javascript – 使用Regex进行Google Maps链接验证

我只是想检查该链接是否是谷歌地图链接

例如 ：

var urls =[
    /// correct urls
    "https://www.google.com/maps",
    "https://www./maps",
    "https:///maps",
    "http:///maps",
    "https://www./maps",
    "https://www.google.com/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4",
    "https://www./maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4",
    "https://www.google.com/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4&layer=t&lci=com.panoramio.all,com.google.webcams,weather",
    "https://www.google.com/maps?ll=37.370157,0.615234&spn=45.047033,93.076172&t=m&z=4&layer=t",

    "https://www.google.com/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4",
    "https://www./maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4",
    "https://www.google.com/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4&layer=t&lci=com.panoramio.all,com.google.webcams,weather",
    "https://www.google.com/maps?ll=37.370157,0.615234&spn=45.047033,93.076172&t=m&z=4&layer=t",

    /// error urls
    "https://www.google.com/",
    "https://zzz.google.com/maps",
    "https://www.google.com/ ",
    "httpsxyz://www.google.com/maps",
    "http://www./ "
    ];

我对Regex非常不满,并尝试使用JavaScript Regex Generator,但它对我来说仍然很难.

我只得到..

Reg = /^http\:\/\/|https\:\/\/|www\.google$/;

它失败了！ 🙁

但是我为你们做了一个现场测试：http:///onuyux/1/edit?javascript,live

编辑：2

在我从@joel harkes&获得帮助之后@ dan1111所以我得到了正则表达式

/^https?\:\/\/(www\.)?google\.[a-z] \/maps\b/

这只是谷歌的正则表达式.{TLD} /地图那么maps.google.{TLD}呢？

我只是想通过这种方式验证Google Maps网址和网址(查看图片)

如果可能,我想验证是否设置了地址或长拉(使用li或q参数检查？)(例如,不仅仅是“maps.google.com”..)

更新列表代码：

/// correct urls
"https://www.google.com/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4",
"https://www./maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4",
"https://www.google.com/maps?ll=37.0625,-95.677068&spn=45.197878,93.076172&t=h&z=4&layer=t&lci=com.panoramio.all,com.google.webcams,weather",
"https://www.google.com/maps?ll=37.370157,0.615234&spn=45.047033,93.076172&t=m&z=4&layer=t",
"https://maps.google.com/maps?q=New York, NY, USA&hl=no&sll=19.808054,-63.720703&sspn=54.337928,93.076172&oq=n&hnear=New York&t=m&z=10",
"https://www.google.com/maps?q=New York, New York, USA&hl=no&ll=40.683762,-73.925629&spn=0.708146,1.454315&sll=41.47566,-72.026367&sspn=11.190693,23.269043&oq=new&hnear=New York&t=m&z=10",

/// error urls
"https://www.google.com/",
"https://zzz.google.com/maps",
"https://www.google.com/ ",
"httpsxyz://www.google.com/maps",
"http://www./ ",
"http://maps.google.com/",
"http://google.com/maps",
"http:///maps",
"?q=New York, New York, USA&hl=no&ll=40.683762,-73.925629&spn=0.708146,1.454315&sll=41.47566,-72.026367&sspn=11.190693,23.269043&oq=new&hnear=New York&t=m&z=10",
"&sll=41.47566,-72.026367&sspn=11.190693,23.269043&oq=new&hnear=New York&t=m&z=10"

现场测试：http:///onuyux/25/edit?javascript,live

解决方法:

完成dan1111的答案这里的模式将匹配指定的域并检查是否设置了ll GET变量：

Reg = /^https?\:\/\/(www\.)?google\.[a-z] \/maps\/?\?([^&] &)*(ll=-?[0-9]{1,2}\.[0-9] ,-?[0-9]{1,2}\.[0-9] |q=[^& ]) ($|&)/;

我无法测试它(jsbin已关闭)但它应匹配任何包含ll或q参数中的至少一个的url.

编辑：

我添加了地图.子域名并修正了一个小错字.在q =之后,应该在字符类之外,让它重复：q = [^&]而不是q = [^& amp; ].
话虽这么说,这是你的正则表达式：

Reg = /^https?\:\/\/(www\.|maps\.)?google\.[a-z] \/maps\/?\?([^&] &)*(ll=-?[0-9]{1,2}\.[0-9] ,-?[0-9]{1,2}\.[0-9] |q=[^&] ) ($|&)/;

EDIT2：

要接受google.co.uk风格域名,请使用以下正则表达式.

Reg = /^https?\:\/\/(www\.|maps\.)?google(\.[a-z] ){1,2}\/maps\/?\?([^&] &)*(ll=-?[0-9]{1,2}\.[0-9] ,-?[0-9]{1,2}\.[0-9] |q=[^&] ) ($|&)/;

希望它现在涵盖所有网址.

来源：https://www./content-1-355151.html