只要把你的有效的JSON內json_decode並將其分配給一個PHP數組如下
//$php_arr = json_decode('YOUR_JSON');
$php_arr = json_decode('[{"id":13},{"id":14},{"id":15,"children":[{"id":16},{"id":17},{"id":18}]},{"id":19},{"id":20},{"id":21,"children":[{"id":22}]}]');
/*comment the following 3 lines when done*/
echo "<pre>";
print_r($php_arr);
echo "</pre>";
/*comment the above 3 lines when done*/
輸出
Array
(
[0] => stdClass Object
(
[id] => 13
)
[1] => stdClass Object
(
[id] => 14
)
[2] => stdClass Object
(
[id] => 15
[children] => Array
(
[0] => stdClass Object
(
[id] => 16
)
[1] => stdClass Object
(
[id] => 17
)
[2] => stdClass Object
(
[id] => 18
)
)
)
[3] => stdClass Object
(
[id] => 19
)
[4] => stdClass Object
(
[id] => 20
)
[5] => stdClass Object
(
[id] => 21
[children] => Array
(
[0] => stdClass Object
(
[id] => 22
)
)
)
)
現在你有PHP數組做什麼你想要什麼
foreach($php_arr as $arr){
if(!isset($arr['children'])){
$q = "insert into tbl(id) values('".$arr['id']."')";
}else{
//your logic
}
}
我認爲他想他的數據存儲在MySQL。無法訪問它。 – aldrin27
我編輯我的回答告訴他,他可以訪問數據,並用它在迭代'foreach'插入它(管理費用,但我不知道他想用它做什麼)。感謝您的反饋! –