Echo 多维 json 和 php

Question

我正在尝试访问具有多维 json 输入的 API，这是数组

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

我想在 Value4 中回显“11”，在 Value6 中回显“12”。我试着回应它

$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;

错误告诉我：

 ! ) 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

有什么想法吗？

Answer 1

首先，您只需要使用一次json_decode，并且您的键有空格，而有空格的键则需要将其作为字符串括在{} 中。试试吧，

$datakota = json_decode(file_get_contents($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

在一行中，直接作为

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

Online Demo

Answer 2

您只需拨打json_decode 一次。

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

或者使用你当前的 json 字符串

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