php - 从特定 Id JSON 获取特定对象

Question

我的 json 文件有点问题，我想做的是从我的 json 文件中获取特定对象，所以在上面的这个 json 中，当我添加此参数时 stream_id= 并为例如我添加这个 id stream_id=200 它应该只显示具有该 id 的对象，所以为了更清楚它应该显示 id:200, name:Ravi Tamada, email:ravi@gmail .com 等等，用 PHP，谢谢

{
        "contacts": [
            {
                    "id": "200",
                    "name": "Ravi Tamada",
                    "email": "ravi@gmail.com",
                    "address": "xx-xx-xxxx,x - street, x - country",
                    "gender" : "male",
                    "url": "http://149.202.196.143:8000/live/djemal/djemal/592.ts"
            },
            {
                    "id": "201",
                    "name": "Johnny Depp",
                    "email": "johnny_depp@gmail.com",
                    "address": "xx-xx-xxxx,x - street, x - country",
                    "gender" : "male",
                    "url":"http://149.202.196.143:8000/live/djemal/djemal/592.ts" 
            },
            {
                    "id": "202",
                    "name": "Leonardo Dicaprio",
                    "email": "leonardo_dicaprio@gmail.com",
                    "address": "xx-xx-xxxx,x - street, x - country",
                    "gender" : "male",


    "url":"http://149.202.196.143:8000/live/djemal/djemal/592.ts" 
            }
        ]
    }

Answer 1

您可以遍历 JSON 并查找所需的 ID。就这么简单。

<?php
$json = /* your json */;
$array = json_decode($json, true);

$result = getInfo(200, $array);

function getInfo($id, $array) {
    foreach($array AS $index=>$json) {
        if($json['id'] == $id) {
            return $json;
        }
    }
}