Question

我有一个数据流作为JSON，我想解析它并希望将结果保存在ini文件中：

{
    "books": [{
        "id": "1",
        "date": "2017-03-12",
        "date_text": "sunday 12 march",
        "title": "title text"
    }, {
        "id": "2",
        "date": "2017-03-12",
        "date_text": "sunday 12 march",
        "title": "title text"
    }]
}

这是我的示例数据，我想知道是否有办法将其保存到文件中，无论它是否包含1个或更多＆＃34; id：s＆＃34; （项）我知道如何解析JSON，而不是如何将它保存为正确格式的文件。

优惠格式：

[Books 0]
id= 1
date= 2017-03-12
date_text=sunday 12 march
title= title text

[Books 1]
id"=2
date=2017-03-12
date_text=sunday 12 march
title=title text

Answer 1

您可以尝试Zend Config执行此任务。打开终端并将zend-config作为依赖项添加到项目中（假设您已经使用composer）：

composer require zendframework/zend-config

现在你可以试试，

$json = <<<JSON
{
    "books": [{
        "id": "1",
        "date": "2017-03-12",
        "date_text": "sunday 12 march",
        "title": "title text"
    }, {
        "id": "2",
        "date": "2017-03-12",
        "date_text": "sunday 12 march",
        "title": "title text"
    }]
}
JSON;

$config = new \Zend\Config\Config(json_decode($json, true), true);
$writer = new \Zend\Config\Writer\Ini();
echo $writer->toString($config);

输出将是：

[books]
0.id = "1"
0.date = "2017-03-12"
0.date_text = "sunday 12 march"
0.title = "title text"
1.id = "2"
1.date = "2017-03-12"
1.date_text = "sunday 12 march"
1.title = "title text"

您的JSON格式应如下所示，以生成您所写的所需输出：

{
    "books 0": {
        "id": "1",
        "date": "2017-03-12",
        "date_text": "sunday 12 march",
        "title": "title text"
    },
    "books 1" : {
        "id": "2",
        "date": "2017-03-12",
        "date_text": "sunday 12 march",
        "title": "title text"
    }
}

如何用PHP将JSON保存到ini文件中？

1 个答案: