blocks|key|2728162|text|如果你在项目树的根目录中有一些模块，假设config_loader.py看起来像这样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2728163|import+os

def+get_config_path():
++++relative_path+=+'conf/constants.cfg'
++++current_dir+=+os.getcwd()
++++return+os.join(current_dir,+relative_path)|code-block|syntax|javascript|2728164|然后在ConfigManager.py或任何其他需要配置的模块中：|2728165|import+config_loader

file_path+=+config_loader.get_config_path()
config_file+=+open(file_path)|2728166|你甚至可以让你的config_loader.py只返回配置文件。|2728167|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

If you had some module in the root of the project tree, say config_loader.py that looked like this:

<pre><code>import os

def get_config_path():
 relative_path = 'conf/constants.cfg'
 current_dir = os.getcwd()
 return os.join(current_dir, relative_path)
</code></pre>

And then in ConfigManager.py or any other module that needs the configs:

<pre><code>import config_loader

file_path = config_loader.get_config_path()
config_file = open(file_path)
</code></pre>

You could even have your config_loader.py just return the config file.

blocks|key|2728252|text|如果conf是一个Python包，那么您可以使用pkgutil.get_data()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2728253|import+pkgutil

data+=+pkgutil.get_data("conf",+"constants.cfg")|code-block|syntax|javascript|2728254|或者如果安装了setuptools+-+pkg_resources.resource_string()|2728255|import+pkg_resources

data+=+pkg_resources.resource_string('conf',+'constants.cfg')|2728256|2728257|如果constants.cfg不在包中，则将其路径作为命令行参数传递，或者在环境变量(例如，CONFIG_MANAGER_CONSTANTS_PATH+)中设置它，或者从一组固定的默认路径(例如，os.path.expanduser("~/.config/ConfigManager/constants.cfg")+)中读取。要找到放置用户数据的位置，可以使用appdirs+module。|2728258|如果您可以从不同的目录运行ConfigManager.py，则不能使用返回当前工作目录的os.getcwd()。出于同样的原因，相对路径"../../..."不会工作。|2728259|如果您确定ConfigManager.py和constants.cfg在文件系统中的相对位置不会改变：|2728260|import+inspect
import+os
import+sys

def+get_my_path():
++++try:
++++++++filename+=+__file__+#+where+we+were+when+the+module+was+loaded
++++except+NameError:+#+fallback
++++++++filename+=+inspect.getsourcefile(get_my_path)
++++return+os.path.realpath(filename)

#+path+to+ConfigManager.py
cm_path+=+get_my_path()
#+go+6+directory+levels+up
sp_path+=+reduce(lambda+x,+f:+f(x),+[os.path.dirname]*6,+cm_path)
constants_path+=+os.path.join(sp_path,+"conf",+"constants.cfg")|2728261|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/pkgutil.html|1|http://packages.python.org/distribute/pkg_resources.html#basic-resource-access|2|https://pypi.python.org/pypi/appdirs^0|2|4|O|I|O|I|0|0|0|7|A|K|V|K|V|1|0|0|0|2|D|1A|T|2Q|1N|50|7|50|E|2|0|D|G|18|B|1W|B|0|5|G|M|D|0|0^^$0|@$1|2|3|4|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|F|3|G|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1I|8|@$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@$9|1N|A|1O|1|1P]]|E|$]]|$1|M|3|N|5|H|7|1Q|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1R|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1S|8|@$9|1T|A|1U|B|C]|$9|1V|A|1W|B|C]|$9|1X|A|1Y|B|C]|$9|1Z|A|20|B|C]]|D|@$9|21|A|22|1|23]]|E|$]]|$1|R|3|S|5|6|7|24|8|@$9|25|A|26|B|C]|$9|27|A|28|B|C]|$9|29|A|2A|B|C]]|D|@]|E|$]]|$1|T|3|U|5|6|7|2B|8|@$9|2C|A|2D|B|C]|$9|2E|A|2F|B|C]]|D|@]|E|$]]|$1|V|3|W|5|H|7|2G|8|@]|D|@]|E|$I|J]]|$1|X|3|-4|5|6|7|2H|8|@]|D|@]|E|$]]]|Y|$Z|$5|10|11|12|E|$13|14]]|15|$5|10|11|12|E|$13|16]]|17|$5|10|11|12|E|$13|18]]]]

If <code>conf</code> is a Python package then you could use <a href="http://docs.python.org/2/library/pkgutil.html" rel="nofollow"><code>pkgutil.get_data()</code></a>:

<pre><code>import pkgutil

data = pkgutil.get_data("conf", "constants.cfg")
</code></pre>

Or if <code>setuptools</code> is installed – <a href="http://packages.python.org/distribute/pkg_resources.html#basic-resource-access" rel="nofollow"><code>pkg_resources.resource_string()</code></a>:

<pre><code>import pkg_resources

data = pkg_resources.resource_string('conf', 'constants.cfg')
</code></pre>

<hr>

If <code>constants.cfg</code> is not in a package then pass its path as a command-line parameter, or set it in an environment variable e.g., <code>CONFIG_MANAGER_CONSTANTS_PATH</code>, or read from a fixed set of default paths e.g., <code>os.path.expanduser("~/.config/ConfigManager/constants.cfg")</code>. To find a place where to put user data, you could use <a href="https://pypi.python.org/pypi/appdirs" rel="nofollow"><code>appdirs</code> module</a>.

You can't use <code>os.getcwd()</code> that returns current working directory if you may run <code>ConfigManager.py</code> from different directories. Relative path <code>"../../..."</code> won't work for the same reason.

If you are certain that the relative position of <code>ConfigManager.py</code> and <code>constants.cfg</code> in the filesystem won't change:

<pre><code>import inspect
import os
import sys

def get_my_path():
 try:
 filename = __file__ # where we were when the module was loaded
 except NameError: # fallback
 filename = inspect.getsourcefile(get_my_path)
 return os.path.realpath(filename)

# path to ConfigManager.py
cm_path = get_my_path()
# go 6 directory levels up
sp_path = reduce(lambda x, f: f(x), [os.path.dirname]*6, cm_path)
constants_path = os.path.join(sp_path, "conf", "constants.cfg")
</code></pre>

blocks|key|43441|text|可以在+3.0%2B中使用pathlib包|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|43442|这将获取跨不同平台的SampleProject文件夹中包含的任何文件的路径。|43443|from+pathlib+import+Path
def+get_file(path):
++++"""
++++returns+the+absolute+path+of+a+file
++++:var
++++str+path
++++++++the+file+path+within+the+working+dir
++++:returns
++++PureWindowsPath+or+PurePosixPath+object
++++++++type+depends+on+the+operating+system+in+use
++++"""
++++def+get_project_root()+->+Path:
++++"""Returns+project+root+folder."""
++++return+Path(__file__).parent.parent

++++return+get_project_root().joinpath(path)|code-block|syntax|javascript|43444|然后，只需使用file_path作为参数调用该函数：|43445|filePath+=+get_file('com/python/example/source/utils/configManager.py')|43446|然后是通常的步骤：|43447|while+open(filePath)+as+f:
++++<DO+YOUR+THING>|43448|entityMap^0|3|F|0|7|2|A|D|0|0|7|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|19|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|1A|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|V|$]]

You can use the pathlib package in Python 3.0+

This gets the path to any file contained in the SampleProject folder
across different platforms. 


<pre><code>from pathlib import Path
def get_file(path):
 """
 returns the absolute path of a file
 :var
 str path
 the file path within the working dir
 :returns
 PureWindowsPath or PurePosixPath object
 type depends on the operating system in use
 """
 def get_project_root() -&gt; Path:
 """Returns project root folder."""
 return Path(__file__).parent.parent

 return get_project_root().joinpath(path)
</code></pre>

Then simply call the function with the file_path as an argument:


<pre><code>filePath = get_file('com/python/example/source/utils/configManager.py')
</code></pre>

And then the usual procedure:


<pre><code>while open(filePath) as f:
 &lt;DO YOUR THING&gt;
</code></pre>

I have the following project structure

<pre><code>SampleProject
 com
 python
 example
 source
 utils
 ConfigManager.py
 conf
 constants.cfg
</code></pre>

How to access constants.cfg from ConfigManager.py. 

I have a limitation

<ol>
<li>I can not give full path(absolute path) of constants.cfg because if I run in different PC it should work with out any modification</li>
<li>Also if I represent something like below, I can access the file. But I don't want to give back slash every time

<pre><code>filename = ..\\..\\..\\..\\..\\..\\constants.cfg`
</code></pre></li>
</ol>

Currently I am doing something like this. But this works only when constants.cfg and ConfigManager.py are in same directory

<pre><code>currentDir = os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))
file = open(os.path.join(currentDir,'constants.cfg')) 
</code></pre>

Python : How to access file from different directory

文件系统

Python

我有以下项目结构SampleProject     com       python          example             source                utils                   ConfigManager.py     conf        constants...

问Python :如何从不同目录访问文件
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