CODEWITHSUNDEEP
javascript
paulalexandru
paulalexandru

Reputation: 9530

Combine 2 URL in javascript

const a = '/sport/football/1';
const b = '/football/1?a=1'

How can I combine these 2 url's to get this one :

const c = '/sport/football/1?a=1';

Note: Basically the first url can have a different starting point and the second one can have multiple query params. But they have in common the middle part.

Upvotes: 1

Views: 554

Answers (3)

Christian
Christian

Reputation: 477

From the way your question looks to me you want to be able to merge two urls in a way the the second one will be appended to the first one if it is not yet in the url or be integrated into the first one if there is a overlap.

For that you need to do some testing and modifying of the strings to get to the result that you want.

Here is one possibility of solving that issue.

Known issue with this solution: it will not handle repitition in a URL.

URL a = asd/dsa/asd/dsa/123?321=123

URL b = /dsa/asd/dsa/123?321=456

result = asd/dsa/asd/dsa/asd/dsa/123?321=456

If you need to cover that case then you have to enhance the index section to find the correct occurence.

// Edit: There was an issue that allowed substrings of b to be found in a, that has been solved by also converting URL a to an array for the check

the error would be as follows:

URL a = asd/dsa/123

URL b = as/321

Result = as/321

because URL b [0] (as) is a substring of asd in URL a and would therefore be found by String.lastIndexOf

function calculateUrl(urla, urlb) {
  // we start by initialising the new url with URL a
  let ret = urla;
  
  // we split the URL a on / and cleaning it up to make sure we dont have an empty first value
  // we need this as a string check might cause a problem when part of aUrlb[0] is found in urla
  const aUrla = urla.split("/").filter(function(elm) {
    return typeof elm === "string" && elm.length > 0;
  });
  
  // we split the URL b on / and cleaning it up to make sure we dont have an empty first value
  const aUrlb = urlb.split("/").filter(function(elm) {
    return typeof elm === "string" && elm.length > 0;
  });
  
  // we generate a new string for URL b without a leeding /
  const appendix = aUrlb.join("/");
  
  // we choose the first element of URL b to check if URL b is part of URL a or if it has to be appended
  const searchTerm = aUrlb[0];
  
  // we perform the check
  const foundAt = aUrla.lastIndexOf(aUrlb[0]);
  
  if(foundAt !== -1) {
    // if we found that URL a and b have a overlap we only use the part of URL a that is unique
    ret = aUrla.slice(0, foundAt).join("/") + "/";
  } else {
    // if we found that URL a and b do not have a overlap we need to make sure that URL a does not contain a ?
    ret = ret.split("?")[0]
    // we also need to make sure URL ends in / as we will append URL b to it
    if(ret.substring(ret.length - 1) !== "/") {
      ret = ret + "/";
    }
  }
  
  // combine URL a and b
  return ret + appendix;
}


// ignore this, it's just so the button will work
document.getElementById("calculate").addEventListener("click", function() {
  let p = document.createElement("p");
  p.textContent = calculateUrl(document.getElementById("urla").value, document.getElementById("urlb").value);
  document.getElementById("output").appendChild(p);
});
<label>
URL A:
<input id="urla">
</label>
<br>
<label>
URL B:
<input id="urlb">
</label>
<br>
<button id="calculate">
Calculate
</button>
<div id="output"></div>

Upvotes: 1

Hanna Rose
Hanna Rose

Reputation: 422

i tried to find the last index of all ducplicate character then merging both of them. let me know if something doesnt work correctly.

const ur1 = '/sport/football/1';
const ur2 = '/football/1?a=1';

// get the last duplicate index
let dupCount = 0, dupMark = 0;
for(let a=0; a<ur1.length; a++){
    for(let b=0; b<ur2.length; b++){
      if( ur1[a+b] === ur2[b] ){
        //console.log( ur1[a+b] +' : '+ ur2[b] );
        dupCount += 1;
        if( ur2[b] !== '/' && dupCount > 0 ){ 
          dupMark = b;
        }
      }else{
        dupCount = 0;
        break;
      }
    }
}

console.log( 'last duplicate url index : '+ dupMark );
console.log( ur1+ur2.substr(dupMark) );

Upvotes: 0

kol
kol

Reputation: 28708

const a = "/sport/football/1";
const b = "/football/1?a=1";

const c = a + b.substring(b.indexOf("?"));
console.log(c);

Upvotes: 1

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