Url URISyntaxException-如何处理带有%

我对Java相当陌生，遇到了这个问题。我试着搜索，但始终没有得到正确的答案

例如，我有一个字符串

String name = anything 10%-20% 04-03-07

现在我需要用这个字符串名构建一个url字符串，如下所示

http://something.com/test/anything 10%-20% 04-03-07

我尝试用%20替换空格，现在我得到的新url为

http://something.com/test/anything%2010%-20%%2004-03-07

当我使用这个url并在firefox中启动它时，它工作得很好，但在Java中处理时，它显然是在抛出

Exception in thread "main" java.lang.IllegalArgumentException
at java.net.URI.create(Unknown Source)
at org.apache.http.client.methods.HttpGet.<init>(HttpGet.java:69)
Caused by: java.net.URISyntaxException: Malformed escape pair at index 39 : 
at java.net.URI$Parser.fail(Unknown Source)
at java.net.URI$Parser.scanEscape(Unknown Source)
at java.net.URI$Parser.scan(Unknown Source)
at java.net.URI$Parser.checkChars(Unknown Source)
at java.net.URI$Parser.parseHierarchical(Unknown Source)
at java.net.URI$Parser.parse(Unknown Source)
at java.net.URI.<init>(Unknown Source)
... 6 more

---

用

%25

对百分号进行编码

http://something.com/test/anything 10%-20%04-03-07

将与

http://something.com/test/anything%2010%25-20%25%2004-03-07

您应该能够使用这个示例—请记住，您需要对路径部分进行URL编码，而不是之前的任何内容，例如

String encodedUrl =
    String.format("http://something.com/%s/%s",
      URLEncoder.encode("test", "UTF-8"),
      URLEncoder.encode("anything 10%-20% 04-03-07", "UTF-8")
    );

注意：URLEncoder将空格编码为

+

而不是

%20

，但它应该同样有效，两者都可以。

您可以使用它从字符串创建uri

String url = "http://something.com/test/anything 10%-20% 04-03-07"

URI uri = new URI(
    url,
    null);
String request = uri.toASCIIString();

HttpClient httpclient = new DefaultHttpClient();
HttpGet httpget = new HttpGet(request);
HttpResponse response = httpclient.execute(httpget);

---

谢谢，这很好用。我正试图逃避它。我没有从替代的角度思考。

String url = "http://something.com/test/anything 10%-20% 04-03-07"

URI uri = new URI(
    url,
    null);
String request = uri.toASCIIString();

HttpClient httpclient = new DefaultHttpClient();
HttpGet httpget = new HttpGet(request);
HttpResponse response = httpclient.execute(httpget);