[英]Opening file by file in a folder
我是使用python编程的新手,但是目前我收到一个编写脚本的任务,该脚本将在type = 0或type = 1出现的所有ID都写下来。 它是一个XML文件,类似于以下示例:
<root>
<bla1 type="0" id = "1001" pvalue:="djdjd"/>
<bla2 type="0" id = "1002" pvalue:="djdjd" />
<bla3 type="0" id = "1003" pvalue:="djdjd"/>
<bla4 type="0" id = "1004" pvalue:="djdjd"/>
<bla5 type="0" id = "1005" pvalue:="djdjd"/>
<bla6 type="1" id = "1006" pvalue:="djdjd"/>
<bla7 type="0" id = "1007" pvalue:="djdjd"/>
<bla8 type="0" id = "1008" pvalue:="djdjd"/>
<bla9 type="1" id = "1009" pvalue:="djdjd"/>
<bla10 type="0" id = "1010" pvalue:="djdjd"/>
<bla11 type="0" id = "1011" pvalue:="djdjd"/>
<bla12 type="0" id = "1009" pvalue:="djdjd"/>
<root>
因此,代码要做的第一件事是将':='基本上替换为'='原因,这导致我的xml上传导致错误。 无论如何,它会写下类型为0的ID和类型为1的ID。这对于一个xml文件非常有效。 不幸的是,我只有一个文件,而且我需要一个像这样的循环:始终打开文件夹中的下一个xml文件(不同名称),并总是将新的ID添加到最后一个xml中找到的ID。 因此,基本上,它总是总是从新的xml文件中添加新找到的ID。
import xml.etree.cElementTree as ET # required import
XmlFile = 'ID3.xml' # insert here the name of the XML-file, which needs to be inside the same folder as the .py file
my_file = open('%s' % XmlFile, "r+") # open the XML-file
Xml2String = my_file.readlines() # convert the file into a list strings
XmlFile_new = [] # new list, which is filled with the modified strings
L = len(Xml2String) # length of the string-list
for i in range(1, L): # Increment starts at 0, therefore, the first line is ignored
if ':=' in Xml2String[i]:
XmlFile_new.append(Xml2String[i].replace(':=', '=')) # get rid of colon
else:
XmlFile_new.append(Xml2String[i])
tree = ET.ElementTree(XmlFile_new)
root = tree.getroot()
id_0 = [] # list for id="0"
id_1 = [] # list for id="1"
id_one2zero = [] # list for ids, that occur twice
for i in range(len(root)):
if 'type="0"' in root[i]: # check for type
a = root[i].index("id") + 5 # search index of id
b = a+6
id_0.append((root[i][a:b])) # the id is set via index slicing
elif 'type="1"' in root[i]: # check for type
a = root[i].index("id") + 5
b = a+6
id_1.append((root[i][a:b]))
else:
print("Unknown type occurred") # If there's a line without type="0" or type="1", this message gets printed
# (Remember: first line of the xml-file is ignored)
for i in range(len(id_0)): # check for ids, that occur twice
for j in range(len(id_1)):
if id_0[i] == id_1[j]:
id_one2zero.append(id_0[i])
print(id_0)
print(id_1)
f = open('write.xml','w')
print >>f, 'whatever'
print('<end>')
解决此问题的一种简单方法是使用os.walk()
函数。 有了它,您可以在一个目录中甚至递归地打开所有文件。
这是一个示例如何使用它:
for root, dirs, files in os.walk("your/path"):
for file in files:
# process your file
如果目录中除xml-files之外还有其他文件,则可以确保仅使用file.endswith(".xml")
处理xml-files。
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