获取Java . io . io异常错误:使用返回400状态代码的url时，服务器为URL:返回HTTP响应代码:400

Question

我正在尝试使用以下代码使用 Groovy 执行 get 请求：

String url = “url of endpoint” def responseXml = new XmlSlurper（）.parse（url）

如果endpoint返回的状态为200，则一切正常，但有一种情况，我们必须验证如下所示的错误响应，返回的状态为400:

    <errors>
    <error>One of the following parameters is required: xyz, abc.</error>
    <error>One of the following parameters is required: xyz, mno.</error>
    </errors>

在这种情况下，parse方法抛出：

    java.io.IOException: Server returned HTTP response code: 400 for URL: "actual endpoint throwing error"
    at sun.net.www.protocol.http.HttpURLConnection.getInputStream0(HttpURLConnection.java:1900)
        at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1498)
        at com.sun.org.apache.xerces.internal.impl.XMLEntityManager.setupCurrentEntity(XMLEntityManager.java:646)
        at com.sun.org.apache.xerces.internal.impl.XMLVersionDetector.determineDocVersion(XMLVersionDetector.java:150)
        at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:831)
        at com.sun.org.apache.xerces.internal.parsers.XML11Configuration.parse(XML11Configuration.java:796)
        at com.sun.org.apache.xerces.internal.parsers.XMLParser.parse(XMLParser.java:142)
        at com.sun.org.apache.xerces.internal.parsers.AbstractSAXParser.parse(AbstractSAXParser.java:1216)
        at com.sun.org.apache.xerces.internal.jaxp.SAXParserImpl$JAXPSAXParser.parse(SAXParserImpl.java:644)
        at groovy.util.XmlSlurper.parse(XmlSlurper.java:205)
        at groovy.util.XmlSlurper.parse(XmlSlurper.java:271)
    
Can anyone pls suggest how to handle if server give error message by throwing 400 status code?

Answer 1

从 HttpURLConnection 类获取响应文本，而不是通过 XmlSlurper 隐式获取响应文本，可以更灵活地处理不成功的响应。尝试这样的事情：

def connection = new URL('https://your.url/goes.here').openConnection()
def content = { ->
    try {
        connection.content as String
    } catch (e) {
        connection.responseMessage
    }
}()

if (content) {
    def responseXml = new XmlSlurper().parseText(content)
    doStuffWithResponseXml(responseXml)
}

更好的方法是使用一个真正的功能齐全的HTTP客户端，比如Spring Framework的Http客户端或别说类。

Answer 2

在问题中，因为我们正在为GET请求获取400状态代码。所以内置的XmlSlurper（）. parse（URI）方法不起作用，因为它抛出io. Excep。Groovy还支持api请求和响应的HTTP方法，以下方法对我有用：

def getReponseBody(endpoint) {     
    URL url = new URL(endpoint)
    HttpURLConnection get = (HttpURLConnection)url.openConnection()
    get.setRequestMethod("GET")
    def getRC = get.getResponseCode()

    BufferedReader br = new BufferedReader(new InputStreamReader(get.getErrorStream()))
    StringBuffer xmlObject = new StringBuffer()
    def eachLine
    while((eachLine = br.readLine()) !=null){
        xmlObject.append(eachLine)
    }
    get.disconnect()
    return new XmlSlurper().parseText(xmlObject.toString())
}