如何在JSON中使用特定项目的名称来接收该项目中的其他数据

Question

for example, I have this JSON file:

[{"name":"Name1",
  "value":24,
  "min":10,
  "max":16,
  "rate":108},

  {"name":"Name2",
    "value":69,
    "min":0,
    "max":6,
    "rate":122}
]

and i have decoded it into a array using:

$json = json_decode($jsondata, true);

now I want to get the value of the 2nd item by using its name for example:

echo $json['Name1']['value'] 

instead of

echo $json[0]['value'] 

any way to do this? Thanks.

Answer 1

You can do something like this :

1 - You create a function that will loop through your array and return the value according to the name

function echoValueByName(array $array, $name) {
    foreach($array as $data) {
        if ($data['name'] == $name) {
            return $data['value'];
        }
    }
    return 'Unknow name : '.$name;
}

2 - Now use it to echo the value by name :

echo echoValueByName($json, 'Name1'); // return 24
echo echoValueByName($json, 'test');  // return Unknow name : test