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Spring 弹簧组件生成器_Spring_Uri - Fatal编程技术网
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Spring 弹簧组件生成器

spring

Spring 弹簧组件生成器,spring,uri,Spring,Uri,我有一个包含编码部分和变量的字符串 https://example.com/foo%2Fbar/{variable} 我正在尝试创建一个URI对象，如下所示： String s = "https://example.com/foo%2Fbar/{0}" URI uri = UriComponentsBuilder.fromHttpUrl(s) .buildAndExpand("baz") .t

我有一个包含编码部分和变量的字符串

https://example.com/foo%2Fbar/{variable}

我正在尝试创建一个URI对象，如下所示：

String s = "https://example.com/foo%2Fbar/{0}"
URI uri = UriComponentsBuilder.fromHttpUrl(s)
                .buildAndExpand("baz")
                .toUri();
结果是
https://example.com/foo%252Fbar/baz
但我想得到
https://example.com/foo%2Fbar/baz

如何确保变量已展开，但避免再次对已编码部分进行编码