獲取 java.io.IOException：服務器返回 HTTP 響應代碼：URL 的 400：使用返回 400 狀態代碼的 url 時

Question

我正在嘗試使用以下代碼使用Groovy執行獲取請求：

字符串 url = “端點的 url” def responseXml = new XmlSlurper().parse(url)

如果端點返回狀態為 200，那么一切正常，但在一種情況下，我們必須驗證如下所示的錯誤響應，返回的狀態為 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>

在這種情況下，解析方法拋出：

    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

在問題中，因為我們正在獲取 GET 請求的 400 狀態代碼。 所以內置的 XmlSlurper().parse(URI) 方法不起作用，因為它拋出 io.Exception。 Groovy 還支持 HTTP 方法用於 api 請求和響應，以下對我有用：

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())
}

Answer 2

從HttpURLConnection class 獲取響應文本而不是通過XmlSlurper隱式獲取響應文本使您可以更靈活地處理不成功的響應。 嘗試這樣的事情：

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)
}

更好的方法是使用實際的全功能 HTTP 客戶端，例如 Spring Framework 的HttpClient或RestTemplate類。

Answer 3

您應該檢查返回碼，然后從 http 請求實例中獲取錯誤 stream 以防出錯。 問題本身與 JsonSlurper 無關，因為如果服務未成功返回返回碼（400、401、500 等），則 http 請求實例不會返回“輸入流”實例。POST 示例如下所示：

http= new URL("yourUrl").openConnection() as HttpURLConnection
http.setRequestMethod('POST')
http.setDoOutput(true)
http.setRequestProperty("Content-Type", 'application/json')
http.setRequestProperty("Accept", 'application/json')
http.setRequestProperty("Authorization", "Bearer $yourTokenVariable")
http.outputStream.write(data.getBytes("UTF-8"))
http.connect()
 
if(http.getResponseCode() != 200 && http.getResponseCode() != 201){ 
     throw new InvalidInputException("There was an error: " + http.getErrorStream().getText("UTF-8"))
} else {
    //You can take input stream here
}