path

问题 I have a cronjob that is the following: */10 * * * * root cd /some/directory && python3 FILE.py >> Output.txt 2>&1 if i run that command from a regular command line, it works fine. But when ran from cronjob, I get /bin/sh: python3: not found But when I run echo $PATH it returns /sbin:/bin:/usr/sbin:/usr/bin:/usr/syno/sbin:/usr/syno/bin:/usr/local/sbin:/usr/local/bin:/usr/local/python3/bin:/opt/bin (Note that /usr/local/python3/bin is included in this directory How do I fix this? 回答1: It seems

问题 I have a cronjob that is the following: */10 * * * * root cd /some/directory && python3 FILE.py >> Output.txt 2>&1 if i run that command from a regular command line, it works fine. But when ran from cronjob, I get /bin/sh: python3: not found But when I run echo $PATH it returns /sbin:/bin:/usr/sbin:/usr/bin:/usr/syno/sbin:/usr/syno/bin:/usr/local/sbin:/usr/local/bin:/usr/local/python3/bin:/opt/bin (Note that /usr/local/python3/bin is included in this directory How do I fix this? 回答1: It seems

问题 I want to write a function that converts regular path to path that includes environment variables: For example: C:\Windows\SomePath convert to: %Windir%\SomePath How would I do that and is this possible? Here is what I'm trying to do, but problem is, I need to check the string for all possible variables, is there some more automatic way? such that -replace operator wont be needed function Format-Path { param ( [parameter(Mandatory = $true)] [string] $FilePath ) if (![System.String]:

问题 I love Python because it comes batteries included, and I use built-in functions, a lot, to do the dirty job for me. I have always been using happily the os.path module to deal with file path but recently I ended up with unexpected results on Python 2.5 under Ubuntu linux, while dealing with string that represent windows file paths : filepath = r"c:\ttemp\FILEPA~1.EXE" print os.path.basename(filepath) 'c:\\ttemp\\FILEPA~1.EXE'] print os.path.splitdrive(filepath) ('', 'c:\ttemp\\FILEPA~1.EXE')

问题 Given a directory as a string, how can I find if ANY file exists in it? os.path.isFile() # only accepts a specific file path os.listdir(dir) == [] # accepts sub-directories My objective is to check if a path is devoid of files only (not sub-directories too). 回答1: To only check one specific directory, a solution like this would suffice: from os import listdir from os.path import isfile, join def does_file_exist_in_dir(path): return any(isfile(join(path, i)) for i in listdir(path)) To dissect

问题 Given a directory as a string, how can I find if ANY file exists in it? os.path.isFile() # only accepts a specific file path os.listdir(dir) == [] # accepts sub-directories My objective is to check if a path is devoid of files only (not sub-directories too). 回答1: To only check one specific directory, a solution like this would suffice: from os import listdir from os.path import isfile, join def does_file_exist_in_dir(path): return any(isfile(join(path, i)) for i in listdir(path)) To dissect

问题 How can I add something to my "Pythonpath". Where exactly are the files located, I have to change to add to my pythonpath? What exactly do I add to my Pythonpath? If Python calls: /Library/Frameworks/Python.framework/Versions/2.6/Resources/Python.app/Contents/ But I want it to call /Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages What do I need to add to make it work. Strange is that I already used the django-admin.py for a startproject ccommand. But now it does

问题 How can I add something to my "Pythonpath". Where exactly are the files located, I have to change to add to my pythonpath? What exactly do I add to my Pythonpath? If Python calls: /Library/Frameworks/Python.framework/Versions/2.6/Resources/Python.app/Contents/ But I want it to call /Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/site-packages What do I need to add to make it work. Strange is that I already used the django-admin.py for a startproject ccommand. But now it does

问题 Here issue in R scripts I am trying to understand how would here() work in a portable way. Found it: See what works later under Final answer - TL;DR - the bottom line, here() is not really that useful running a script.R from commandline. The way I understand it with help from JBGruber: here() looks for the root directory of a project (e.g., an RStudio project, Git project or other project defined with a .here file) starting at the current working directory and moving up until it finds any

问题 I started to use Kubernetes few month ago and I actually migrate my microServices to my rancher cluster (RKE). Everything works good, my deployments are good and services too. I would like use ingress. Everything looks good, services are find by ingress and pods are find by services. However when I try to go to the website, I have a 404 error page from ingress controller. You can see my configuration for juste two paths : one nginx and on grafana. Someone knows how can i fix it and use