What is the best way to loop through each child element and retrieve a specific element?

I am working on a find() function that searches for an element by tag name within a nested structure. I have attempted to iterate through all the children elements to locate the matching element.

function find(tag: string) {

    let children = data.children;

    for(let i = 0; i <= children.length; i++) {
       
                if (children[i].type == 0 && children[i].tag == tag) {
            return children[i];
        } else {
            
            if (children[i].children) {
                children = children[i].children;
            }
        }
     
    }
}

However, I have encountered two issues. Firstly, the found variable has a different scope inside the loop and is not accessible outside it.

The children variable in the loop body is also causing syntax errors.

Any suggestions on how to simplify this code?

let data = {
  "children": [
     {children: [{tag: 1, children: [{tag: 3}]}]},
   ]
}

I also tried the following approach:

function find(tag: string) {

let children = data.children;
let found;

  return function() {
    for(let i = 0; i < children.length; i++) {
         if (children[i].tag == tag) {
             return;
         } 

         if (children[i].children) {
             children = children[i].children;
         }
    }
  }
}


let a = find('tag');
let result = a();
console.log(result);

Answer №1

key issues:

  • You should use < instead of <= to avoid looping on non-existent array elements (which causes a syntax error)
  • To make your result accessible, simply return it from the function

Additionally, there are other problems:

  • When checking for sub-elements, use children[i].children instead of children[i]
  • Avoid reusing the variable children when iterating over sub-elements as it loses reference to the top level children. Consider using recursion for unlimited level search and avoiding confusion between levels

NOTE: Type declarations for variables have been omitted for snippet testing purposes

function find(tag, optionalData) {

    let children = optionalData ? optionalData : data.children;
    let found;

    for(let i = 0; i < children.length; i++) {
        if (children[i].type == 0 && children[i].tag == tag) {
            return children[i];
        } else if (children[i].children) {
            found = find(tag, children[i].children);
            if(found){
                return found;
            }
        }
    }
    
    return false;
}

let data = {
  "children": [
     {children: [{type: 0, tag: 1, children: [{type: 0, tag: 3}]}]},
   ]
}

console.log(find('1'));

BONUS: It's recommended to use 2 arguments for this type of function and pass data as an argument. This approach increases reusability and eliminates the need for ternary operators:

function find(children, tag) {
    
    let found;

    for(let i = 0; i < children.length; i++) {
        if (children[i].type == 0 && children[i].tag == tag) {
            return children[i];
        } else if (children[i].children) {
            found = find(children[i].children, tag);
            if(found){
                return found;
            }
        }
    }
    
    return false;
}

let data = {
  "children": [
     {children: [{type: 0, tag: 1, children: [{type: 0, tag: 3}]}]},
   ]
}

console.log(find(data.children, '1'));

Answer №2

By consolidating the loop into a single element when it is not found using

if (children[i]) { children = children[i]; }
, you are simplifying the process. However, it might be more efficient to remove this step. Additionally, in the for loop, consider using i<children.length instead of i<=children.length.

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