通过纯javascript更好的方法来做到这一点

Question

if(window.location.href.indexOf("=38805") > -1
  || window.location.href.indexOf("=38807") > -1
  || window.location.href.indexOf("=38816") > -1
  || window.location.href.indexOf("=38815") > -1
  || window.location.href.indexOf("=38814") > -1
  || window.location.href.indexOf("=38813") > -1
  || window.location.href.indexOf("=38811") > -1

  ){

    do something

  }

Basically, I am using a separate css for the pages that contain these strings. I might over 50 pages. Wondering if there is a cleaner way to write this. Put them into an array?

Answer 1

JS some function , is exactly for stuff like that:

let options = ["=38805","=38807","=38816"]; //...and the others
let link = window.location.href;

if( options.some( option => link.includes(option))){
    console.log('yay! =)');
}

You're actually going through your array of options, and asking a question about each of the elements in the array : "Are you present at my URL?".

Then, the some method will return true (and in this case- active the if statment ) only if one or more elements in the options array is answering true to your includes question.

---

And by the way- JS have another method that cover a similar set of mind. the every metohd .

This method, as you can understand by its name, will return true only if all the elements in the array is answering true to your question.

Answer 2

How about something like this to clean it up a bit:

const ids = [38805, 38807, 38811, 38813, 38814, 38815, 38816];

let windowHrefHasId = false;
for (let id of ids) {
    if (window.location.href.indexOf('=' + id.toString()) > -1) {
        windowHrefHasId = true;
        break;
    }
}

if (windowHrefHasId) {
    // do something
}