使用Java API / util查找不安全字符并将其替换为百分比编码形式？

Question

有谁知道像样的Java API / util来用其百分比编码形式查找和替换URL中的不安全字符？

Answer 1

"http://google.com?" + URLEncoder.encode("...", "UTF-8");

See javadocs .

One should add the expected character encoding.

Answer 2

See Java - Convert String to valid URI object for pure Java style:

String urlStr = "http://abc.dev.domain.com/0007AC/ads/800x480 15sec h.264.mp4";
URL url = new URL(urlStr);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();

Beware that it handles protocols available to Java natively. Otherwise you may need to register your own one like "s3" for Amazon S3 URIs.