“无法从...初始化默认的wsdl". - 为什么? [英] "Can not initialize the default wsdl from..." -- Why?

问题描述

我的pom.xml包含以下内容，以自动为具有下面指定的WSDL的正常工作的Web服务生成客户端:

My pom.xml contains the following to auto generate a client for a working web service having the WSDL specified below:

        <plugin>
            <groupId>org.apache.cxf</groupId>
            <artifactId>cxf-codegen-plugin</artifactId>
            <version>2.3.1</version>
            <executions>
                <execution>
                    <id>generate-sources</id>
                    <configuration>
                        <sourceRoot>${basedir}/target/generated/src/main/java</sourceRoot>
                        <wsdlOptions>
                            <wsdlOption>
                                <wsdl>${basedir}/src/main/wsdl/myclient.wsdl</wsdl>
                                <extraargs>
                                    <extraarg>-client</extraarg>
                                    <extraarg>-verbose</extraarg>
                                </extraargs>
                                <wsdlLocation>wsdl/myclient.wsdl</wsdlLocation>
                            </wsdlOption>
                        </wsdlOptions>
                    </configuration>
                    <goals>
                        <goal>wsdl2java</goal>
                    </goals>
                </execution>
            </executions>
        </plugin>

项目构建良好，没有任何错误或警告，我可以在wsdl文件夹下看到JAR文件中的文件myclient.wsdl.

The project builds fine, without any errors or warnings and I can see the the file myclient.wsdl in the JAR file right under a wsdl folder.

但是当我尝试运行该JAR时:

But when I try running that JAR:

  java -Xmx1028m -jar myclient-jar-with-dependencies.jar

它抱怨无法从wsdl/myclient.wsdl初始化默认的wsdl"

It complains that "Can not initialize the default wsdl from wsdl/myclient.wsdl"

为什么?

我想念什么?

如何找到pom.xml中的wsdl/myclient.wsdl 转换为的路径，使客户端的JAR在运行时抱怨?

How can I find out what path that wsdl/myclient.wsdl in pom.xml translates into, that makes the client's JAR complain at run time?

更新:我知道一些涉及修改自动生成代码的解决方案/解决方法:

Update: I am aware of some solutions/workarounds that involve modifying the auto-generated code:

1. 为wsdl URL传递"null"，然后使用((BindingProvider)port).getRequestContext().放入(BindingProvider. ENDPOINT_ADDRESS_PROPERTY ，" http://example.com/ ....)设置地址
2. 将WSDL作为Java资源加载，并将其位置传递到服务的构造函数中.

1. Pass "null" for the wsdl URL and then use the ((BindingProvider)port).getRequestContext().put(BindingProvider.ENDPOINT_ADDRESS_PROPERTY, "http://example.com/....") to set the address.
2. load the WSDL as a Java resource and pass its location into your service's constructor.

但是，我对这样的解决方案更感兴趣，该解决方案要求在pom.xml中输入正确的值，例如 classpath方法 (但不幸的是，由于某种原因，classpath对我不起作用).

But I am more interested in a solution that requires entering the right values into the pom.xml like the classpath approach (but unfortunately classpath didn't work for me for some reason).

有什么主意我应该在那输入什么吗?显然，这是为该特定插件找出正确路径规则的一种非常简单的情况，但我遗漏了一些东西，我不知道它是什么.

Any ideas what I should be typing there instead? Apparently this is a very simply case of figuring out the correct path rules for that particular plugin, but I am missing something and I don't know what it is.

推荐答案

我注意到

I notice the cfx examples use slightly different locations for sourceRoot, wsdl and wsdlLocation.

请记住，通常，src/main/resources中的文件包含在生成的工件中.为了包含src/main/wsdl中的文件，需要将其作为资源添加到pom.xml中:

Remember that typically, files in src/main/resources are included in the produced artifact. In order for files in src/main/wsdl to be included, it needs to be added as a resource in the pom.xml:

<resources>
    <resource>
        <directory>src/main/wsdl</directory>
    </resource>
</resources>

提示:

• 将您怀疑的路径设置为已知的错误路径，并查看是否收到相同的错误消息.
• 解压缩生成的*.jar文件，并检查是否包含wsdl，以及路径是什么.

• Set the paths you suspect to known bad paths and see if you get the same error-message.
• Unzip the produced *.jar-file(s) and check if the wsdl is included, and what the path is.

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