Java URI 类 : constructor determines whether or not query is encoded?

Question

这种行为是故意的吗？

//create the same URI using two different constructors

URI foo = null, bar = null;
try { 
    //constructor: URI(uri string)
    foo = new URI("http://localhost/index.php?token=4%2F4EzdsSBg_4vX6D5pzvdsMLDoyItB");
} catch (URISyntaxException e) {} 
try { 
    //constructor: URI(scheme, authority, path, query, fragment) 
    bar = new URI("http", "localhost", "/index.php", "token=4%2F4EzdsSBg_4vX6D5pzvdsMLDoyItB", null);
} catch (URISyntaxException e) {}

//the output:
//foo.getQuery() = token=4/4EzdsSBg_4vX6D5pzvdsMLDoyItB
//bar.getQuery() = token=4%2F4EzdsSBg_4vX6D5pzvdsMLDoyItB

URI(string uri) 构造函数似乎正在解码 URI 的查询部分。我认为查询部分应该被编码？为什么另一个构造函数不解码查询部分？

Answer 1

来自 URI JavaDoc :

The single-argument constructor requires any illegal characters in its argument to be quoted and preserves any escaped octets and other characters that are present.

The multi-argument constructors quote illegal characters as required by the components in which they appear. The percent character ('%') is always quoted by these constructors. Any other characters are preserved.

因此 URI(String) 希望您正确编码所有内容，并假设 %2F 是这样一个编码的八位字节，它将被解码为 /。

其他构造函数会对 % 字符进行结束编码(对于输入 %2F 产生 %252F)，因此在解码后您仍然会得到 %2F。

我假设构造函数之间存在偏差的目的是允许诸如 new URI(otherUri.toString()) 和 toString() 返回完全编码的 URI 之类的内容。