以下测试读取文件,并使用lxml.html为页面生成DOM / Graph的叶节点。
但是,我也试图弄清楚如何从“字符串”获取输入。使用
lxml.html.fromstring(s)
不起作用,因为这会生成“元素”而不是“ElementTree”。
所以,我正在试图弄清楚如何将元素转换为ElementTree。
思想
import lxml.html
from lxml import etree # trying this to see if needed
# to convert from element to elementtree
#cmd='cat osu_test.txt'
cmd='cat o2.txt'
proc=subprocess.Popen(cmd, shell=True,stdout=subprocess.PIPE)
s=proc.communicate()[0].strip()
# s contains HTML not XML text
#doc = lxml.html.parse(s)
doc = lxml.html.parse('osu_test.txt')
doc1 = lxml.html.fromstring(s)
for node in doc.iter():
if len(node) == 0:
print "aaa ",node.tag, doc.getpath(node)
#print "aaa ",node.tag
nt = etree.ElementTree(doc1) <<<<< doesn't work.. so what will??
for node in nt.iter():
if len(node) == 0:
print "aaa ",node.tag, doc.getpath(node)
#print "aaa ",node.tag
===============================
更新:::
(解析html而不是xml) 添加了Abbas建议的更改。得到以下错误:
doc1 = etree.fromstring(s)
File "lxml.etree.pyx", line 2532, in lxml.etree.fromstring (src/lxml/lxml.etree.c:48621)
File "parser.pxi", line 1545, in lxml.etree._parseMemoryDocument (src/lxml/lxml.etree.c:72232)
File "parser.pxi", line 1424, in lxml.etree._parseDoc (src/lxml/lxml.etree.c:71093)
File "parser.pxi", line 938, in lxml.etree._BaseParser._parseDoc (src/lxml/lxml.etree.c:67862)
File "parser.pxi", line 539, in lxml.etree._ParserContext._handleParseResultDoc (src/lxml/lxml.etree.c:64244)
File "parser.pxi", line 625, in lxml.etree._handleParseResult (src/lxml/lxml.etree.c:65165)
File "parser.pxi", line 565, in lxml.etree._raiseParseError (src/lxml/lxml.etree.c:64508)
lxml.etree.XMLSyntaxError: Entity 'nbsp' not defined, line 48, column 220
UPDATE :::
管理以使测试正常运行。我不确定为什么。如果有人需要提供解释,这将有助于未来的人偶然发现这一点。
from cStringIO import StringIO
from lxml.html import parse
doc1 = parse(StringIO(s))
for node in doc1.iter():
if len(node) == 0:
print "aaa ", node.tag, doc1.getpath(node)
似乎StringIO模块/类实现了IO功能,它满足了解析包需要继续处理测试html的输入字符串。类似于铸造在其他语言中提供的......或者
感谢
答案 0 :(得分:9)
要从_Element
(使用lxml.html.fromstring
生成)获取根树,您可以使用getroottree
方法:
doc = lxml.html.parse(s)
tree = doc.getroottree()
答案 1 :(得分:2)
etree.fromstring
方法解析XML字符串并返回根元素。 etree.ElementTree
类是元素周围的树包装器,因此需要一个元素进行实例化。
因此,将根元素传递给etree.ElementTree()
构造函数应该可以提供您想要的内容:
root = etree.fromstring(s)
nt = etree.ElementTree(root)
答案 2 :(得分:1)
_Element
,通过如下调用返回:
tree = etree.HTML(result.read(), etree.HTMLParser())
可以像_ElementTree
那样:
tree = tree.getroottree() # convert _Element to _ElementTree
希望这就是你所期望的。