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How to extract a directory from a file path?
Shell Programming and Scripting hello, i have a configuration file app.conf under /tmp, containing values like : param1=/data/something param2=/data/somethingelse i have a bash script that has to list the files under the path that corresponds to param2 : #!/bin/bash dir=$ (cat tmp/app.conf | grep param2 | sed 8. Shell Programming and Scripting
How to get last 2 folder names from a directory path?
I need to locate last two folder name from a directory path. Say, the path is “E:\\ROSC10\\ROSC_Image”. I need the ROSC10 and ROSC_Image i two different variable o that I can marge them into a single string. My code given below. Can anyone help me out. Thanks in advance. One way to achieve this is:
How to extract the parent directory in Python?
I need to extract the name of the parent directory of a certain path. This is what it looks like: I would like to extract directory_i_need.
How to get root directory of given path in Bash?
At this point basedirectory will be the string home Then you write: PS: Never use PATH variable in your script as it will overrider default PATH and you script won’t be able to execute many system utilities
How to obtain the path of a file via shell?
Writing a custom function and exporting it is too much work and risk for error for something like this, here are two lines: This makes up for the shortcomings of realpath, store it in a shell script fullpath. You can now call:
How to extract path from below given string?
Closed 6 years ago. how to extract path location from below given string. expected output. In general, $ {parameter%word} removes word from the end of parameter. In our case, we want to remove the final slash and all characters which follow: /*.
Where is the file path stored in PowerShell?
In this code the path is stored in the %1 argument (which is written in the registry under quote that are escaped: \\”%1\\” ). To retrieve it, we need the $arg (inbuilt arg). Don’t forget the quote around $FilePath.