python-如何逃避实际命名的BeautifulSoup ISO标记中的父属性？

好的,这很有趣.这是XML：

<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/maven-v4_0_0.xsd">

    <parent>
        <groupId>com.parent</groupId>
        <artifactId>parent</artifactId>
        <version>1.0-SNAPSHOT</version>
        <relativePath>../pom.xml</relativePath>
    </parent>

    <build>
        <sourceDirectory>src</sourceDirectory>
    </build>

我想使用简单的BeautifulSoup分层表示法到达实际上名为< parent>的节点.但parent实际上是此API中的保留属性标签.

with open(pom) as pomHandle:
    soup = BeautifulSoup(pomHandle)

#this returns the proper build node
buildNode = soup.project.build
#this does not return the proper parent node but the XML parent of the project node
#(which is the whole doc) because 'parent' is reserved
parentNode = soup.project.parent

如何克服此限制？

解决方法:

您可以使用find()代替：

soup.project.find('parent')

从本质上讲,这是同一件事,因为BeautifulSoup在Tag类的__getattr __()方法中使用了find-the-under.

希望能有所帮助.

标签：dom,beautifulsoup,xml,python,xml-parsing
来源： https://codeday.me/bug/20191122/2058184.html