Question

此问题与提及的重复答案/问题完全无关......

我有一个包含JavaScript的超过1,000个项目的大型列表我这种格式...

var plugins = [
    {
        name: "Roundabout - Interactive, turntable-like areas",
        url: "http:\/\/fredhq.com/projects/roundabout/",
        tag: "slide"
    },
    {
        name: "Slides - Simple slideshow plugin for jQuery",
        url: "http://slidesjs.com/",
        tag: "slide"
    }

    ......lots more
]

我的目标是将它们导入到像这样的PHP数组中....

Array
(
    [0] => Array
        (
            [name] => Roundabout - Interactive, turntable-like areas/
            [url] => http://fredhq.com/projects/roundabout/
            [tag] => slide
        )

    [1] => Array
        (
            [name] => Slides - Simple slideshow plugin for jQuery
            [url] => http://slidesjs.com/
            [tag] => slide
        )

    [2] => Array
        (
            [name] => Orbit - jQuery Image Slider Plugin
            [url] => http://www.zurb.com/playground/jquery_image_slider_plugin/
            [tag] => slide
        )

)

我尝试使用PHP的json_decode()函数，但源JS没有包含在"中的密钥，所以它不起作用。

当我在键和一些反斜杠周围手动添加"时，PHP json_decode()正常工作....

$json = '[{
    "name": "Roundabout - Interactive, turntable-like areas/",
    "url": "http:\/\/fredhq.com/projects/roundabout/",
    "tag": "slide"
},
{
    "name": "Slides - Simple slideshow plugin for jQuery",
    "url": "http:\/\/slidesjs.com\/",
    "tag": "slide"
},
{
    "name": "Orbit - jQuery Image Slider Plugin",
    "url": "http:\/\/www.zurb.com/playground/jquery_image_slider_plugin/",
    "tag": "slide"
}]';

echo '<pre>';
print_r(json_decode($json, true));
echo '</pre>';

问题是我的大型列表没有用引号括起来的键，所以我需要一种方法来自动化它。

有人帮助我实现最终目标，因为我的尝试到目前为止都失败了吗？

Answer 1

看起来问题可以通过采用这种方法手动解决，因为已经说过这里介绍的JSON是有效的，最好使用控制台或其他东西来使JSON对所有地方都有效，不仅仅是在JavaScript中使用：

plugins = JSON.stringify(plugins);

这会使plugins看起来像：

"[{"name":"Roundabout - Interactive, turntable-like areas","url":"http://fredhq.com/projects/roundabout/","tag":"slide"},{"name":"Slides - Simple slideshow plugin for jQuery","url":"http://slidesjs.com/","tag":"slide"}]"

删除引号并将其放入JSONLint将提供有效的JSON，如下所示：

[
    {
        "name": "Roundabout - Interactive, turntable-like areas",
        "url": "http://fredhq.com/projects/roundabout/",
        "tag": "slide"
    },
    {
        "name": "Slides - Simple slideshow plugin for jQuery",
        "url": "http://slidesjs.com/",
        "tag": "slide"
    }
]

现在最好的方法是复制此输出并将其保存在文本中。

将JavaScript Object / JSON转换为PHP数组麻烦

1 个答案: