php json decode

Question

我想解碼一些JSON並在MySQL中插入值。這裏是我的代碼：

$json = ' 
{"d":[{"Id":1059,"Name":"Alfa Romeo - 145"},{"Id":20020,"Name":"Alfa Romeo - 146"},{"Id":1060,"Name":"Alfa Romeo - 147"},{"Id":20021,"Name":"Alfa Romeo - 155"},{"Id":1061,"Name":"Alfa Romeo - 156"},{"Id":20022,"Name":"Alfa Romeo - 159"},{"Id":20023,"Name":"Alfa Romeo - 164"},{"Id":20024,"Name":"Alfa Romeo - 166"},{"Id":20025,"Name":"Alfa Romeo - 33"},{"Id":20026,"Name":"Alfa Romeo - Brera"},{"Id":20027,"Name":"Alfa Romeo - GT"},{"Id":20028,"Name":"Alfa Romeo - GTV"},{"Id":239,"Name":"Alfa Romeo - Misc"},{"Id":20029,"Name":"Alfa Romeo - Spider"},{"Id":20030,"Name":"Alfa Romeo - Sportwagon"}]} 
'; 

$json_dec = json_decode($json); 

$nr = 0; 
while($nr 14) { 
$id_nr = $json_dec['d']['$nr']['Id']; 
$make = $json_dec['d']['$nr']['Name']; 
} 

，我得到的錯誤是

Fatal error: Cannot use object of type stdClass as array in 
C:\wamp\www\get_cat.php on line 18 

Answer 1

把true作爲第二個參數用於json_decode

$json_dec = json_decode($json,true); 

否則，它會返回一個對象，而不是一個數組。

在這裏看到更多的信息：http://php.net/manual/en/function.json-decode.php