How can I get a specific parameter from location.search?

A non-regex approach, you can simply split by the character '&' and iterate through the key/value pair:

function getParameter(paramName) {
  var searchString = window.location.search.substring(1),
      i, val, params = searchString.split("&");

  for (i=0;i<params.length;i++) {
    val = params[i].split("=");
    if (val[0] == paramName) {
      return val[1];
    }
  }
  return null;
}

2020 EDIT:

Nowadays, in modern browsers you can use the URLSearchParams constructor:

const params = new URLSearchParams('?year=2020&month=02&day=01')

// You can access specific parameters:
console.log(params.get('year'))
console.log(params.get('month'))

// And you can iterate over all parameters
for (const [key, value] of params) {
  console.log(`Key: ${key}, Value: ${value}`);
}

function gup( name ) {
    name = name.replace(/[\[]/,"\\\[").replace(/[\]]/,"\\\]");
    var regexS = "[\\?&]"+name+"=([^&#]*)";
    var regex = new RegExp( regexS );
    var results = regex.exec( window.location.href );
    if( results == null )
        return "";
    else
        return results[1];
}
var year = gup("year"); // returns "2008"

A Simple One-Line Solution:

let query = Object.assign.apply(null, location.search.slice(1).split('&').map(entry => ({ [entry.split('=')[0]]: entry.split('=')[1] })));

Expanded & Explained:

// define variable
let query;

// fetch source query
query = location.search;

// skip past the '?' delimiter
query = query.slice(1);

// split source query by entry delimiter
query = query.split('&');

// replace each query entry with an object containing the query entry
query = query.map((entry) => {

   // get query entry key
   let key = entry.split('=')[0];

   // get query entry value
   let value = entry.split('=')[1];

   // define query object
   let container = {};

   // add query entry to object
   container[key] = value;

   // return query object
   return container;
});

// merge all query objects
query = Object.assign.apply(null, query);

I used a variant of Alex's - but needed to to convert the param appearing multiple times to an array. There seem to be many options. I didn't want rely on another library for something this simple. I suppose one of the other options posted here may be better - I adapted Alex's because of the straight forwardness.

parseQueryString = function() {
    var str = window.location.search;
    var objURL = {};

    // local isArray - defer to underscore, as we are already using the lib
    var isArray = _.isArray

    str.replace(
        new RegExp( "([^?=&]+)(=([^&]*))?", "g" ),
        function( $0, $1, $2, $3 ){

            if(objURL[ $1 ] && !isArray(objURL[ $1 ])){
                // if there parameter occurs more than once, convert to an array on 2nd
                var first = objURL[ $1 ]
                objURL[ $1 ] = [first, $3]
            } else if(objURL[ $1 ] && isArray(objURL[ $1 ])){
                // if there parameter occurs more than once, add to array after 2nd
                objURL[ $1 ].push($3)
            }
            else
            {
                // this is the first instance
                objURL[ $1 ] = $3;
            }

        }
    );
    return objURL;
};

Grab the params from location.search with one line:

const params = new Map(this.props.location.search.slice(1).split('&').map(param => param.split('=')))

Then, simply:

if(params.get("year")){
  //year exists. do something...
} else {
  //year doesn't exist. do something else...
}

ES6 answer:

const parseQueryString = (path = window.location.search) =>
  path.slice(1).split('&').reduce((car, cur) => {
   const [key, value] = cur.split('=')
   return { ...car, [key]: value } 
  }, {})

for example:

parseQueryString('?foo=bar&foobar=baz')
// => {foo: "bar", foobar: "baz"}

This is what I like to do:

window.location.search
    .substr(1)
    .split('&')
    .reduce(
        function(accumulator, currentValue) {
            var pair = currentValue
                .split('=')
                .map(function(value) {
                    return decodeURIComponent(value);
                });

            accumulator[pair[0]] = pair[1];

            return accumulator;
        },
        {}
    );

Of course you can make it more compact using modern syntax or writing everything into one line...

I leave that up to you.