我有以下XML文件,我必须解析该文件并将其提取为csv文件中的数据。在此文件中,我有两个框(box_id),它们包装在两个不同的父对象(parent_box_id)上,并且每个框的内容的详细信息(元素sgtin-> info_sgtin)。
<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<doc>
<info id_reference="2">
<data_down>
<tree>
<box_id>046071598600870568</box_id>
<parent_box_id>046071598600875594</parent_box_id>
</tree>
<tree>
<box_id>046071598600870575</box_id>
<parent_box_id>046071598600875594</parent_box_id>
</tree>
<tree>
<sgtin>
<info_sgtin>
<sgtin>04607008133585B0SE1HVHBGR3A</sgtin>
<box_id>046071598600870568</box_id>
<gtin>04607008133585</gtin>
<series_number>026A</series_number>
</info_sgtin>
</sgtin>
<parent_box_id>046071598600870568</parent_box_id>
</tree>
<tree>
<sgtin>
<info_sgtin>
<sgtin>046070081335856F7P78HBVBEH2</sgtin>
<box_id>046071598600870568</box_id>
<gtin>04607008133585</gtin>
<series_number>026A</series_number>
</info_sgtin>
</sgtin>
<parent_box_id>046071598600870568</parent_box_id>
</tree>
<tree>
<sgtin>
<info_sgtin>
<sgtin>046070081335854T61H7CSXDE9W</sgtin>
<box_id>046071598600870575</box_id>
<gtin>04607008133585</gtin>
<series_number>026A</series_number>
</info_sgtin>
</sgtin>
<parent_box_id>046071598600870575</parent_box_id>
</tree>
</data_down>
</info>
</doc>
为此,我决定在Python中使用Elementtree,但是问题是在我的XML文件中,我有两个tag变体。
首先,我遍历所有详细信息并捕获box_id的值,但是在那之后,我必须转到父项并获取其中包装了box_id的parent_box_id。
换句话说,我想通过以下方式获取数据:
parent_box_id box_id sgtin series_number
046071598600875594 046071598600870568 04607008133585B0SE1HVHBGR3A 026A
046071598600875594 046071598600870568 046070081335856F7P78HBVBEH2 026A
046071598600875595 046071598600870575 046070081335854T61H7CSXDE9W 026A
但是我不知道如何获取parent_box_id值。感谢社区的任何支持。
这是我的代码:
import csv
import xml.etree.ElementTree as ET
csv.writer(open('result.csv','w'),delimiter=';', quotechar='"', quoting=csv.QUOTE_MINIMAL))
tree = ET.parse('test.xml')
root = tree.getroot()
with open('result.csv','a',newline='') as myfile:
writer = csv.writer(myfile, delimiter=';', quotechar='"', quoting=csv.QUOTE_MINIMAL)
for alist in root.iter('info_sgtin'):
sgtin = alist.find('sgtin').text
box_id = alist.find('box_id').text
series = alist.find('series_number').text
writer.writerow([sgtin,box_id,series])
这是使用XPATH的解决方案:
root = etree.parse(...) # assuming this is your ElementTree
for elem in root.iter("info_sgtin"):
sgtin = elem.xpath("sgtin")[0].text # you can use .find as well
series_number = elem.xpath("series_number")[0].text
box_id = elem.xpath("box_id")[0].text
# first get the parent and then the following sibling which is the parent_box_id
parent_box_id = elem.xpath("parent::sgtin/following-sibling::parent_box_id")[0].text
print(parent_box_id, box_id, sgtin, series_number)
输出:
046071598600870568 046071598600870568 04607008133585B0SE1HVHBGR3A 026A
046071598600870568 046071598600870568 046070081335856F7P78HBVBEH2 026A
046071598600870575 046071598600870575 046070081335854T61H7CSXDE9W 026A