json_decode + json_encode 组合不创建原始 JSON

Question

我有一些通过 API 调用获得的 JSON，我在其上运行 json_decode，从中获取一个数组，然后使用 json_encode 对其进行重新编码。然而，结果不是相同的 JSON；它弄乱了 URL。如何让它正确编码？

原创

{"created_at":"Mon, 19 Mar 2012 01:34:41 +0000","entities":{"hashtags":[{"text":"stanford","indices":[23,32]}],"urls":[{"url":"http:\/\/t.co\/Of4z6jKG","expanded_url":"http:\/\/360.io\/5sZc2T","display_url":"360.io\/5sZc2T","indices":[33,53]}],"user_mentions":[]},"from_user":"rayfk","from_user_id":335143881,"from_user_id_str":"335143881","from_user_name":"Raymond Kennedy","geo":{"coordinates":[37.4227,-122.1753],"type":"Point"},"id":181554251733020673,"id_str":"181554251733020673","iso_language_code":"en","metadata":{"result_type":"recent"},"profile_image_url":"http:\/\/a0.twimg.com\/profile_images\/1468102095\/image_normal.jpg","profile_image_url_https":"https:\/\/si0.twimg.com\/profile_images\/1468102095\/image_normal.jpg","source":"<a href="http:\/\/www.occipital.com\/360\/app" rel="nofollow">360 Panorama<\/a>","text":"View from mid lake log #stanford http:\/\/t.co\/Of4z6jKG","to_user":null,"to_user_id":null,"to_user_id_str":null,"to_user_name":null}

解码/编码组合后

{"created_at":"Mon, 19 Mar 2012 01:34:41 +0000","entities":{"hashtags":[{"text":"stanford","indices":[23,32]}],"urls":[{"url":"http:\/\/t.co\/Of4z6jKG","expanded_url":"http:\/\/360.io\/5sZc2T","display_url":"360.io\/5sZc2T","indices":[33,53]}],"user_mentions":[]},"from_user":"rayfk","from_user_id":335143881,"from_user_id_str":"335143881","from_user_name":"Raymond Kennedy","geo":{"coordinates":[37.4227,-122.1753],"type":"Point"},"id":181554251733020673,"id_str":"181554251733020673","iso_language_code":"en","metadata":{"result_type":"recent"},"profile_image_url":"http:\/\/a0.twimg.com\/profile_images\/1468102095\/image_normal.jpg","profile_image_url_https":"https:\/\/si0.twimg.com\/profile_images\/1468102095\/image_normal.jpg","source":"<a href="http:\/\/www.occipital.com\/360\/app" rel="nofollow">360 Panorama<\/a>","text":"View from mid lake log #stanford http:\/\/t.co\/Of4z6jKG","to_user":null,"to_user_id":null,"to_user_id_str":null,"to_user_name":null}

这些是完整的 sn-ps，但罪魁祸首是：

原创 "source":"<a href="http:\/\/www.occipital.com\/360\/app" rel="nofollow">360 Panorama<\/a>"

之后 "source":"<a href="http:\/\/www.occipital.com\/360\/app" rel="nofollow">360 Panorama<\/a>"

Answer 1

我不确定是什么原因造成的，但您可以通过将html_entity_decode() 函数应用于 after 版本来纠正它。这会将 < 或 " 等内容更改回其原始形式。

根据它如何影响您的引用，您也可以传递一些标志以获得所需的结果。

• ENT_COMPAT：将转换双引号并保留单引号。
• ENT_QUOTES：将转换双引号和单引号。
• ENT_NOQUOTES：不转换双引号和单引号。

[编辑]

通过这个函数运行你损坏的 JSON：

function fixDoubleQuotedJSON($broken_json)
{
   return str_replace('"','\\"',$broken_json);
}

Answer 2

http://codepad.org/DMkAS2iR

他们似乎是平等的，除了在一个不例外的地方：

before: "id":181554251733020673,"id_str"
after:  "id":181554251733020000,"id_str"

在“无损” json 转换后，这些 id 不会真正匹配，并且从 PHP 5.4 开始支持 JSON_BIGINT_AS_STRING 选项。

顺便说一下Codepad的php版本是5.2.5；