PHP/JSON 解析

Question

Array
(
    [0] => Array
    (
        [faq_id] => 14
        [faq_faqcat_id] => 3
        [faq_order] => 8
        [faq_subject] => 800018
        [faq_content] => 800019
        [faq_datecreated] => 1213382678
        [faq_dateupdated] => 1215547984
        [faq_views] => 22
    )

    [1] => Array
    (
        [faq_id] => 15
        [faq_faqcat_id] => 3
        [faq_order] => 9
        [faq_subject] => 800020
        [faq_content] => 800021
        [faq_datecreated] => 1213382698
        [faq_dateupdated] => 1215547986
        [faq_views] => 13
    )

    [2] => Array
    (
        [faq_id] => 16
        [faq_faqcat_id] => 3
        [faq_order] => 10
        [faq_subject] => 800022
        [faq_content] => 800023
        [faq_datecreated] => 1213382711
        [faq_dateupdated] => 1215547988
        [faq_views] => 15
    )
)

以上数据是 PHP 脚本的输出。我必须将其转换为 JSON 对象并在 iPhone 上显示。谁能告诉我怎么做？

Answer 1

看看json_encode和 json_decode职能

对于 iOS 上的解析部分，您可以使用 native NSJSONSerialization类(class)。