从特定 ID JSON 中获取特定对象

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，当我添加这个参数 stream_id= 和例如我添加了这个 id stream_id=200 它应该只显示具有该 id 的对象，所以更清楚地说它应该显示 id:200, name:Ravi Tamada, email:ravi@gmail .com 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

您可能需要更频繁地通过 id 获取数据，而不是每次都循环遍历数组。所以你可以

foreach($users_array as $user_object){
    $users[$user_object->id]=$user_object;
}

然后访问：

echo $users[202]->email;

Answer 2

您可以遍历您的 JSON 并查找您需要的 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;
        }
    }
}

Answer 3

您应该能够使用如下所示的简单 js/jquery 函数循环遍历 JSON，您只需传递指定的 id。

function getJsonById(id){
var json = {
        "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" 
            }
        ]
    };

var l= json['contacts'].length;
var c = 0;
while(c<l){
  if(json['contacts'][c]['id']==id){
    console.log(json['contacts'][c]);
     return json['contacts'][c];
  } 
  c++;
}
}
return "";
});

Answer 4

也许您可以将 json_decode 与 $assoc=true 一起使用（此选项将 json 转换为数组，甚至转换为嵌套数组），然后循环遍历数组以找到所需的元素。