我如何使用ElementTree在xml文件中搜索具有特定值的特定“父”标签的标签？ （蟒蛇） [英] How do I search for a Tag in xml file using ElementTree where i have a certain "Parent"tag with a specific value? (python)

问题描述

我刚刚开始学习Python，必须编写一个解析xml文件的程序。我必须在2个不同的文件中找到一个称为OrganisationReference的标签并将其返回。实际上，有多个使用此名称的Tag，但只有一个，我要返回的Tag，具有带有值为DEALER的Tag OrganisationType作为父Tag的Tag（不确定该术语是否正确）。我试图为此使用ElementTree。下面是代码：

 将xml.etree.ElementTree导入为ET 
 
 tree1 = ET.parse （'Master1.xml'）
 root1 = tree1.getroot（）
 
 tree2 = ET.parse（'Master2.xml'）
 root2 = tree2.getroot（）
 
用于root1.findall（ ./ Organisation / OrganisationId / [@ OrganisationType ='DEALER'] / OrganisationReference）中的OrganisationReference ：) 
 print（OrganisationReference.attrib）
 $ b rootb.findall（ ./ Organisation / OrganisationId / [@ OrganisationType ='DEALER'] / OrganisationReference）中的OrganisationReference的$ b：
 print（OrganisationReference.attrib）
  

但这不会返回任何内容（也没有错误）。有人可以帮我吗？

我的文件如下：

  < MessageOrganisationCount> a< / MessageOrganisationCount> 
< MessageVehicleCount> x< / MessageVehicleCount> 
< MessageCreditLineCount> y< / MessageCreditLineCount> 
< MessagePlanCount> z< / MessagePlanCount> 
< OrganisationData> 
< Organization> 
< OrganisationId> 
< OrganisationType>经销商< / OrganisationType> 
< OrganizationReference> WHATINEED< / OrganisationReference> 
< / OrganisationId> 
< OrganisationName> XYZ。< / OrganisationName> 
 .... 
  

由于OrganisationReference在此出现了多次文件在开始和结束标签之间具有不同的文本，我想得到的正是在第9行中看到的：它具有OrganisationId作为父标签，而DEALER也是OrganisationId的子标签。

解决方案

您与原始尝试非常接近。您只需要对xpath进行一些更改，并对python进行微小的更改。

xpath的第一部分以开头。 /组织。由于您是从根目录开始执行xpath，因此它期望 Organization 是孩子。不是;它是一个后代。

尝试将 ./组织更改为 .//组织。 （ // 是 / descendant-or-self :: node（）/ 的缩写。有关详细信息，请参见此处。）

第二个问题是 OrganisationId / [@ OrganisationType ='DEALER'] 。那是无效的xpath。 / 应该从 OrganisationId 和谓词。

此外， @ 是属性的缩写语法：: 轴和 OrganisationType 是元素，而不是属性。

尝试将 OrganisationId / [@ OrganisationType ='DEALER'] 更改为 OrganisationId [OrganisationType ='经销商'] 。

python问题与 print（OrganisationReference.attrib）。 OrganisationReference 没有任何属性；

尝试将 print（OrganisationReference.attrib）更改为 print（OrganisationReference。文本）。

下面是一个示例，仅出于演示目的使用一个XML文件...

XML输入（Master1.xml；添加了 doc 元素以使其格式正确）

 < doc> 
< MessageOrganisationCount> a< / MessageOrganisationCount> 
< MessageVehicleCount> x< / MessageVehicleCount> 
< MessageCreditLineCount> y< / MessageCreditLineCount> 
< MessagePlanCount> z< / MessagePlanCount> 
< OrganisationData> 
< Organization> 
< OrganisationId> 
< OrganisationType>经销商< / OrganisationType> 
< OrganizationReference> WHATINEED< / OrganisationReference> 
< / OrganisationId> 
< OrganisationName> XYZ。< / OrganisationName> 
< / Organisation> 
< / OrganisationData> 
< / doc> 
  

Python

< pre class = lang-py prettyprint-override> 将xml.etree.ElementTree导入为ET

tree1 = ET.parse（'Master1.xml'）
root1 = tree1.getroot（）

用于root1.findall（ .// Organisation / OrganisationId [OrganisationType ='DEALER'] / OrganisationReference）中的OrganisationReference：
print（OrganisationReference.text ）


打印输出

 已处理的
  

---

还要注意，似乎根本不需要使用 getroot（）。您可以直接在树上使用 findall（） ...

