/(cat|directory|product)_([^?]*)\??(.*)$/gi"

colbydauph Over a year ago

Yep, the solution has been modified to add an explanation.

Manu Over a year ago

if isPartial is present how to get just that query parameter www.mysite.com/22222/directory_22222?categoryId=12328&isPartial=true

colbydauph Over a year ago

The function will return an object with the properties: name, value, and params. You can access the isPartial parameter like this

var obj = parseUrl('www.mysite.com/22222/directory_22222?categoryId=12328&isPartial=true'); var isPartial = obj.params.isPartial

Manu Over a year ago

i have url like staples.com/Duracell-DL2430-30-Volt-Lithium-Battery/…', and my regex is /^.*\/(cat|directory|product)_([^?#]*)\??(.*)$/gi i am getting result like {"name":"product","value":"448696","params":{"#id":"'dropdown_37439'"}}, my expected result is {"name":"product","value":"448696","params":{"id":"'dropdown_37439'"}}

|

Matyas · Accepted Answer · 2014-01-22 14:28:41Z

0

The regex below would have in its match groups 1 & 2 the desired values

/^\/[^\/]+\/([^_]+)_([^\/_?]+).*$/

Explained piece by peace on the string /HP-ENVY-TouchSmart-m7-j010dx-173-Touc‌h-Screen-Refurbished-Laptop/product_8000:

^: from beginning
\/: match a /
[^\/]+: match everything until a / (HP-ENVY-TouchSmart-m7-j010dx-173-Touc‌h-Screen-Refurbished-Laptop)
\/: match a /
([^_]+) match and capture the value before the _ (product)
_: match a _
([^\/_?]+) match and capture the value after the _ stopped by a ?, _ or / (8000)
.* match until the end - if there is anything
$ end

Example:

var re = /^[^\/]+\/[^\/]+\/([^_]+)_([^\/_?]+).*$/;
var matches = re.exec('www.mysite.com/22222/directory_22222?categoryId=12328');
console.log(matches.splice(1));

output:

["directory", "22222"]

edited Jan 22, 2014 at 14:28

answered Jan 22, 2014 at 13:55

Matyas

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7 Comments

Manu Over a year ago

i tried the example you have given. var match = req.url.match(/^[^\/]+\/[^\/]+\/([^_]+)_([^\/_?]+).*$/); console.log(match); i am getting null in console, what could be the possible reason?

Matyas Over a year ago

I should see the url that did not match as well to give you an appropriate answer

Manu Over a year ago

this is the url "/HP-ENVY-TouchSmart-m7-j010dx-173-Touch-Screen-Refurbished-Laptop/product_8000" which i get from req.url in nodejs request object.

Matyas Over a year ago

well of course, since in your question the provided url examples were starting with www.mysite.com and my regex takes that into consideration as well (see example)

Matyas Over a year ago

Here, updated the regex:

/^\/[^\/]+\/([^_]+)_([^\/_?]+).*$/.exec("/HP-ENVY-TouchSmart-m7-j010dx-173-Touch-Screen-Refurbished-Laptop/product_8000" ).splice(1)

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user578895 · Accepted Answer · 2014-01-22 14:33:00Z

0

use the url module to help you, not everything needs to be done with a regex :)

var uri = require( 'url' ).parse( 'www.mysite.com/22222/directory_22222?categoryId=12328', true );

which yields (with other stuff):

{ 
  query: { categoryId: '12328' },
  pathname: 'www.mysite.com/22222/directory_22222'
}

now to get your last part:

uri.pathParams = {};
uri.pathname.split('/').pop().split('_').forEach( function( val, ix, all ){
    (ix&1) && ( uri.pathParams[ all[ix-1] ] = val );
} );

which yields:

{ 
  query: { categoryId: '12328' },
  pathParams: { directory: '22222 },

  ... a bunch of other stuff you don't seem to care about
}

answered Jan 22, 2014 at 14:33

user578895

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