解码无效JSON在PHP

我有看起来像这样的解码无效JSON在PHP

USER SAVED IN DB.{"user":{"_token":"9ylCAviuuCaNlCtXNya5pEXkY8vkJepZAohsG5VI","submit":"engageiq_post_data","affiliate_id":"","campaign_id":"","offer_id":"","s1":"","s2":"","s3":"","s4":"","s5":"","address":"","phone1":"","phone2":"","phone3":"","phone":"","source_url":"http:\/\/pfr_laravel.dev\/registration","ip":"192.168.10.1","screen_view":"1","first_name":"fff","last_name":"ff","email":"[email protected]","zip":"00501","birthdate":"","dobmonth":"04","dobday":"12","dobyear":"1965","gender":"M","chk_agree":"","submitBtn":"Submit","state":"NY","city":"Holtsville","revenue_tracker_id":1},"revenue_tracker_id":1,"path_type":2,"campaigns":[[1,15,25,48,38,23,44],[245],[249],[27,4,19,181,18],[16],[246],[51,52,151],[10],[26,2,185,180,45,184,182]],"creatives":[]}

外部源的无效JSON的回报，如果我试图解码

json_decode($myjson, true)

我只希望path_type这将返回null项并将其值

所以在我的代码，我需要这个

if (path_type == 2){}

有什么建议吗？

来源

2017-02-20 Ian Adem

难道这_USER保存在DB._应该在那里？ – MrDarkLynx

是的。我认为这是结果的原因null –

那就是无效的JSON。你必须以'{“user”开头：'... – MrDarkLynx

首先你想从你的JSON中去掉不必要的字符串。
然后要解码JSON存储在一个变量，并检查PATH类型：

$myjson = substr($myjson, 17); // remove bulk from JSON 

$data = json_decode($myjson, true); 

if ($data['path_type'] == 2) { 
    // code 
}

来源

2017-02-20 10:02:22 MrDarkLynx

扩大对@ MrDarkLynx的答案我会用这个表达式： ^(.+?){"user" 使用正则表达式{"user"之前删除一切，一定要只删除第一个捕获的组。

您现在有可以使用有效的JSON =）

来源

2017-02-20 11:02:20 g3mini

我认为这是更好的解决方案，但现在我使用MrDarkLynx答案，因为它为我工作。 –

解码无效JSON在PHP

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