'Recursion on subarrays with "child" key
I have array of arrays - tree structure of main menu.
I try to find in this tree one node with needed slug and return this node with all it's childs.
I write little recurcive function
<?php
$tree = '[{"id":1,"structure":1,"parent":0,"slug":"medicinskaya-ge","child":{"2":{"id":2,"structure":1.1,"parent":1,"slug":"dnk-diagnostika","child":{"3":{"id":3,"structure":"1.1.1","parent":2,"slug":"dnk-diagnostika","datafile":"ssz","template":"ssz"},"4":{"id":4,"structure":"1.1.2","parent":2,"slug":"dnk-diagnostika","child":{"5":{"id":5,"structure":"1.1.2.1","parent":4,"slug":"dnk-diagnostika"},"6":{"id":6,"structure":"1.1.2.2","parent":4,"slug":"testirovanie-ge"},"7":{"id":7,"structure":"1.1.2.3","parent":4,"slug":"dnk-diagnostika"}}},"8":{"id":8,"structure":"1.1.3","parent":2,"slug":"dnk-diagnostika"},"9":{"id":9,"structure":"1.1.4","parent":2,"slug":"texnologiya-kol"}}}}}]';
$tree = json_decode($tree, true);
function getSlugData(string $slug, array $data)
{
foreach ($data as $row) {
if ($row['slug'] == $slug) {
return $row;
}
if (isset($row['child'])) {
//return $this->getSlugData($slug, $row['child']);
}
}
return [];
}
$result = getSlugData('testirovanie-ge', $tree);
print_r($result);
But as a result I have an empty array. If I print_r($row) when $row['slug'] == $slug - It appears on screen.
if ($row['slug'] == $slug) {
exit(print_r($row));
return $row;
}
What's my mistake?
Solution 1:[1]
In the recursive function, you must not break the loop early unless you find your slug match. If a non-slug-match has a child element, you must iterate it and potentially pass up a match in subsequent recursive calls.
Code: (Demo)
function getSlugData(string $slug, array $data): array
{
foreach ($data as $row) {
if ($row['slug'] === $slug) {
return $row;
}
if (isset($row['child'])) {
$deeper = getSlugData($slug, $row['child']);
if ($deeper) {
return $deeper;
}
}
}
return [];
}
P.s. you aren't calling a class method, so $this-> is inappropriate.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | mickmackusa |
