使用 php 从 JSON 对象中选择数据

Question

我有json数据我想要select一些数据在我告诉你之前请仔细阅读数据。

{
    "response": {
        "status": 1,
        "httpStatus": 200,
        "data": [{
            "offer_id": "6912",
            "Thumbnail": {
                "10116": {
                    "id": "10116",
                    "offer_id": "6912",
                    "display": "Icandytv_IN_Call-30-19-20-51).gif",
                    "thumbnail": "https:\/\/media.go2speed.org\/brand\/files\/mobvista\/6912\/thumbnails_100\/Icandytv_IN_Call(04-30-19-20-51).gif"
                }
            }
        }],
        "errors":[] ,
        "errorMessage": null
    }
}

从上面的数据我想收集缩略图的值请帮助我使用 PHP

找到它

Answer 1

也许这是一个 copy/paste 问题，但是您的 JSON-String 缺少一些 }。但这不是问题。如果您想在 json 中访问 thumbnail（小写），请尝试以下操作：

<?php

$json = '{"response": {"status":1,"httpStatus":200,"data":[{"offer_id":"6912","Thumbnail":{"10116":{"id":"10116","offer_id":"6912","display":"Icandytv_IN_Call (04-30-19-20-51).gif","thumbnail":"https://media.go2speed.org/brand/files/mobvista/6912/thumbnails_100/Icandytv_IN_Call(04-30-19-20-51).gif"}}}],"errors":[],"errorMessage":null}}';

// Make JSON accessible for PHP
$data = json_decode($json);

// Get access to Thumbnail object
$thumbnail0 = $data->response->data[0]->Thumbnail;

// Get access to thumbnail object "10116"
$thumbnail_10116 = $thumbnail0->{"10116"};

// Thumbnail Url
$thumbnailUrl = $thumbnail_10116->thumbnail;

echo $thumbnailUrl . "\n";

// or in one swoop
$thumbnailUrl2 = $data->response->data[0]->Thumbnail->{"10116"}->thumbnail;

echo $thumbnailUrl2 . "\n";

?>

希望，这有帮助

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