我需要将文件路径名传递给一个模块。如何从目录名、基本文件名和文件格式字符串构建文件路径?
该目录在调用时可能存在，也可能不存在。
例如:
dir_name='/home/me/dev/my_reports'
base_filename='daily_report'
format = 'pdf'
我需要创建一个字符串'/home/me/dev/my_reports/daily_report.pdf'
手动连接这些片段似乎不是一个好方法。我尝试了os.path.join:
join(dir_name,base_filename,format)
但是它给出了
/home/me/dev/my_reports/daily_report/pdf
当前回答
为了更好地理解，添加以下代码:
import os
def createfile(name, location, extension):
print(name, extension, location)
#starting creating a file with some dummy contents
path = os.path.join(location, name + '.' + extension)
f = open(path, "a")
f.write("Your contents!! or whatever you want to put inside this file.")
f.close()
print("File creation is successful!!")
def readfile(name, location, extension):
#open and read the file after the appending:
path = os.path.join(location, name + '.' + extension)
f = open(path, "r")
print(f.read())
#pass the parameters here
createfile('test','./','txt')
readfile('test','./','txt')
其他回答
如果你足够幸运地运行Python 3.4+，你可以使用pathlib:
>>> from pathlib import Path
>>> dirname = '/home/reports'
>>> filename = 'daily'
>>> suffix = '.pdf'
>>> Path(dirname, filename).with_suffix(suffix)
PosixPath('/home/reports/daily.pdf')
在基本文件名中添加格式不是更好吗?
dir_name='/home/me/dev/my_reports/'
base_filename='daily_report.pdf'
os.path.join(dir_name, base_filename)
为什么不直接:
>>> import os
>>> os.path.join(dir_name, base_filename + "." + format)
'/home/me/dev/my_reports/daily_report.pdf'
为了更好地理解，添加以下代码:
import os
def createfile(name, location, extension):
print(name, extension, location)
#starting creating a file with some dummy contents
path = os.path.join(location, name + '.' + extension)
f = open(path, "a")
f.write("Your contents!! or whatever you want to put inside this file.")
f.close()
print("File creation is successful!!")
def readfile(name, location, extension):
#open and read the file after the appending:
path = os.path.join(location, name + '.' + extension)
f = open(path, "r")
print(f.read())
#pass the parameters here
createfile('test','./','txt')
readfile('test','./','txt')
from pathlib import Path
# Build paths inside the project like this: BASE_DIR / 'subdir'.
BASE_DIR = Path(__file__).resolve().parent.parent
TEMPLATE_PATH = Path.joinpath(BASE_DIR,"templates")
print(TEMPLATE_PATH)
