Question

这是我的代码：

   echo '<br/>';
   echo 'Json data from DB   '.json_encode($output);
   $data=array();
   $array=json_decode($output,true);
   echo '<br/>';
   echo 'Concerted into an array   '.json_encode($array);

这是输出：

Json data from DB  [{"0":"1","key-1":"1","1":"1","key-2":"1","2":"1","key-3":"1","3":"1","key-4":"1"}] 
Concerted into an array null

为什么json_devode返回null？如果我这样尝试：

$data = '[{"0":"1","key-1":"1","1":"1","key-2":"1","2":"1","key-3":"1","3":"1","key-4":"1"}]';

// convert to an array
$data = json_decode($data, true);

然后正常打印出来：

Json data from DB  [{"0":"1","key-1":"1","1":"1","key-2":"1","2":"1","key-3":"1","3":"1","key-4":"1"}] 
Concerted into an array  {"0":"1","key-1":"1","1":"1","key-2":"1","2":"1","key-3":"1","4":"1","key-4":"1"}

Answer 1

因为json_decode采用字符串而$output不是字符串（由json_encode证明：它是一个数组）。

Answer 2

你好像在混淆。在您的示例中$output似乎已经是一个数组，您想再次解码它？怎么样？

执行此操作时：

json_encode($output);

它返回一个正确的JSON对象，这意味着$output已经是一个数组。你不能json_decode非JSON对象。您似乎可以直接使用$output，或者需要更清楚地说明您的问题。

json_decode在有效值php上返回null

2 个答案: