Array highest value or create next value? - php

Question

I have an array of names and id's.
Array(
0=>array(
name=>New York,
id=>45763448349,
can=0
),
1=>array(
name=>New York2,
id=>45763464566,
can=0
),
3=>array(
name=>New York3,
id=>45763464854,
can=0
),
4=>array(
name=>New York4,
id=>45763464955,
can=0
),
5=>array(
name=>New York5,
id=>45763464745,
can=0
),
6=>array(
name=>New York6,
id=>45763464235,
can=1
)
)
For example (there will be different names (not all New York)) and I have an x variable with a name (for example x=New York) what I need to do is find the highest name and get the id and then make sure can is 1 otherwise specific the next highest name.
For example let's assume that I have x=New York so I look through the array and find that the highest array name is New York6. I see that can=1 so I just get the id. Now if can != 1 then I could specific New York7.
NOTE: This array will contain many different values (not all New York, I just used those for an example)
Now how on earth would I even get started?

Answer 1

In code:
function city_cmp($record_a, $record_b) {
return strcmp($record_a["name"], $record_b["name"];
}
usort($cities, 'city_cmp');
end($cities);
$city = current($cities);
do {
if ($city["can"] === 1) {
$can_city = $city;
break;
}
} while ($city = prev($cities));
Note: optionally, you could skip the break and make $can_city part of your while condition. Some prefer that as it can make the loop exit condition clearer.
In quick steps:
usort() the array
use end() to point at the end of the array and prev() until you find a can=1.

Answer 2

Since it's a multi-dimensional array (I'll call it $array) you probably have to loop through it once to find relevant data.
$highest_id = array('id' => 0, 'count' => 0, 'can' => 0);
for each ( $array AS $key => $data )
{
if ( substr($data['name'],0,strlen($x)) == $x )
{
$current_count = (int) substr($data['name'],strlen($x));
if ( $current_count > $highest_id['count'] )
{
$highest_id = array('id' => $data['id'], 'count' => $current_count, 'can' => $data['can']);
}
}
}
if ( $highest_id['can'] == 1 )
{
$id = $highest_id['id'];
}
else
{
// can is not 1 => specify new $x
}

Answer 3

Start by dividing your task into smaller ones. The first filter criterion seems to be can=1. Let's check that with array_filter.
$ar = array(array('name'=>'New York', 'id'=>45763448349, 'can'=>1), array('name'=>'New York2', 'id'=>45763464566, 'can'=>0), array('name'=>'New York3', 'id'=>45763464854, 'can'=>0), array('name'=>'New York4', 'id'=>45763464955, 'can'=>0), array('name'=>'New York5', 'id'=>45763464745, 'can'=>0), array('name'=>'New York6', 'id'=>45763464235, 'can'=>1), array('name'=>'New Yorkistan', 'id'=>0, 'can'=1));
$ar = array_filter($ar, function($el) {return $el['can'];});
// Now $ar contains the entries 'New York', 'New York2', and 'New Yorkistan'
Now, filter out the entries which do not match New York(and a number).
$ar = array_filter($ar, function($el) {
$search = 'New York';
$name = $ar['name'];
$potentialNumber = substr($name, strlen($search));
return ((substr($name, 0, strlen($search)) == $search) && // starts with NY
($potentialNumber == '' || is_numeric($potentialNumber)); // [0-9]*
});
// $ar contains only 'New York' and 'New York6' now
Now, you just need some sorting with usort and pick the first element:
usort($ar, function($el1, $el2) {
$id1 = $el1['id']; $id2 = $el2['id'];
if ($id1 == $id2) return 0;
return $id1 < $id2 ? 1 : -1;
});
if (count($ar) > 0) {
$result = $ar[0];
} else {
// Nothing matched your criteria
}

Answer 4

Something like this:
//$my_array is your array of values
//$string = 'New York'; your search term
$id = false;
for($i = count($my_array) - 1; $i >= 0; $i++) {
if($id === false) {
if(strpos($my_array[$i]['name'], $string) !== false && $my_array[$i]['can'] == 1) {
$id = $my_array[$i]['id'];
}
}
}
//Insert it if no id was found
if($id === false) {
$id = 'something';
$can = 1;
array_push($my_array, array('name' => $string, 'id' => $id, 'can' => $can));
}
Does that work?
edit
(In case it's not obvious, $string is whatever you are searching for)