回聲多維JSON和PHP

Question

我試圖訪問具有與多維JSON輸入的API，這是陣列

[{ 
"Value 1":"a", 
"Value 2": 
    { 
     "Value 3": 
     { 
      "Value4":11, 
      "Value5":"C", 
     }, 
     "Value 4": 
     { 
      "Value6":12, 
     } 
    } 
}] 

欲呼應「11」值4和「12」中Value6。我已經嘗試呼應它

$varkota = 'url to json output'; 
$datakota = json_decode(file_get_contents($varkota, true)); 
$data1 = json_decode($datakota[0]->Value2); 
$data2 = json_decode($data1[0]->Value3); 
echo $data2[0]->Value4; 

錯誤告訴我：

!) SCREAM: Error suppression ignored for 
(!) Warning: json_decode() expects parameter 1 to be string, object given in debug.php on line 6 
Call Stack 
# Time Memory Function Location 
1 0.0020 145280 {main}() ..\debug.php:0 
2 4.1329 188648 json_decode () ..\debug.php:6 

(!) SCREAM: Error suppression ignored for 
(!) Notice: Trying to get property of non-object in debug.php on line 7 
Call Stack 
# Time Memory Function Location 
1 0.0020 145280 {main}() ..\debug.php:0 

(!) SCREAM: Error suppression ignored for 
(!) Notice: Trying to get property of non-object in on line 8 
Call Stack 
# Time Memory Function Location 
1 0.0020 145280 {main}() ..\debug.php:0 

任何想法？

Answer 1

首先，你需要使用json_decode只有一次，而你的密鑰具有被有空間空間和鍵，那麼你需要附上它作爲字符串和{}。試着像，

$datakota = json_decode(file_get_contents($varkota, true)); 
$data1 = $datakota[0]->{'Value 2'}; 
$data2 = $data1->{'Value 3'}; // no need to use array [0], as Value 2 is object 
echo $data2->Value4; // no need to use array [0], as Value 3 is also an object 

在單行線，只把它作爲

echo $datakota[0]->{'Value 2'}->{'Value 3'}->Value4; 

Online Demo

Answer 2

您只需要撥打json_decode一次。

$varkota = 'url to json output'; 
$datakota = json_decode(file_get_contents($varkota, true)); 
$data1 = $datakota[0]->Temperature->Metric; 
// ... 

或與您當前的JSON字符串

$data1 = $datakota[0]->{"Value 2"}->{"Value 3"};