我知道我的JSON是有效的,我想要將所有的KEY從數組中取出並放入一個對象中。但是,似乎我可以訪問一個對象鍵或值,整個數組或一個鍵值對。我還沒有想出如何解析出所有的鍵或數組中的所有值。從一個有效的JSON數組創建一個對象用PHP
這裏是我試過:
的print_r($ json_obj)產量:
Array ([0] => Array ([0] => uploads/featured/doublewm-4097.jpg [1] => featured) [1] => Array ([0] => uploads/featured/moon-5469.jpg [1] => featured))
的print_r($ json_obj [0] [1])產量:
featured
print_r($ json_obj [1] [0])得出:
uploads/featured/moon-5469.jpg
的print_r($ json_obj [1] [1])yeilds:
featured
的print_r($ json_obj [0] [0])的產率:
uploads/featured/doublewm-4097.jpg
PHP代碼:
<?php
$resultSet = '[["uploads/featured/doublewm-4097.jpg","featured"],
["uploads/featured/moon-5469.jpg","featured"]]';
$json_obj = json_decode($resultSet);
// print_r($json_obj);
print_r($json_obj[0][1]);
?>
JSON驗證每個JSONLint
[
[
"uploads/featured/doublewm-4097.jpg",
"featured"
],
[
"uploads/featured/moon-5469.jpg",
"featured"
]
]
我想結束智慧公頃,在json_obj所有鍵...即對象:
json_obj = array(
'uploads/featured/moon-5469.jpg',
'uploads/featured/doublewm-4097.jpg'
);
究竟什麼是你的問題? 'json_decode'正確解碼給定的JSON - 如果你想要一些不同的東西,你可以輕鬆地遍歷數組並創建新的結果。 – kero
您的輸入和預期輸出是什麼?講清楚。 – xdazz
我的輸入被硬編碼到resultSet =「」; 我的預期輸出是來自JSON的文件路徑(鍵)的數組....類似於json_obj = array();示例在結尾 – user3161804