正则表达式以匹配Node.js中的url模式

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.

Answer 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

Answer 2

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-Touc‌​h-Screen-Refurbished-Laptop/product_8000 :

• ^ : from beginning
• \\/ : match a /
• [^\\/]+ : match everything until a / ( HP-ENVY-TouchSmart-m7-j010dx-173-Touc‌​h-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"]

Answer 3

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
}