Question

REST Client的HTTP Builder会返回HttpResponseDecorator。如何从中获取原始响应（用于记录目的）？

编辑（某些代码可能很方便）：

    withRest(uri: domainName) {
        def response = post(path: 'wsPath', query: [q:'test'])
        if (!response.success) {
            log.error "API call failed. HTTP status: $response.status"
            // I want to log raw response and URL constructed here
        }

Answer 1

我一直在做同样问题的噩梦。这是我使用HTTPBuilder的解决方案： -

response.failure = {resp ->
    println "request failed with status ${resp.status}, response body was [${resp.entity.content.text}]"
    return null
}

希望有所帮助！

Answer 2

我使用了XmlUtil，它返回了漂亮的xml：

    def data = respXml.data
    assert data instanceof groovy.util.slurpersupport.GPathResult

    println "${XmlUtil.serialize(data)}"

如果您的数据是来自groovyx.net.http.HttpResponseDecorator的解析响应

希望有帮助。

Answer 3

试试这个：

println response.data

如何从HttpResponseDecorator获取原始响应和URL

3 个答案: