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Java 如何从URI中删除片段_Java - Fatal编程技术网

Java 如何从URI中删除片段

Java 如何从URI中删除片段,java,Java,我需要关于URI API的建议 我是这样做的URI new URI("http", "", "localhost", 9000, "/action", "param=1","") res0: java.net.URI = http://@localhost:9000/action?param=1# 正如您所看到的,它将@放在localhost前面,我也看到URI的末尾。我怎样才能摆脱这些?他们为什么要来 URI构造函数的第二个参数是userInfo。。。将其设置为null,而不是空字符串 ne

我需要关于URI API的建议

我是这样做的URI

new URI("http", "", "localhost", 9000, "/action", "param=1","")
res0: java.net.URI = http://@localhost:9000/action?param=1#

正如您所看到的,它将@放在localhost前面,我也看到URI的末尾。我怎样才能摆脱这些?他们为什么要来

URI构造函数的第二个参数是userInfo。。。将其设置为null,而不是空字符串

new URI("http", null, "localhost", 9000, "/action", "param=1", null)

您正在设置userInfo和片段,它们分别用“@”和“@”正确分隔

URI u = new URI("https", null, "localhost", 8080, "/myapi", "q=all", null);
System.out.print(u);
给予

https://localhost:8080/myapi?q=all

也许取消最后一个attr也能解决这个问题?是的,我试过这个建议,你是对的。非常感谢。