查找具有特定数据字符串的数组并返回其具有的另一个数据字符串

I have this JSON(Cant change it) every array in bbData contains["Username","ID","Age"]

我有这个JSON(不能改变它)bbData中的每个数组都包含[“Username”,“ID”,“Age”]

{"bbData":[
 ["Peter","/id/5423","42.4"],
 ["Bob","/id/5355","32.1"],
 ["Dolan","/id/5113","22.6"]
]]}

I know an id for a user, let's say "/id/5423". How can i then make PHP find the array with that id and return the age data which lays in the same array?

我知道用户的id,让我们说“/ id / 5423”。然后,我如何让PHP找到具有该id的数组并返回位于同一数组中的年龄数据?

2 个解决方案

#1

PHP >= 5.5.0

PHP> = 5.5.0

$data = json_decode($myJsonString);
$searchId = "/id/5355";

$key = array_search($searchId, array_column($data->bbData, 1));
if ($key === false) {
    $found = $searchId, ' not found';
} else {
    $found = $data->bbData[$key];
}
var_dump($found);

EDIT

For earlier versions of PHP, replace

对于早期版本的PHP,请替换

$key = array_search($searchId, array_column($data->bbData, 1));

with

$key = array_search(
    $searchId, 
    array_map(
        function($value) {
            return $value[1];
        },
        $data->bbData
    )
);

2 个解决方案

#1

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