在 Javascript 中将 URL 作为 GET 参数传递
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Passing a URL as a GET parameter in Javascript
提问by MikeD
I am trying to make a bookmarklet that uses the user's current URL, kind of like the tinyURL bookmarklet that uses this javascript code
我正在尝试制作一个使用用户当前 URL 的书签，有点像使用此 javascript 代码的 tinyURL 书签
javascript:void(location.href='http://tinyurl.com/create.php?url='+location.href)
So I copied that same thing and made
所以我复制了同样的东西并制作了
javascript:void(location.href='http://mywebsite.com/create.php?url='+location.href)
Then I use:
然后我使用：
$url=$_GET['url'];
to retrieve it. The problem is, if I am on a url that already has some get style info in the url, it messes everything up.
取回它。问题是，如果我在一个已经在 url 中有一些 get 样式信息的 url 上，它会把一切搞砸。
Example, If I am on:
例如，如果我在：
http://www.google.ca/webhp?um=1&hl=en&safe=off
The '_GET' code sets $url to be
'_GET' 代码将 $url 设置为
http://www.google.ca/webhp?um=1
So I think the info in google URL is messing up all of my URL parsing, I imagine I am doing something very incorrectly or someone has a very elegant solution for this. What should I do? please help
所以我认为 google URL 中的信息弄乱了我所有的 URL 解析，我想我做的事情非常不正确，或者有人对此有一个非常优雅的解决方案。我该怎么办？请帮忙
回答by Crozin
URLhas a specified format. That part after ?, or to be more exactly between ?and #if exists, is called query string. It contains a list of key-value pairs - a variable name, =character and the value. Variables are separated by &:
URL具有指定的格式。之后的部分?，或者更确切地说是介于?和#如果存在，称为查询字符串。它包含一个键值对列表——一个变量名、=字符和值。变量由 分隔&：
key1=value1&key2=value2&key3=value3&key4=value4
You should escape location.hrefas it cancontains some special characters like ?, &or #.
您应该转义，location.href因为它可能包含一些特殊字符，例如?,&或#。
To escape string in JavaScript use encodeURIComponent()function like so:
要在 JavaScript 中转义字符串，请使用如下encodeURIComponent()函数：
location.href = "http://tinyurl.com/create.php?url=" + encodeURIComponent(location.href)
It will replace characters like &into %26. That sequence of characters isn't treated as a variable separator so it will be attached as a variable's value.
它将替换像&into 之类的字符%26。该字符序列不被视为变量分隔符，因此它将作为变量值附加。
回答by Greg
Try
尝试
javascript:void(location.href='http://mywebsite.com/create.php?url='+encodeURIComponent(location.href));
回答by Crayon Violent
so what are you wanting, just the url without the query string?
所以你想要什么，只是没有查询字符串的网址？
$url = explode('?',$_GET['url']);
$url = $url[0];
回答by Alec Gorge
javascript:void(location.href='http://mywebsite.com/create.php?url='+encodeURIComponent(location.href))
You need to escape the characters.
您需要对字符进行转义。
回答by T.Todua
To pass url in $_GET parameter like this:
要像这样在 $_GET 参数中传递 url：
http://yoursite.com/page.php?myurl=http://google.com/bla.php?sdsd=123&dsada=4323
then you need to use encode function:
那么你需要使用编码功能：
echo 'http://yoursite.com/page.php?url='.urlencode($mylink);
//so, your output (url parameter) will get like this
//http://yoursite.com/page.php?url=http%3A%2F%2Fgoogle.com%2Flink.php%3Fname%3Dsta%26car%3Dsaab
after that, you need to decode the parameter with:
之后，您需要使用以下方法解码参数：
$variab = $_GET['url'];
$variab = preg_replace("/%u([0-9a-f]{3,4})/i","&#x\1;",urldecode($variab));
$variab = html_entity_decode($variab,null,'UTF-8');
echo $variab;
such way, you can pass the correct link as a parameter.
这样，您可以将正确的链接作为参数传递。
