问如何使用JAX-WS在AXL客户机中使用GetLineReq
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Stack Overflow用户

提问于 2014-07-18 05:17:04

回答 1查看 1.5K关注 0票数 1

我想要做的是使用Cisco的AXL API从CUCM获得一个目录号。以下是相关守则：

private void getNumber(){
    AXLAPIService axlService = new AXLAPIService();
    AXLPort axlPort = axlService.getAXLPort();

    String validatorUrl = "https://mycucm:8443/axl/";

    ((BindingProvider) axlPort).getRequestContext().put(BindingProvider.ENDPOINT_ADDRESS_PROPERTY, validatorUrl);
    ((BindingProvider) axlPort).getRequestContext().put(BindingProvider.USERNAME_PROPERTY, Demo.ucAdmin);
    ((BindingProvider) axlPort).getRequestContext().put(BindingProvider.PASSWORD_PROPERTY, Demo.ucPswd);


    GetLineReq axlParams = new GetLineReq();


    axlParams.setPattern("7491817");
    axlParams.setUuid("{48a6ff28-cea7-fc3e-3d82-8cc245ed4ea3}");


    GetLineRes getLineResponse = axlPort.getLine(axlParams);


    Demo.informUser("Line Information: \n" 
            + "Alerting Name: " + getLineResponse.getReturn().getLine().getAlertingName() + "\n" 
            + "Dial Number: " + getLineResponse.getReturn().getLine().getPattern() + "\n" 
            + "Description: " + getLineResponse.getReturn().getLine().getDescription() + "\n"
            + " " + getLineResponse.getReturn().getLine().getShareLineAppearanceCssName());
}

从这个图表中可以看出，只需要指定数字的uuid或模式：

但是，只有当我指定uuid (这不是我想要实现的)时，代码才能工作。我给出的唯一东西是模式，我想使用它来获取所有其他信息。我已经从思科：如何..。使用JAX-WS创建AXL Java客户端上看过这个网站了

当我省略uuid时，会得到以下错误：

Exception in thread "main" javax.xml.ws.soap.SOAPFaultException: Item not valid: The specified Line was not found

我还检查了如何使用AXLSqlToolkit将目录号存储在CUCM数据库中：

<?xml version="1.0" encoding="UTF-8" standalone="no"?><SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/"><SOAP-ENV:Header/><SOAP-ENV:Body><axl:executeSQLQueryResponse xmlns:axl="http://www.cisco.com/AXL/API/7.1" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" sequence="1405672933992"><return><row><dnorpattern>7491817</dnorpattern><pkid>48a6ff28-cea7-fc3e-3d82-8cc245ed4ea3</pkid></row></return></axl:executeSQLQueryResponse></SOAP-ENV:Body></SOAP-ENV:Envelope>

有谁知道，只有使用pattern-value?才能获得目录号？

jax-ws

cisco-axl

java

soap

单节点MySQL

低成本高体验，解决您的基础业务数据需求

回答 1

Stack Overflow用户

回答已采纳

发布于 2014-07-25 00:06:24

我自己想出来的。routePartitionName也是一个必须指定的强制参数。下面是我的方法的工作代码：

private void getNumber(){
AXLAPIService axlService = new AXLAPIService();
AXLPort axlPort = axlService.getAXLPort();

String validatorUrl = "https://mycucm:8443/axl/";

((BindingProvider) axlPort).getRequestContext().put(BindingProvider.ENDPOINT_ADDRESS_PROPERTY, validatorUrl);
((BindingProvider) axlPort).getRequestContext().put(BindingProvider.USERNAME_PROPERTY, Demo.ucAdmin);
((BindingProvider) axlPort).getRequestContext().put(BindingProvider.PASSWORD_PROPERTY, Demo.ucPswd);


GetLineReq axlParams = new GetLineReq();
ObjectFactory objectFactory = new ObjectFactory(); //This is new

XFkType routePartition = new XFkType(); 
routePartition.setValue("PHONES");     // This is where you specify your route partition name

axlParams.setPattern("7491817");
axlParams.setRoutePartitionName(objectFactory.createGetLineReqRoutePartitionName(routePartition));


GetLineRes getLineResponse = axlPort.getLine(axlParams);


Demo.informUser("Line Information: \n" 
        + "Alerting Name: " + getLineResponse.getReturn().getLine().getAlertingName() + "\n" 
        + "Dial Number: " + getLineResponse.getReturn().getLine().getPattern() + "\n" 
        + "Description: " + getLineResponse.getReturn().getLine().getDescription() + "\n"
        + " " + getLineResponse.getReturn().getLine().getShareLineAppearanceCssName());
}