我在期待

System.out.println(java.net.URLEncoder.encode("Hello World", "UTF-8"));

输出:

你好%20世界

(20是ASCII十六进制空格码)

然而,我得到的是:

你好+世界

我用错方法了吗?我应该使用的正确方法是什么?


当前回答

查看uri类。

其他回答

该类执行application/x-www-form- urlenencoded -type编码,而不是百分比编码,因此替换为+是正确的行为。

从javadoc:

When encoding a String, the following rules apply: The alphanumeric characters "a" through "z", "A" through "Z" and "0" through "9" remain the same. The special characters ".", "-", "*", and "_" remain the same. The space character " " is converted into a plus sign "+". All other characters are unsafe and are first converted into one or more bytes using some encoding scheme. Then each byte is represented by the 3-character string "%xy", where xy is the two-digit hexadecimal representation of the byte. The recommended encoding scheme to use is UTF-8. However, for compatibility reasons, if an encoding is not specified, then the default encoding of the platform is used.

URLEncoder使用字符集“ISO-8859-1”

只是在Android上也在挣扎,设法偶然发现了Uri。encode(字符串,字符串)而特定于android (android.net.Uri)可能对一些有用。

静态字符串编码(字符串s,字符串允许)

以https://developer.android.com/reference/android/net/Uri.html编码(以)

查询参数

org.apache.commons.httpclient.util.URIUtil
    URIUtil.encodeQuery(input);

如果你想转义URI中的字符

public static String escapeURIPathParam(String input) {
  StringBuilder resultStr = new StringBuilder();
  for (char ch : input.toCharArray()) {
   if (isUnsafe(ch)) {
    resultStr.append('%');
    resultStr.append(toHex(ch / 16));
    resultStr.append(toHex(ch % 16));
   } else{
    resultStr.append(ch);
   }
  }
  return resultStr.toString();
 }

 private static char toHex(int ch) {
  return (char) (ch < 10 ? '0' + ch : 'A' + ch - 10);
 }

 private static boolean isUnsafe(char ch) {
  if (ch > 128 || ch < 0)
   return true;
  return " %$&+,/:;=?@<>#%".indexOf(ch) >= 0;
 }

如果您正在使用jetty,那么org.eclipse.jetty.util.URIUtil将解决这个问题。

String encoded_string = URIUtil.encodePath(not_encoded_string).toString();