創建zipfile：TypeError：類型'ElementTree'的對象沒有len（）

Question

我正在寫一個XML數據到zip。

from xml.etree.ElementTree import Element, SubElement, ElementTree 
from zipfile import ZipFile 

def create_tree(): 
    root = Element("root") 
    doc = SubElement(root, "doc") 
    SubElement(doc, "field", name="blah").text = "text" 

    return ElementTree(root) 


def test(): 
    """ 
    Create zip 
    """ 
    with ZipFile("xml.zip", 'w') as ziparc: 
     element_tree = create_tree() 
     ziparc.writestr("file.xml", element_tree) 


if __name__ == "__main__": 
    test() 

錯誤：

File "main_test2_2.py", line 168, in test 
    ziparc.writestr('file.xml', element_tree) 
    File "/usr/lib/python2.7/zipfile.py", line 1127, in writestr 
    zinfo.file_size = len(bytes)   # Uncompressed size 
TypeError: object of type 'ElementTree' has no len() 

請告訴我，我怎麼可以將XML數據寫入到歸檔？

Answer 1

寫的元素放入一個假文件（緩衝區）

from xml.etree.ElementTree import Element, SubElement, ElementTree 
from zipfile import ZipFile 
from io import BytesIO 

def create_tree(): 
    root = Element("root") 
    doc = SubElement(root, "doc") 
    SubElement(doc, "field", name="blah").text = "text" 

    return ElementTree(root) 


def test(): 
    """ 
    Create zip 
    """ 
    with ZipFile("xml.zip", 'w') as ziparc: 
     element_tree = create_tree() 
     outbuf = BytesIO() 
     element_tree.write(outbuf) 
     ziparc.writestr("file.xml", outbuf.getvalue()) 


if __name__ == "__main__": 
    test() 

編輯：另一個用戶試圖表明tostring方法，但它並沒有完成&沒有可能的工作，因爲第一個參數必須是一個Element不是ElementTree，其次是因爲進口（ElementTree是一個包裝和一個子類別，因此存在歧義）。

但是，我修改了完整的源代碼，它也適用，我認爲它甚至更好的解決方案（歡呼誰刪除了他的留言本等用戶！）

from xml.etree.ElementTree import Element, SubElement 
from zipfile import ZipFile 
import xml.etree.ElementTree 

def create_tree(): 
    root = Element("root") 
    doc = SubElement(root, "doc") 
    SubElement(doc, "field", name="blah").text = "text" 

    return root 


def test(): 
    """ 
    Create zip 
    """ 
    with ZipFile("xml.zip", 'w') as ziparc: 
     element_tree = create_tree() 

     ziparc.writestr("file.xml", xml.etree.ElementTree.tostring(element_tree)) 


if __name__ == "__main__": 
    test()