Question

我正在运行AndroidHive中的应用程序，该应用程序有feed.json该应用加载下面的内容。

http://api.androidhive.info/feed/feed.json这是帖子网址http://www.androidhive.info/2014/06/android-facebook-like-custom-listview-feed-using-volley/

{
    "feed": [{
                "id": 1,
                "name": "National Geographic Channel",
                "image": "A URL",
                "status": "Science and etc"
                And Cosmos is all about making science an experience.
                ","
                profilePic ": "
                A URL ","
                timeStamp ": "
                1403375851930 ","
                url ": null
    }]
}

以上是默认内容我想知道如何从这个内容或使用PHP脚本添加更多内容。我很想知道怎么做，因为我注意到在www_androidhive_info中创建的大多数应用程序源代码都使用.json来加载内容，这是webview的替代方法

Answer 1

您可以使用json_decode()函数将JSON字符串转换为对象，添加变量，然后将对象转换回JSON

$obj = json_decode($json, true);
$obj->feed[0]->url = "google.com"
:

Answer 2

加载JSON
解码JSON
添加内容
编码为JSON
将其发送到您的应用

代码：

<?php

function failure($reason) {
    header('HTTP/1.1 500 Internal Server Error');
    echo json_encode(['failure' => $reason]);
    exit;
}

// 1. Load the JSON
$content = file_get_contents('http://api.androidhive.info/feed/feed.json');
if (!$content) {
    return failure('Failed to load upstream JSON');
}

//  2. Decode the JSON
$decodedContent = json_decode($content, true);
if ($decodedContent === false) {
    return failure('Failed to parse upstream JSON');
} elseif (empty($decodedContent)) {
    return failure('Upstream JSON empty');
}

// 3. Add your content
$decodedContent['feed'][] = [
    "id": 12,
    "name": "An example name",
    "image": "https://example.com/test_image.jpg",
    "status": "Hello!",
    "profilePic": "https://example.com/test_profile_pic.jpg",
    "timeStamp" => time(),
    "url" => "https://example.com"
];

// 4. Encode to JSON
$responseContent = json_encode($decodedContent);

// 5. Send it to your app
echo $responseContent;

Answer 3

解码JSON并为更改创建新函数，然后进行编码。

<?php
$JsonVar = file_get_contents('http://api.androidhive.info/feed/feed.json');
$myjson = json_decode($JsonVar,true);

function Changes() {
    $myjson->feed[2]->name = "FeedID2";
    $myjson->feed[2]->id = "2";
    $myjson->feed[2]->image = "link to image.png";
    $myjson->feed[2]->status = "I am here.";
    $myjson->feed[2]->profilePic = "link to image.png";
    $myjson->feed[2]->timeStamp = time();
    $myjson->feed[2]->url = 'http://url.com';
}

json_encode(Changes());
?>

Answer 4

这是一些与我正在寻找的东西有多么相关，这个答案发现@ Is there a JSON api based CMS that is hosted locally?

最佳答案我的问题

或者如果您感到困惑，可以查看

Update JSON value if exists otherwise add it in PHP

如何更新或添加JSON的值

4 个答案: