Evaluate for multiple checks for items in an array

Question

I have 3 values whose value is to be compared to a key, like:

array=["test", "test1", "test2"];
array.find((ele)=>{
        return (ele === "test" || ele === "test1" || ele === "test2");
      });

I'm already doing this, however I wanted to check if there is better way to do this? I tried:

if (["test", "test1", "test2"].every(function(x){
     {return true;}
    }))

this does not work, any ideas?

Answer 1

You could use Array#Includes to check to check if you array has strings you are looking for!

Demo:

 let arr = ["test", "test1", "test2"]; if(arr.includes("test") || arr.includes("tes1") || arr.includes("test2")) { console.log(true); }

You could also use ArrayMap and startswith function to look for strings that matches the required condition.

Demo:

 let arr = ["test", "test1", "test2"]; let checkArr = arr.map(x => x.startsWith('test')).join(',') console.log(checkArr)

Answer 2

How about using startsWith() ?

 let array = ["test","test1","test2"]; array.forEach(key => console.log(key.startsWith("test")));

Answer 3

You can use Array#some with Array#includes to check if the first array contains any of the elements in the second.

 array=["test", "test1", "test2"]; if (["test", "test1", "test2"].some(x=>array.includes(x))){ console.log("true"); }