2017-10-18 94 views
2

我需要解析這個數據集我已經但我嘗試了幾種方法,但它不起作用。Laravel數據集解析錯誤

{"id":2,"area_id":"1","payment_term_id":"1","payment_type_id":"1","trade_channel_id":"1","salesman_id":1,"customer_name":"xxx","contact_number":"1231212","fax_number":"1231232","home_address":"asd","delivery_address":"dasd","email_address":"[email protected]","created_by":"4","updated_by":"4","is_active":"1","created_at":"2017-10-17 06:02:59","updated_at":"2017-10-17 06:13:22"} 

我想這一個,但這個錯誤說

(1/1)ErrorException 試圖讓非對象的屬性

$details = $request->get('details'); 
return $details->id; 

我也試過這個

$details = $request->get('details'); 
return $details['id']; 

(1/1)ErrorException 非法字符串偏移'id '

我使用laravel 5.4

回答

1

由於數據來自請求時,它一JSON和不是一個集合或對象。所以,你需要做這樣的事情:

return json_decode($request->details)->id; 
+1

這工作感謝@alexey –

1

要在控制器檢索JSON有效載荷,可使用:的

$details = json_decode($request->details)->id); 
dd($details); 

代替

$details = $request->get('details'); 
return $details['id']; 
+1

你會得到'試着獲得非對象屬性'與此代碼。您需要先解碼並獲取ID。 –

+1

@AlexeyMezenin謝謝!我已經更新了我的答案! –

1

要爲陣列狀$details['id']使用試試這個:

$details = json_decode($request->details, true); 
echo $details['id'];