我正在嘗試通過讀取 Excel 工作表使用 python 腳本創建 XML 文件。使用 yattag 我能夠完成此任務，盡管不完全是我需要的格式。我已經粘貼了下面的代碼，并且已經驗證了沒有空格/制表符的混合。目標是將整個項目包裝在“node”標簽中，并為兩個“category”標簽再添加 2 個子類別。我收到錯誤，因為在“節點”標簽之后，“位置”選項卡之前有 2 個選項卡。如果我修復了錯誤，我就會得到第一組代碼?；旧现恍枰獙ⅰ?lt;/node”拉到底部（如果有任何意義的話）。<node type="document" action="create"> <location>TempCD</location> <title>doc1</title> <file>E:\Doc1.docx</file> <mime>application</mime> </node> <category name="Content"> <attribute name="Function">asd</attribute> <attribute name="Commodity">sf</attribute> <attribute name="Sub-Commodity">qw</attribute> <attribute name="Contract/Document Owner">e</attribute> <subites>reapply</subitems> </category> <category name="Content Server Categories:LYB:LYB-GSC-Contracts"> <attribute name="Supplier">Altom Transport</attribute> <attribute name="Pricing Terms">Fixed</attribute> <attribute name="Term Type">Fixed</attribute> <subitems name="Commodity">reapply</subitems> </category> from openpyxl import load_workbook from yattag import Doc, indent wb = load_workbook("input_sample.xlsx") ws = wb.worksheets[0] # Create Yattag doc, tag and text objects doc, tag, text = Doc().tagtext() xml_header = '<?xml version="1.0" encoding="UTF-8"?>' xml_schema = '<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema"></xs:schema>' doc.asis(xml_header) doc.asis(xml_schema) for row in ws.iter_rows(min_row=2): row = [cell.value for cell in row] with tag('node', type=row[0], action=row[1]): with tag("location"): text(row[2]) with tag("title"): text(row[3]) with tag("file"): text(row[4]) with tag("mime"): text(row[5]) with tag('category', name=row[6]): )
1 回答
慕運維8079593
TA貢獻1876條經驗 獲得超5個贊
這應該會給你你正在尋找的 xml：
from openpyxl import load_workbook
from yattag import Doc, indent
wb = load_workbook("input_sample.xlsx")
ws = wb.worksheets[0]
# Create Yattag doc, tag and text objects
doc, tag, text = Doc().tagtext()
xml_header = '<?xml version="1.0" encoding="UTF-8"?>'
xml_schema = '<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema"></xs:schema>'
doc.asis(xml_header)
#doc.asis(xml_schema) # invalid
with tag('root'): # required for valid xml
for row in ws.iter_rows(min_row=2):
row = [cell.value for cell in row]
with tag('node', type=row[0], action=row[1]):
with tag("location"): text(row[2])
with tag("title"): text(row[3])
with tag("file"): text(row[4])
with tag("mime"): text(row[5])
with tag('category', name=row[6]):
with tag("attribute", name='Function'): text(row[7])
with tag("attribute", name='Commodity'): text(row[8])
with tag("attribute", name='Sub-Commodity'): text(row[9])
with tag("attribute", name='Contract/Document Owner'): text(row[10])
with tag("subitems"): text("reapply")
with tag('category', name=row[11]):
with tag("attribute", name='Supplier'): text(row[12])
with tag("attribute", name='Pricing Terms'): text(row[13])
with tag("attribute", name='Term Type'): text(row[14])
with tag("subitems"): text("reapply")
result = indent(
doc.getvalue(),
indentation = ' ',
indent_text = False
)
with open("test_resulted.xml", "w") as f:
f.write(result)
輸出
<?xml version="1.0" encoding="UTF-8"?>
<root>
<node type="2" action="2">
<location>2</location>
<title>2</title>
<file>2</file>
<mime>2</mime>
<category name="2">
<attribute name="Function">2</attribute>
<attribute name="Commodity">2</attribute>
<attribute name="Sub-Commodity">2</attribute>
<attribute name="Contract/Document Owner">2</attribute>
<subitems>reapply</subitems>
</category>
<category name="2">
<attribute name="Supplier">2</attribute>
<attribute name="Pricing Terms">2</attribute>
<attribute name="Term Type">2</attribute>
<subitems>reapply</subitems>
</category>
</node>
<node>
..........
</node>
..............
</root>
添加回答
舉報
0/150
提交
取消