urlencode

From Whats the java equivalent of Python’s urllib.urlencode? Like >>> urllib.urlencode({'abc':'d f', 'def': '-!2'}) 'abc=d+f&def=-%212' Where I can pass a HashMap of key values and it encodes and gives me the url string .. Edit: I wanted to avoid this scenario String data = URLEncoder.encode("key1", "UTF-8") + "=" + URLEncoder.encode("value1", "UTF-8"); data += "&" + URLEncoder.encode("key2", "UTF-8") + "=" + URLEncoder.encode("value2", "UTF-8"); of manually concatenating the strings with = and & java.net.URLEncoder should work for you - though you would have to extend it to accept the hashmap

This works fine: GET /mvc/Movies/TitleIncludes/Lara%20Croft When I submit a request that contains a colon, like this: GET /mvc/Movies/TitleIncludes/Lara%20Croft:%20Tomb ...it generates a 400 error. The error says ASP.NET detected invalid characters in the URL. If I try url-escaping, the request looks like this: GET /mvc/Movies/TitleIncludes/Lara%20Croft%3A%20Tomb ...and this also gives me a 400 error. If I replace the colon with a | : GET /mvc/Movies/TitleIncludes/Lara%20Croft|%20Tomb ..that was also rejeted as illegal, this time with a 500 error. The message: Illegal characters in path. URL

I have an ASP.NET MVC 4 Web Application running on IIS 7.5 for a few weeks now and I noticed that just recently I have been getting quite a lot of the following errors: System.Web.HttpException (0x80004005): A potentially dangerous Request.Path value was detected from the client (?). at System.Web.HttpRequest.ValidateInputIfRequiredByConfig() at System.Web.HttpApplication.PipelineStepManager.ValidateHelper(HttpContext context) I know that this exception is caused when the URL path contains illegal characters but the strange thing is that in my case the URL is valid. Here is one example of the

Here is a test application that writes out XML Files. Why are the spaces in my path being converted to %20 ? public class XmlTest { public static void main(String[] args) { String filename = "C:\\New Folder\\test.xml"; try { DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance(); DocumentBuilder docBuilder = docFactory.newDocumentBuilder(); Document doc = docBuilder.newDocument(); TransformerFactory transformerFactory = TransformerFactory.newInstance(); Transformer transformer = transformerFactory.newTransformer(); transformer.setOutputProperty(OutputKeys.INDENT, "yes");

I have a strange issue with urlencoding a plus sign + as a query param for a request against an API. The API's documentation states: The date has to be in the W3C format, e.g. '2016-10-24T13:33:23+02:00'. So far so good, so I'm using this code (minimalized) to generate the url, using Spring's UriComponentBuilder: DateTimeFormatter formatter = DateTimeFormatter.ofPattern("yyyy-MM-dd'T'HH:mm:ssX"); ZonedDateTime dateTime = ZonedDateTime.now().minusDays(1); String formated = dateTime.format(formatter); UriComponentsBuilder uriComponentsBuilder = UriComponentsBuilder.fromUriString(baseUrl);

当url地址含有中文，或者参数有中文的时候，这个算是很正常了，但是把这样的url作为参数传递的时候（最常见的callback），需要把一些中文甚至'/'做一下编码转换。 一、urlencode urllib库里面有个urlencode函数，可以把key-value这样的键值对转换成我们想要的格式，返回的是a=1&b=2这样的字符串，比如： >>> from urllib import urlencode >>> data = { ... 'a': 'test', ... 'name': '魔兽' ... } >>> print urlencode(data) a=test&name=%C4%A7%CA%DE 如果只想对一个字符串进行urlencode转换，怎么办？urllib提供另外一个函数：quote() >>> from urllib import quote >>> quote('魔兽') '%C4%A7%CA%DE' 二、urldecode 当urlencode之后的字符串传递过来之后，接受完毕就要解码了——urldecode。urllib提供了unquote()这个函数，可没有urldecode()！ >>> from urllib import unquote >>> unquote('%C4%A7%CA%DE') '\xc4\xa7\xca\xde' >>>