我有一个网址,其中我有很少的宏,我不想编码,但剩下的部分应该是。例如 -
https://example.net/adaf/${ABC}/asd/${WSC}/
应编码为
https%3A%2F%2Fexample.net%2Fadaf%2F${ABC}%2Fasd%2F${WSC}%2F
URLEncoder.encode(string, encoding)
编码整个字符串。我需要一种类型的功能 - encode(string, start, end, encoding)
有没有现有的图书馆这样做?
AFAIK没有标准库提供这样的重载方法。但是您可以围绕标准API构建自定义包装函数。
为了实现您的目标,代码可能如下所示:
public static void main(String[] args) throws UnsupportedEncodingException {
String source = "https://example.net/adaf/${ABC}/asd/${WSC}/";
String target = "https%3A%2F%2Fexample.net%2Fadaf%2F${ABC}%2Fasd%2F${WSC}%2F";
String encodedUrl = encode(source, 0, 25, StandardCharsets.UTF_8.name()) +
source.substring(25, 31) +
encode(source, 31, 36, StandardCharsets.UTF_8.name()) +
source.substring(36, 42) +
encode(source, 42, 43, StandardCharsets.UTF_8.name());
System.out.println(encodedUrl);
System.out.println(encodedUrl.equals(target));
}
static String encode(String s, int start, int end, String encoding) throws UnsupportedEncodingException {
return URLEncoder.encode(s.substring(start, end), StandardCharsets.UTF_8.name());
}
https%3A%2F%2Fexample.net%2Fadaf%2F${ABC}%2Fasd%2F${WSC}%2F
true
但这将是非常混乱的。
作为替代方案,您可以在编码后简单地替换您不想使用其原始值转义的字符集:
public static void main(String[] args) throws UnsupportedEncodingException {
String source = "https://example.net/adaf/${ABC}/asd/${WSC}/";
String target = "https%3A%2F%2Fexample.net%2Fadaf%2F${ABC}%2Fasd%2F${WSC}%2F";
String encodedUrl = URLEncoder.encode(source, StandardCharsets.UTF_8.name())
.replaceAll("%24", "\\$")
.replaceAll("%7B", "{")
.replaceAll("%7D", "}");
System.out.println(encodedUrl);
System.out.println(encodedUrl.equals(target));
}
https%3A%2F%2Fexample.net%2Fadaf%2F${ABC}%2Fasd%2F${WSC}%2F
true