I am working with an array of urls and for each url i wanna find a image corresponding to the site domain. my first attempt was
const url = new URL(props.url);
const platform = url.hostname.split(".")[1];
console.log(platform)
const platform_logos = {
"codechef": "images/chef.png",
"withgoogle": "images/google.png",
.
.
.
"codeforces": "images/codeforces.png",
}
let platform_logo = platform_logos[platform];
but it doesnt work with url of type 'https://momo2022fr.hackerearth.com' so i had to resort to
let platform_logo = "images/code.png"
if (url.includes("hackerearth")) {
platform_logo = "images/hackerearth.png"
}
else if (url.includes("hackerrank")) {
platform_logo = "images/hackerrank.png"
}
else if (url.includes("codeforces")) {
platform_logo = "images/codeforces.png"
}
else if (url.includes("codechef")) {
platform_logo = "images/chef.png"
}
else if (url.includes("atcoder")) {
platform_logo = "images/atcoder.png"
}
else if (url.includes("leetcode")) {
platform_logo = "images/leetcode.png"
}
else if (props.url.includes("withgoogle")) {
platform_logo = "images/google.png"
}
Is there any better way of writing the code below, it just feels like it violates DRY
You could just do the same thing as in your first solution and store the mapping from the substring to the image path in an ocject:
const platform_logos = {
"hackerearth": "images/hackerearth.png",
"hackerrank": "images/hackerrank.png",
"codeforces": "images/codeforces.png",
"codechef": "images/chef.png",
"atcoder": "images/atcoder.png",
"leetcode": "images/leetcode.png",
"withgoogle": "images/google.png"
};
Then you could iterate over the key-value pairs in your object to find the key that is part of the URL and return it once it matches:
function getLogo(url) {
for(const [key, value] of Object.entries(platform_logos)) {
if(url.contains(key)) {
return value;
}
}
}
let platform_logo = getLogo(url);
You could change how you're reading the url to only get the root domain.
location.hostname.split('.').reverse().splice(0,2).reverse().join('.').split('.')[0]
This code would give hackerearth for https://momo2022fr.hackerearth.com/.
So there are several ways of achieving this. These are just two from the top of my head.
Parsing the url and using a switch() to determine the outcome, with fallback if none is found.
const url = new URL("https://www.withgoogle.com/search?q=test");
const sites = [
"hackerearth",
"hackerrank",
"codeforces",
"codechef",
"atcoder",
"leetcode",
"withgoogle",
];
console.info(url.hostname);
const site = url.hostname.match(new RegExp(`${sites.join("|")}`));
let logo = "";
switch (site[0]) {
case "hackerearth":
logo = "images/hackerearth.png";
break;
case "hackerrank":
logo = "images/hackerrank.png";
break;
case "codeforces":
logo = "images/codeforces.png";
break;
case "codechef":
logo = "images/chef.png";
break;
case "atcoder":
logo = "images/atcoder.png";
break;
case "leetcode":
logo = "images/leetcode.png";
break;
case "withgoogle":
logo = "images/google.png";
break;
default:
logo = "images/code.png";
break;
}
console.info(logo);Then there is the modern way, with less code and programming the fallback.
// const url = new URL("https://eee.com/test");
const url = new URL("https://www.withgoogle.com/search?q=test");
const sites = {
hackerearth: "images/hackerearth.png",
hackerrank: "images/hackerrank.png",
codeforces: "images/codeforces.png",
codechef: "images/chef.png",
atcoder: "images/atcoder.png",
leetcode: "images/leetcode.png",
withgoogle: "images/google.png",
default: "images/code.png",
};
let site = url.hostname.match(new RegExp(`${Object.keys(sites).join("|")}`));
if (site === null) {
site = "default";
}
console.info(site, sites[site]);