RestTemplate to NOT escape url

Question

我正在成功使用Spring RestTemplate:

String url = "http://example.com/path/to/my/thing/{parameter}";
ResponseEntity<MyClass> response = restTemplate.postForEntity(url, payload, MyClass.class, parameter);

这很好.

但是,有时候parameter是%2F.我知道这不是理想的,但事实就是如此.正确的URL应该是:http://example.com/path/to/my/thing/%2F但是当我设置parameter为"%2F"它时,双重转义为http://example.com/path/to/my/thing/%252F.我该如何防止这种情况？

Answer 1

而不是使用StringURL,用a 构建URI一个UriComponentsBuilder.

String url = "http://example.com/path/to/my/thing/";
String parameter = "%2F";
UriComponentsBuilder builder = UriComponentsBuilder.fromHttpUrl(url).path(parameter);
UriComponents components = builder.build(true);
URI uri = components.toUri();
System.out.println(uri); // prints "http://example.com/path/to/my/thing/%2F"

使用UriComponentsBuilder#build(boolean)指示

是否所有在此构建器中设置的组件都是encoded(true)或不是(false)

这或多或少等同于自己替换{parameter}和创建URI对象.

String url = "http://example.com/path/to/my/thing/{parameter}";
url = url.replace("{parameter}", "%2F");
URI uri = new URI(url);
System.out.println(uri);

然后,您可以将此URI对象用作方法的第一个参数postForObject.

Answer 2

您可以告诉其余模板您已经编码了uri.这可以使用UriComponentsBuilder.build(true)完成.这样休息模板将不会重新尝试逃避uri.其余大多数模板api都会接受URI作为第一个参数.

String url = "http://example.com/path/to/my/thing/{parameter}";
url = url.replace("{parameter}", "%2F");
UriComponentsBuilder builder = UriComponentsBuilder.fromUriString(url);
// Indicate that the components are already escaped
URI uri = builder.build(true).toUri();
ResponseEntity<MyClass> response = restTemplate.postForEntity(uri, payload, MyClass.class, parameter);