Question

我有这个代码，我搜索一个值（inputCategory），它通常位于路径的末尾（例如/blah/blah/<category/blah或有时/blah/blah/<category>

def category_search(inputDirectory, inputCategory, root):
    categorySearch = os.path.split(os.path.normpath(inputDirectory)) 
    if categorySearch[1] == inputName:
        Logger.info("SEARCH: Files appear to be in their own directory")
        categorySearch2 = os.path.split(os.path.normpath(categorySearch[0]))
        if categorySearch2[1] == twoCategory or categorySearch2[1] == oneCategory:
            if not inputCategory:
            Logger.info("SEARCH: Determined Category to be: %s", categorySearch2[1])
                inputCategory = categorySearch2[1]
        elif not inputCategory:
            Logger.error("SEARCH: Could not identify category from the directory structure.")
            sys.exit(-1)
        else:
            pass

    elif categorySearch[1] == twoCategory or categorySearch[1] == oneCategory:
        if os.path.isdir(os.path.join(inputDirectory, inputName)):
            inputDirectory = os.path.join(inputDirectory, inputName)
        else:
            Logger.info("SEARCH: The directory passed is the root directory for category %s", categorySearch[1])
            Logger.info("SEARCH: We will try and determine which files to process, individually")
            root = 1
        if not inputCategory:
            Logger.info("SEARCH: Determined Category to be: %s", categorySearch[1])
            inputCategory = categorySearch[1]
    elif not inputCategory:
        Logger.error("SEARCH: Could not identify category from the directory structure")
        sys.exit(-1)
    else:
        Logger.info("SEARCH: The directory passed does not appear to include a category")
        root = 1
    return inputDirectory, inputCategory, root

但问题是代码不是很动态，例如，如果我得到像/blah/blah/<category>/blah/blah

这样的路径，就会被抛弃

我正在寻找有关如何优化此代码的输入，捕获上述情况以及可能更多的情况？

我在python上很新，所以要温柔:)感谢您的帮助！

Answer 1

我使用这样的东西进行路径检查，但不确定它是否适合你... 我也是非常绿色的蟒蛇！

# deconstruct the folder: /a/b/c --> /a/b --> /a --> /
pathHead, pathTail = os.path.split(inputDirectory)  
while pathTail and ( pathTail != inputCategory ): 
    pathHead,pathTail = os.path.split(pathHead)

if not pathTail:
    print 'didnt find category'
else:
    print 'found category: ',pathTail , ' in path: ' , pathHead

编辑：修复elif到其他地方......

使用python中的值匹配目录路径

1 个答案: