這裏返回null是我的代碼:json_decode上一個有效值PHP
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);
這裏是輸出:
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
爲什麼json_devode返回null?如果我嘗試相同的這樣的:
$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);
然後它通常打印出來:
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"}
三重檢查您嘗試解碼的變量! – deceze