Php 如何转换谷歌地图'；s"；“嵌入地图”；链接至；共享链接“；

例如，下面是一个例子

如果按“共享链接”按钮，则有两个选项卡“共享链接”和“嵌入地图”

共享链接的数据：

https://www.google.com/maps/place/Lexington,+KY/@38.0279975,-84.751751,10z/data=!3m1!4b1!4m5!3m4!1s0x88424429cc9ceb25:0x84f08341908c4fdd!8m2!3d38.0394389!4d-84.5013428

<iframe src="https://www.google.com/maps/embed?pb=!1m18!1m12!1m3!1d402280.8397603136!2d-84.75175098821046!3d38.02799750322416!2m3!1f0!2f0!3f0!3m2!1i1024!2i768!4f13.1!3m3!1m2!1s0x88424429cc9ceb25%3A0x84f08341908c4fdd!2sLexington%2C+KY!5e0!3m2!1sen!2sus!4v1484200230503" width="600" height="450" frameborder="0" style="border:0" allowfullscreen></iframe>

嵌入地图的数据：

https://www.google.com/maps/place/Lexington,+KY/@38.0279975,-84.751751,10z/data=!3m1!4b1!4m5!3m4!1s0x88424429cc9ceb25:0x84f08341908c4fdd!8m2!3d38.0394389!4d-84.5013428

<iframe src="https://www.google.com/maps/embed?pb=!1m18!1m12!1m3!1d402280.8397603136!2d-84.75175098821046!3d38.02799750322416!2m3!1f0!2f0!3f0!3m2!1i1024!2i768!4f13.1!3m3!1m2!1s0x88424429cc9ceb25%3A0x84f08341908c4fdd!2sLexington%2C+KY!5e0!3m2!1sen!2sus!4v1484200230503" width="600" height="450" frameborder="0" style="border:0" allowfullscreen></iframe>

我只能访问嵌入地图的数据。有没有办法将嵌入地图的iframe的src更改为共享链接的src？有人知道它使用什么算法吗

---

示例：我想更改

”https://www.google.com/maps/embed?pb=！1m18！1m12！1m3！1d402280.8397603136！2d-84.75175098821046！3d38.02799750322416！2m3！1f0！2f0！3m2！1i1024！2i768！4f13.1！3m3！1m2！1s0x88424429cc9ceb25%3A0x84f08341908c4fdd！2Lexington%2C+KY！5e0！3m2！1sen！2sus！4v1484200230503“

"https://www.google.com/maps/place/Lexington，+KY/@38.0279975，-84.751751,10z/数据=！3m1！4b1！4m5！3m4！1s0x88424429cc9ceb25:0x84f08341908c4fdd！8m2！3d38.0394389！4d-84.5013428“by PHP

您需要全部数据吗？因为如果您在浏览器中复制粘贴链接结果，这将起作用

$array = explode('=', $src);
$data = array_filter(explode('!', $array[1]));

$location = $x = $y = '';
foreach($data as $s)
{
    if(substr($s, 0, 2) == "2s" and strlen($s) > 5)
    {
        $location = substr($s, 2);
    }
    elseif(substr($s, 0, 2) == "3d" and strlen($s) > 5)
    {
        $x = substr($s, 2);
    }
    elseif(substr($s, 0, 2) == "2d" and strlen($s) > 5)
    {
        $y = substr($s, 2);
    }
}
if($location != "" and $x != "" and $y != "")
    $result = 'https://www.google.com/maps/place/'.urldecode($location).'/@'.$x.','.$y;
else
{
    // parse error
}

---

虽然不完美，但经过了一点测试，效果很好。strlen是为了防止获取其他变量，因为我不确定它们可能是什么，但模式显示的长度不超过5个字符。

您需要全部数据吗？因为如果您在浏览器中复制粘贴链接结果，这将起作用

$array = explode('=', $src);
$data = array_filter(explode('!', $array[1]));

$location = $x = $y = '';
foreach($data as $s)
{
    if(substr($s, 0, 2) == "2s" and strlen($s) > 5)
    {
        $location = substr($s, 2);
    }
    elseif(substr($s, 0, 2) == "3d" and strlen($s) > 5)
    {
        $x = substr($s, 2);
    }
    elseif(substr($s, 0, 2) == "2d" and strlen($s) > 5)
    {
        $y = substr($s, 2);
    }
}
if($location != "" and $x != "" and $y != "")
    $result = 'https://www.google.com/maps/place/'.urldecode($location).'/@'.$x.','.$y;
else
{
    // parse error
}

虽然不完美，但经过了一点测试，效果很好。strlen是为了防止获得其他变量，因为我不确定它们可能是什么，但模式显示长度不超过5个字符