【发布时间】:2017-10-02 10:56:06
【问题描述】:
<project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
<dependencyManagement>
<dependencies>
<dependency>
<groupId>javax</groupId>
<artifactId>javaee-api</artifactId>
<version>7.0</version>
</dependency>
<dependency>
<groupId>telnet</groupId>
<artifactId>telnet.service</artifactId>
<version>1.10.0-SNAPSHOT</version>
</dependency>
<dependency>
<groupId>xmlthing</groupId>
<artifactId>xmlthing.service</artifactId>
<version>1.9.0-SNAPSHOT</version>
</dependency>
<dependency>
<groupId>parser</groupId>
<artifactId>parser.service</artifactId>
<version>1.4.0-SNAPSHOT</version>
</dependency>
</dependencies>
</dependencyManagement>
还有什么比这更实质性的?
我从version 属性中提取值时遇到问题groupId=telnet。如何获取1.10.0-SNAPSHOT 的xmlpath 值,其中groupId=telnet?
抱歉之前没提过: 它应该是 linux/unix 格式(xmllint、grep、sed ...)任何东西 :) 非常感谢! 兄弟, S.
【问题讨论】:
标签: for-xml-path