检查2D关联数组中的字符串，返回信息（PHP）

I am learning PHP and I am looking to search an array for a given string and return some values if it matches. I have tried using in_array() and array_search but both haven't been able to match/return the value correctly.

print_r($names);

Array
(
    [0] => Array
        (
            [name] => A
            [age] => 42
            [location] => Australia
        )

    [1] => Array
        (
            [name] => B
            [age] => 21
            [location] => France
        )

    [2] => Array
        (
            [name] => Z
            [age] => 50
            [location] => Japan
        )
)

I would like to check to see if a name exists in the array and return the information for the same index location.

$name="A";
if (in_array($name, $names)) {
    $age=$names['age'];
    $location=$names['location'];
    echo "$name was found! Their Age is: $age, They are based in: $location";
}
else{
    echo"$name was not found";
}

However, the name is not found and the age and location data is not loaded.

Output:

A was not found

Expected Output:

A was found! Their Age is: 42, They are based in: Australia

Testing using array_search()

$result = array_search("$name",$names);
echo $result;

No value printed.

Any feedback on how I can fix my (in_array) or how I would go about using array_search() to list the information would be greatly appreciated.

You can try this -

$position = array_search($name, array_column($your_array, 'name'));
$info = $your_array[$position];

Explanation

array_column($your_array, 'name')

Will return the array with all names with respective indexes.

array_search($name, array_column($your_array, 'name'))

Will return the index where the name is found.

$info = $your_array[$position]

Will store the particular sub-array in $info.

So it should be something like-

echo $info['name'] . " was found! Their Age is: " . $info['age'] . ", They are based in: " . $info['location'];

array_column() & array_search()

Working code