urlencode

While learning the concept of Cookies in PHP, I come across the following statement from w3schools PHP Tutorial : The value of the cookie is automatically URLencoded when sending the cookie, and automatically decoded when received (to prevent URLencoding, use setrawcookie() instead) I did not get the meaning of this statement. I have following doubts regarding the above statement : From where the cookie is being sent to whom and at where the cookie is being received from whom? What actually does happen by means of "Value of the cookie is automatically URLencoded when sending the cookie, and

目录 转换规则 那哪些字符是需要转化的呢？ 和编码无关 结论 参考 转换规则 urlencode：返回字符串，此字符串中除了 -_. 之外的所有非字母数字字符都将被替换成百分号（ % ）后跟 两位十六进制数 ，空格则编码为加号（ + ） 按照每个字符对应的字符编码，不是符合我们范围的，统统的转化为%的形式也就是了。自然也是16进制的形式。 那哪些字符是需要转化的呢？ ASCII 的控制字符 这些字符都是不可打印的，自然需要进行转化。 一些非ASCII字符 这些字符自然是非法的字符范围。转化也是理所当然的了。 一些保留字符 很明显最常见的就是“&”了，这个如果出现在url中了，那你认为是url中的一个字符呢，还是特殊的参数分割用的呢？ 就是一些不安全的字符了。 例如：空格。为了防止引起歧义，需要被转化为“+”。 和编码无关 通过urlencode的转化规则和目的，我们也很容易的看出，urleocode是基于 字符编码 的。同样的一个汉字，不同的编码类型，肯定对应不同的urleocode的串。gbk编码的有gbk的encode结果。 apache等服务器，接受到字符串后，可以进行decode，但是还是无法解决编码的问题。编码问题，还是需要靠约定或者字符编码的判断解决。 因此，urleocode只是为了url中一些非ascii字符，可以正确无误的被传输，至于使用哪种编码

I'm trying to do a somewhat simple thing with no luck - I want to display Hebrew/Arabic characters in my URL. For example: I want the URL to display a file named: aאm.php So I've percent encoded the middle UTF8 characters and the result is: a%D7%90m.php . I've uploaded a%D7%90m.php to my server (Apache) and tried to request the pages www.example.com/a%D7%90m.php & www.example.com/aאm.php but my server responded: The requested URL /a%D7%90m.php was not found on this server. So I tried to upload aאm.php (without the percent encoding) instead, but again no luck when browsing www.example.com/a%D7

I'm running into a problem while providing Hadoop a directory that contains spaces. e.g inputDir = /abc/xyz/folder name/abc.txt Hadoop somehow doesn't know about the "folder name" being the name of the folder with spaces between words. I get the below error while doing that java.io.FileNotFoundException: File does not exist: /abc/xyz/folder Also, I tried providing with URL encoded. java.io.FileNotFoundException: File does not exist: /abc/xyz/folder%20name/abc.txt But still throws me the same error. Does anybody know the workaround for this ? Any help is appreciated. Replacing the space with

I'm running into some issues trying to guess how to handle URLs with parameters on Symfony+Twig. I have this route: <route id="artist.front" path="/artist/{kw}/"> <default key="_controller">App\Web\Controllers\Front::homePage</default> </route> And this code on the Twig template: {% autoescape false %} <a href="{{ path('artist.front',{'kw':a.urlkeyword|url_encode} )}}">{{ a.name }}</a> {% endautoescape %} urlkeyword parameter is passed as is (i.e. not URL encoded previously) both on the entity constructor and on its getter (getURLKeyword). However, when rendering the template I get this HTML

I've got a resource on my server named: some image.png There's a space in the name. When I type the url into the browser (chrome), it's transforming the space into %20: some%20.png WHen I use URLEncoder.encode("some image.png") from my application, I'm getting: some+image.png which causes a 404. What's the right way to encode? http://www.permadi.com/tutorial/urlEncoding/ Note that because the character is very commonly used, a special code ( the "+" sign) has been reserved as its URL encoding. Thus the string "A B" can be URL encoded as either "A%20B" or "A+B" However, your code is probably

I get a url string from a user and would like to transform it to a legal http url: " http://one.two/three?four five " should turn into " http://one.two/three?four%20five " however, URLEncoder does not help, as it encodes the entire string (including the legal "://"). help? With external library: import org.apache.commons.httpclient.util.URIUtil; String myUrl_1= "http://one.two/three?four five"; System.out.println(URIUtil.encodeQuery(myUrl_1)); And the output: http://one.two/three?four%20five Or String webResourceURL = "http://stackoverflow.com/search?q=<script>alert(1)</script> s"; System.out