在Javascript中将URL作为GET参数传递

Question

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

javascript:void(location.href='http://tinyurl.com/create.php?url='+location.href)

So I copied that same thing and made

javascript:void(location.href='http://mywebsite.com/create.php?url='+location.href)

Then I use:

$url=$_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.

Example, If I am on:

http://www.google.ca/webhp?um=1&hl=en&safe=off

The '_GET' code sets $url to be

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

Answer 1

URL has 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 & :

key1=value1&key2=value2&key3=value3&key4=value4

You should escape location.href as it can contains some special characters like ? , & or # .

To escape string in JavaScript use encodeURIComponent() function like so:

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.

Answer 2

尝试

javascript:void(location.href='http://mywebsite.com/create.php?url='+encodeURIComponent(location.href));

Answer 3

To pass url in $_GET parameter like this:

http://yoursite.com/page.php?myurl=http://google.com/bla.php?sdsd=123&dsada=4323

then you need to use encode function:

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:

$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.

Answer 4

javascript:void(location.href='http://mywebsite.com/create.php?url='+encodeURIComponent(location.href))

你需要逃避角色。

Answer 5

so what are you wanting, just the url without the query string?

$url = explode('?',$_GET['url']);
$url = $url[0];