本文实例讲述了Python实现对象转换为xml的方法。分享给大家供大家参考,具体如下:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
|
# -*- coding:UTF-8 -*- ''''' Created on 2010-4-20 @author: 忧里修斯 ''' import xml.etree.ElementTree as ET import xml.dom.minidom as minidom from addrbook.domain import Person class Converter( object ): ''''' 实现Python对象与xml之间的相互转换 ''' root = None #根节点 def __init__( self ): pass @staticmethod def createRoot(rootTag): ''''' 创建根节点 ''' root = ET.Element(rootTag) return root @staticmethod def getXmlString(element,defaultEncoding = 'utf-8' ): ''''' 根据节点返回格式化的xml字符串 ''' try : rough_string = ET.tostring(element, defaultEncoding) reparsed = minidom.parseString(rough_string) return reparsed.toprettyxml(indent = " " , encoding = defaultEncoding) except : print 'getXmlString:传入的节点不能正确转换为xml,请检查传入的节点是否正确' return '' @staticmethod def classToElements(classobj,rootTag = None ): ''''' 根据传入的对象的实例,根据对象的属性生成节点,返回由节点组成的列表 classobj:对象的实例 rootTag:根节点名称 ''' attrs = None #保存对象的属性集 elelist = [] #节点列表 try : attrs = classobj.__dict__.keys() #获取该对象的所有属性(即成员变量) except : print 'classToElements:传入的对象非法,不能正确获取对象的属性' if attrs ! = None and len (attrs) > 0 : #属性存在 for attr in attrs: attrvalue = getattr (classobj, attr) #属性值 #属性节点 attrE = ET.Element(attr) attrE.text = attrvalue #加入节点列表 elelist.append(attrE) return elelist @staticmethod def classToXML(classobj,rootTag = None ): ''''' Python自定义模型类转换成xml,转换成功返回的是xml根节点,否则返回None classobj:对象的实例 rootTag:根节点名称 ''' try : classname = classobj.__class__.__name__ #类名 if rootTag ! = None : root = Converter.createRoot(rootTag) else : root = Converter.createRoot(classname) elelist = Converter.classToElements(classobj, rootTag) for ele in elelist: root.append(ele) return root except : print 'classToXML:转换出错,请检查的传入的对象是否正确' return None @staticmethod def collectionToXML(listobj,rootTag = 'list' ): ''''' 集合(列表、元组、字典)转换为xml,转换成功返回的是xml根节点,否则返回None ''' try : classname = listobj.__class__.__name__ #类名 root = Converter.createRoot(rootTag) if isinstance (listobj, list ) or isinstance (listobj, tuple ): #列表或元组 if len (listobj) > = 0 : for obj in listobj: #迭代列表中的对象 itemE = Converter.classToXML(obj) root.append(itemE) elif isinstance (listobj, dict ): #字典 if len (listobj) > = 0 : for key in listobj: #迭代字典中的对象 obj = listobj[key] itemE = Converter.classToXML(obj) itemE. set ( 'key' , key) root.append(itemE) else : print 'listToXML:转换错误,传入的对象:' + classname + '不是集合类型' return root except : print 'collectionToXML:转换错误,集合转换成xml失败' return None if __name__ = = '__main__' : p1 = Person( 'dredfsam' , '男' , '133665' ) p2 = Person( 'dream' , '女' , 'r' ) p3 = Person( '得分' , '男' , 'sdf' ) personList = {} personList[ 'p1' ] = p1 personList[ 'p2' ] = p2 personList[ 'p3' ] = p3 # personList.append(p1) # personList.append(p2) # personList.append(p3) root = Converter.collectionToXML(personList) print Converter.getXmlString(root) # plist = (p1,p2,p3)#{'name':'sdf'} # print type(plist) # print type(plist),isinstance(plist, list) |
希望本文所述对大家Python程序设计有所帮助。