JavaScript syntax declare object inside function

Question

I wrote a code:

 const search = (what, arr) => arr.filter((el) => 
 (onde(what,el).tag).includes((onde(what,el).input)));

but it call function onde(), two times.

But it is not needed.

How to set part = onde(what,el);

inside const search ?

so it can be called one time only, as part.tag and part.input

 const search = (what, arr) => arr.filter((el) => 
(onde(what,el).tag).includes((onde(what,el).input)));
 
 function exibe(el, index, array) {console.log(index+" = " + el.name +" " + el.value+" " + el.other);}
 function onde(what,where) { array1=[]; tags =what.split(':')[0]; tag =tags.split(','); 
 tag.forEach(element => { array1.push(where[element]);});
 array['tag']=array1.join(' ').toLowerCase();
 array['input']=what.split(':')[1].toLowerCase();
 return array;
 }
 
 var array = [
 { name:"string 1", value:"this A", other: "that 10" },
 { name:"string 2 this", value:"those B", other: "that 20" },
 { name:"string 3", value:"this C", other: "that 30" }
 ];
 const serached = search("name,value:this",array);
 if (serached) {
 serached.forEach(exibe);
 } else {
 console.log('No result found');
 }

Answer 1

Save it in a variable first.

const search = (what, arr) => arr.filter((el) => {
 const { tag, input } = onde(what, el);
 return tag.includes(input);
});

or

const search = (what, arr) => arr.filter((el) => {
 const result = onde(what, el);
 return result.tag.includes(result.input);
});