递归文件编辑 [英] recursive file editing
问题描述
我是一个蟒蛇新手;这里有一些与我试图解决的问题相关的问题。如果python是最好的
乐器，或者如果awk或混合使用bash和sed会更好，我会徘徊：
1）我怎么会得到递归下降
通过所有子目录中的所有文件
我的脚本被调用的那个？
2）每个审查的文件应该编辑;如果找到这种类型的字符串
foo.asp = dev？bar（其中bar可以是数字或空格）
它应该总是用这个字符串代替
foo-bar.html（如果bar是一个空格，新字符串是foo-.html）
3）读取的文件名称本身可以是foo.asp = dev？bar;
编辑后的输出文件也应该按照相同的规则重命名
如上所述......或者这可以通过bash脚本更好地处理？
任何提示赞赏
-
Michele Alzetta
这是一个Perl单行：
perl -p -i -e's / foo / bar / gi''`find。/`
问候，
- -
Leif Biberg Kristensen
http://solumslekt.org/
Validare necesse est
>我是一个蟒蛇新手;这里有一些与问题有关的问题我试图解决;如果python是最好的工具，或者如果awk或bash和sed的组合会更好，我会徘徊：
1）我将如何递归下降
通过
我的脚本调用的所有子目录中的所有文件？
查看os.path.walk。
2）每个被检查的文件都应该被编辑;如果找到这种类型的字符串
foo.asp = dev？bar（其中bar可以是数字或空格）
它应该总是用这个代替字符串
foo-bar.html（如果bar为空格，则新字符串为foo-.html）
查看re模块。 />
3）读取的文件名称本身可以是foo.asp = dev？bar;
编辑后的输出文件也应该根据相同的规则重命名。如上所述......或者这会更好地通过bash脚本来处理吗？
不管你感觉更舒服，Python都可以做到。
- Josiah
Il Sat，2004年4月3日22:35:30 +0200，Leif B. Kristensen ha scritto：
这是一个Perl单行：
perl -p -i -e'/ foo / bar / gi''`find ./`
没用;然而我意识到我可以反复运行各种变化
sed -is / foo / bar / g *
然后重命名文件一个名为renna的bash脚本。
我肯定python可以做到但我被卡住了 -
for os.walk中的文件（''mydir''）：
打印文件[2]
给我所有文件的名称
但我如何用r +模式打开每个？
for os.walk（''mydir''）：
file（thing， mode = r +）
语法无效
-
Michele Alzetta
I''m a python newbie; here are a few questions relative to a
problem I''m trying to solve; I''m wandering if python is the best
instrument or if awk or a mix of bash and sed would be better:
1) how would I get recursively descend
through all files in all subdirectories of
the one in which my script is called ?
2) each file examined should be edited; IF a string of this type is found
foo.asp=dev?bar (where bar can be a digit or an empty space)
it should always be substituted with this string
foo-bar.html (if bar is an empty space the new string is foo-.html)
3) the names of files read may themselves be of the sort foo.asp=dev?bar;
the edited output file should also be renamed according to the same rule
as above ... or would this be better handled by a bash script ?
Any hints appreciated
--
Michele Alzetta
This is a Perl one-liner:
perl -p -i -e ''s/foo/bar/gi'' `find ./`
regards,
--
Leif Biberg Kristensen
http://solumslekt.org/
Validare necesse est
> I''m a python newbie; here are a few questions relative to aproblem I''m trying to solve; I''m wandering if python is the best
instrument or if awk or a mix of bash and sed would be better:
1) how would I get recursively descend
through all files in all subdirectories of
the one in which my script is called ?
Check out os.path.walk.
2) each file examined should be edited; IF a string of this type is found
foo.asp=dev?bar (where bar can be a digit or an empty space)
it should always be substituted with this string
foo-bar.html (if bar is an empty space the new string is foo-.html)
Check out the re module.
3) the names of files read may themselves be of the sort foo.asp=dev?bar;
the edited output file should also be renamed according to the same rule
as above ... or would this be better handled by a bash script ?
Do it however you feel more comfortable, Python can do it.
- Josiah
Il Sat, 03 Apr 2004 22:35:30 +0200, Leif B. Kristensen ha scritto:
This is a Perl one-liner:
perl -p -i -e ''s/foo/bar/gi'' `find ./`
Didn''t work; however I realized I could just repeatedly run variations of
sed -i s/foo/bar/g *
and then rename the files with a bash script called renna.
I''m sure python could do it but I got stuck -
for file in os.walk(''mydir''):
print file[2]
gives me the names of all files
but how do I open each with r+ mode ?
for thing in os.walk(''mydir''):
file(thing,mode=r+)
is invalid syntax
--
Michele Alzetta
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