使用Scalaxb生成代码时出错

Question

为了其中一个实现，我不得不篡改WSDL文档，我遇到了Scalaxb！我现在正试图从我所拥有的WSDL文件中生成一些scala类，正如我所预期的那样，我遇到了一些问题：

下面是WSDL文件的一个片段：

<?xml version="1.0" encoding="UTF-8" ?>
<wsdl:definitions
 xmlns:soap="http://schemas.xmlsoap.org/wsdl/soap/"
 xmlns:tns="http://www.myservice.com/MyServices/2012/06/18/"
 xmlns:wsdl="http://schemas.xmlsoap.org/wsdl/"
 xmlns:mime="http://schemas.xmlsoap.org/wsdl/mime/"
 xmlns:s="http://www.w3.org/2001/XMLSchema"
 name="MyServices"
 targetNamespace="http://www.myservice.com/MyServices/2012/06/18/">
<wsdl:types>
 <s:schema elementFormDefault="qualified" targetNamespace="http://www.myservice.com/MyServices/2012/06/18/">
  <s:complexType name="UserCredentials">
   <s:sequence>
    <s:element name="UserName" type="s:string" />
    <s:element name="Password" type="s:string" />
   </s:sequence>
  </s:complexType>
  <s:element name="UserCredentials" type="tns:UserCredentials" />
  <s:complexType name="AnotherComplexType" >
   <s:sequence>
    <s:element name="Index" type="s:int" />
    <s:element name="Name" type="s:string" />
    <s:element name="Status" type="s:boolean" />
   </s:sequence>
  </s:complexType>
  ....

假设WSDL文件的其余部分完全没有问题，当我试图编译项目时，我遇到了以下错误：

[error] /Users/joe/Desktop/scalaxb-soap-example/target/scala-2.11/src_managed/main/sbt-scalaxb/myservice/xmlprotocol.scala:1542: not found: value userCredentials
[error]             scalaxb.toXML(userCredentials, Some("http://www.myservice.com/MyServices/2012/06/18/"), "UserCredentials", defaultScope), defaultScope, baseAddress, "POST", Some(new java.net.URI("http://1.1.1.1/cgi-bin/cgi.cgi?WebService=SetGPTimerChannel"))).transform({ case (header, body) => 
[error]                           ^
[error] /Users/joe/Desktop/scalaxb-soap-example/target/scala-2.11/src_managed/main/sbt-scalaxb/myservice/xmlprotocol.scala:1544: value toSeq is not a member of Any
[error]               scala.xml.Elem(null, "Body", scala.xml.Null, defaultScope, true, body.toSeq: _*)
[error]                                                                                     ^
[error] /Users/joe/Desktop/scalaxb-soap-example/target/scala-2.11/src_managed/main/sbt-scalaxb/myservice/xmlprotocol.scala:1551: not found: value userCredentials
[error]             scalaxb.toXML(userCredentials, Some("http://www.myservice.com/MyServices/2012/06/18/"), "UserCredentials", defaultScope), defaultScope, baseAddress, "POST", Some(new java.net.URI("http://1.1.1.1/cgi-bin/cgi.cgi?WebService=SomeServiceCall"))).transform({ case (header, body) => 
[error]                           ^
[error] /Users/joe/Desktop/scalaxb-soap-example/target/scala-2.11/src_managed/main/sbt-scalaxb/myservice/xmlprotocol.scala:1553: value toSeq is not a member of Any
[error]               scala.xml.Elem(null, "Body", scala.xml.Null, defaultScope, true, body.toSeq: _*)
[error]                                                                                     ^

知道我为什么要面对这个问题吗？这是我的build.sbt：

import ScalaxbKeys._

val scalaXml = "org.scala-lang.modules" %% "scala-xml" % "1.0.2"
val scalaParser = "org.scala-lang.modules" %% "scala-parser-combinators" % "1.0.1"
val dispatchV = "0.11.1" // change this to appropriate dispatch version
val dispatch = "net.databinder.dispatch" %% "dispatch-core" % dispatchV

organization := "com.eon"

name := "scalaxb-myservice-sample"

scalaVersion := "2.11.6"

scalaxbSettings

packageName in (Compile, scalaxb) := "rdmservice"

dispatchVersion in (Compile, scalaxb) := dispatchV

async in (Compile, scalaxb) := true

sourceGenerators in Compile <+= scalaxb in Compile

libraryDependencies ++= Seq(scalaXml, scalaParser, dispatch)

Answer 1

如果您只需要生成WSDL代码，那么使用scalaxb可能有点过了。作为一种选择，您可以将wsimport包装为一个简单的SBT任务，在主代码编译之前执行。除了一个更少的依赖之外，它还保持了存储库的原始状态，并使其免于提交生成的样板Java代码。如果有人感兴趣，这里有一个示例模板项目：https://github.com/sainnr/sbt-scala-wsdl-template。