java - Tomcat 无法解析 web.xml，在提到的代码点没有错误

Question

我正在为我的网络服务使用 Jersey，这就是我的 web.xml 的样子:

<?xml version="1.0" encoding="UTF-8"?>
<web-app>
    <servlet>
        <servlet-name>jersey</servlet-name>
        <servlet-class>com.sun.jersey.spi.container.servlet.ServletContainer</servlet-class>

        <init-param>
            <param-name>com.sun.jersey.config.property.packages</param-name>
            <param-name>com.rohanprabhu.external.interfaces.service.web</param-name>
        </init-param>**

        <init-param>
            <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
            <param-name>true</param-name>
        </init-param>

        <load-on-startup>5</load-on-startup>
    </servlet>
    <servlet-mapping>
        <servlet-name>jersey</servlet-name>
        <url-pattern>*</url-pattern>
    </servlet-mapping>
</web-app>

我收到一个错误Occurred at line 10 column 22，这是我在我的文件中标记为“**”的地方(它实际上不在文件中，我只是把它在这里)。这是我得到的堆栈跟踪的(一部分):

    at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)
    at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
    at java.lang.reflect.Method.invoke(Method.java:483)
    at org.gradle.launcher.bootstrap.ProcessBootstrap.runNoExit(ProcessBootstrap.java:54)
    at org.gradle.launcher.bootstrap.ProcessBootstrap.run(ProcessBootstrap.java:35)
    at org.gradle.launcher.GradleMain.main(GradleMain.java:23)
Caused by: java.lang.IllegalArgumentException: Can't convert argument: null
    at org.apache.tomcat.util.IntrospectionUtils.convert(IntrospectionUtils.java:889)
    at org.apache.tomcat.util.digester.CallMethodRule.end(CallMethodRule.java:476)
    at org.apache.tomcat.util.digester.Digester.endElement(Digester.java:1057)
    ... 188 more
Occurred at line 10 column 22
Marking this application unavailable due to previous error(s)

这是整个堆栈跟踪，以防有帮助:http://pastebin.com/EX4bMGex

Answer 1

我同意错误消息不是最理想的，但我也确定您想要一个 <param-name>和一个<param-value>每<init-param> . :-)