 将xml.etree.ElementTree导入为ET 
 
 tree1 = ET.parse（'Master1.xml'）
 
 for tree1.findall（  .//Organisation/OrganisationId[OrganisationType='DEALER']/OrganisationReference）：
打印（OrganisationReference.text）
  

I just started learning Python and have to write a program, that parses xml files. I have to find a certain Tag called OrganisationReference in 2 different files and return it. In fact there are multiple Tags with this name, but only one, the one I am trying to return, that has the Tag OrganisationType with the value DEALER as a parent Tag (not quite sure whether the term is right). I tried to use ElementTree for this. Here is the code:

    import xml.etree.ElementTree as ET

    tree1 = ET.parse('Master1.xml')
    root1 = tree1.getroot()

    tree2 = ET.parse('Master2.xml')
    root2 = tree2.getroot()

    for OrganisationReference in root1.findall("./Organisation/OrganisationId/[@OrganisationType='DEALER']/OrganisationReference"):
        print(OrganisationReference.attrib)

    for OrganisationReference in root2.findall("./Organisation/OrganisationId/[@OrganisationType='DEALER']/OrganisationReference"):
        print(OrganisationReference.attrib)

But this returns nothing (also no error). Can somebody help me?

My file looks like this:

  <MessageOrganisationCount>a</MessageOrganisationCount>
  <MessageVehicleCount>x</MessageVehicleCount>
  <MessageCreditLineCount>y</MessageCreditLineCount>
  <MessagePlanCount>z</MessagePlanCount>
  <OrganisationData>
      <Organisation>
          <OrganisationId>
              <OrganisationType>DEALER</OrganisationType>
              <OrganisationReference>WHATINEED</OrganisationReference>
          </OrganisationId>
          <OrganisationName>XYZ.</OrganisationName>
 ....

Due to the fact that OrganisationReference appears a few more times in this file with different text between start and endtag, I want to get exactly the one, that you see in line 9: it has OrganisationId as a parent tag, and DEALER is also a child tag of OrganisationId.

解决方案

You were super close with your original attempt. You just need to make a couple of changes to your xpath and a tiny change to your python.

The first part of your xpath starts with ./Organization. Since you're doing the xpath from root, it expects Organization to be a child. It's not; it's a descendant.

Try changing ./Organization to .//Organization. (// is short for /descendant-or-self::node()/. See here for more info.)

The second issue is with OrganisationId/[@OrganisationType='DEALER']. That's invalid xpath. The / should be removed from between OrganisationId and the predicate.

Also, @ is abbreviated syntax for the attribute:: axis and OrganisationType is an element, not an attribute.

Try changing OrganisationId/[@OrganisationType='DEALER'] to OrganisationId[OrganisationType='DEALER'].

The python issue is with print(OrganisationReference.attrib). The OrganisationReference doesn't have any attributes; just text.

Try changing print(OrganisationReference.attrib) to print(OrganisationReference.text).

Here's an example using just one XML file for demo purposes...

XML Input (Master1.xml; with doc element added to make it well-formed)

<doc>
    <MessageOrganisationCount>a</MessageOrganisationCount>
    <MessageVehicleCount>x</MessageVehicleCount>
    <MessageCreditLineCount>y</MessageCreditLineCount>
    <MessagePlanCount>z</MessagePlanCount>
    <OrganisationData>
        <Organisation>
            <OrganisationId>
                <OrganisationType>DEALER</OrganisationType>
                <OrganisationReference>WHATINEED</OrganisationReference>
            </OrganisationId>
            <OrganisationName>XYZ.</OrganisationName>
        </Organisation>
    </OrganisationData>
</doc>

Python

import xml.etree.ElementTree as ET

tree1 = ET.parse('Master1.xml')
root1 = tree1.getroot()

for OrganisationReference in root1.findall(".//Organisation/OrganisationId[OrganisationType='DEALER']/OrganisationReference"):
    print(OrganisationReference.text)

Printed Output

WHATINEED

---

Also note that it doesn't appear that you need to use getroot() at all. You can use findall() directly on the tree...

import xml.etree.ElementTree as ET

tree1 = ET.parse('Master1.xml')

for OrganisationReference in tree1.findall(".//Organisation/OrganisationId[OrganisationType='DEALER']/OrganisationReference"):
    print(OrganisationReference.text)

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