解碼無效JSON在PHP

我有看起來像這樣的解碼無效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":[]}

外部源的無效JSON的回報，如果我試圖解碼

json_decode($myjson, true)

我只希望path_type這將返回null項並將其值

所以在我的代碼，我需要這個

if (path_type == 2){}

有什麼建議嗎？

來源

2017-02-20 Ian Adem

難道這_USER保存在DB._應該在那裏？ – MrDarkLynx

是的。我認爲這是結果的原因null –

那就是無效的JSON。你必須以'{「user」開頭：'... – MrDarkLynx

首先你想從你的JSON中去掉不必要的字符串。
然後要解碼JSON存儲在一個變量，並檢查PATH類型：

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

$data = json_decode($myjson, true); 

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

來源

2017-02-20 10:02:22 MrDarkLynx

擴大對@ MrDarkLynx的答案我會用這個表達式： ^(.+?){"user" 使用正則表達式{"user"之前刪除一切，一定要只刪除第一個捕獲的組。

您現在有可以使用有效的JSON =）

來源

2017-02-20 11:02:20 g3mini

我認爲這是更好的解決方案，但現在我使用MrDarkLynx答案，因爲它爲我工作。 –

解碼無效JSON在PHP

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