我有这个:
Background:
* url 'http://localhost:15672/api/exchanges/%2F/my_exchange'
Scenario:
Given path 'publish'
这里的问题是 url 被解析为:
http://localhost:15672/api/exchanges///my_exchange/publish
代替:
http://localhost:15672/api/exchanges/%2F/my_exchange/publish
Thanks
EDITED:
我会改进这个问题。
我在这里上传了一个非常简单的项目:https://github.com/italktothewind/karate-encoding https://github.com/italktothewind/karate-encoding
它有一个 Wiremock 正在监听/bar/%2F/foo
此功能正在运行:
Feature: Working example
Scenario:
Given url 'http://localhost:1081/bar/%2F/foo'
When method get
Then status 200
但这个功能不起作用(我在项目中放置了一个@ignore标志,这样它就可以成功构建):
Feature: Non working example
Background:
* url 'http://localhost:1081/bar/%2F'
Scenario:
Given path 'foo'
When get
Then status 200
这两个功能的区别在于使用url
and path
.
对我有用:
* url 'https://httpbin.org/anything/%2F/my_exchange'
* method get
并在日志中:
1 > GET https://httpbin.org/anything/%2F/my_exchange
1 > Accept-Encoding: gzip,deflate
1 > Connection: Keep-Alive
1 > Host: httpbin.org
1 > User-Agent: Apache-HttpClient/4.5.11 (Java/1.8.0_231)
编辑:当您有非常规的 URL 时,空手道中的一般建议是仅使用url
(不要使用path
): https://stackoverflow.com/a/53638335/143475 https://stackoverflow.com/a/53638335/143475
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