Obviously I know the difference between a file and a directory :)
I also know that both a file and a directory have a path.
But when writing code, what will you use when?
E.g.:
I want to clear all *.js files in a directory...
$filesystem->removeJsFiles($jsDir|$jsPath);
Copy an XML file...
$filesystem->copy($xmlFile|$xmlPath);
So when will you typically use $path as a variable instead of $dir or $file (or the other way around)?
If you don't need to distinguish between filenames, paths and file handles I would use simply jsDir and xmlFile. Otherwise I would use jsDirName or jsDirPath and xmlFileName or xmlFilePath, and jsDir and xmlFile for file handles.
Related
I'm trying to use external script with variables, but in result I get only "no such file or directory".
1st.ksh
#!bin/ksh
PATHNAME = `dirname $0`
. $PATHNAME/2nd.ksh
Echo $EXTVAR
2nd.ksh
#!bin/ksh
EXTVAR=1
I tried to use "Source" instead of "." (Source $PATHNAME/2nd.ksh) and I get the same result.
To run script I'm using full path to the script - cygdrive/e/Folder/1st.ksh.
2nd.ksh in this path too (cygdrive/e/Folder/).
All rights was granted for both files (chmod u=rwx,g=rwx,o=rwx filename).
If I put files in cygwin home path (/home/username/) I have the same.
Please help to understand what I'm doing wrong.
Thanks in advance!
$() should be used in ksh instead of `` (link)
. should be user instead of source (link)
"=" must not be surrounded with spaces. You should write: PATHNAME=$(dirname $0)
you should be aware of case-sensitiveness: echo, source
Is it possible to rename a file after building it in Sublime Text 3? By default, the output is the same as the input; so filename.scss is built to filename.css. But what if I want filename.scss.css by default (to indicate that this file is based off of a scss file). Is this possible?
Yes. Read through the Build Systems Reference for details. First, there are several variables you can use. $file is the reference to the full path of the current file, say /home/foo/test.php. There is also $filepath (/home/foo), $file_name (test.php), $file_extension (php), $file_base_name (test), and some others. You can also use regexes just about anywhere inside curly braces ($file is the same as ${file}). For example, ${file/\.php/\.txt/} will rename its suffix from .php to .txt (/home/foo/test.txt). ${filepath/testing/production} changes the directory. Here are several combined in a contrived example:
"cmd": ["myprocessor", "--infile", "$file", "--outfile", "/mnt/${project_name}/var/www/assets/${file_base_name/test/final}.css"],
...
For your particular case, this should work:
"cmd": ["myprocessor", "--infile", "$file", "--outfile", "$filepath/$file_name.css"],
should take /path/to/yourfile.scss and spit out the processed /path/to/yourfile.scss.css if that's what you want.
is it possible to load module from file with extension other than .lua?
require("grid.txt") results in:
module 'grid.txt' not found:
no field package.preload['grid.txt']
no file './grid/txt.lua'
no file '/usr/local/share/lua/5.1/grid/txt.lua'
no file '/usr/local/share/lua/5.1/grid/txt/init.lua'
no file '/usr/local/lib/lua/5.1/grid/txt.lua'
no file '/usr/local/lib/lua/5.1/grid/txt/init.lua'
no file './grid/txt.so'
no file '/usr/local/lib/lua/5.1/grid/txt.so'
no file '/usr/local/lib/lua/5.1/loadall.so'
no file './grid.so'
no file '/usr/local/lib/lua/5.1/grid.so'
no file '/usr/local/lib/lua/5.1/loadall.so'
I suspect that it's somehow possible to load the script into package.preaload['grid.txt'] (whatever that is) before calling require?
It depends on what you mean by load.
If you want to execute the code in a file named grid.txt in the current directory, then just do dofile"grid.txt". If grid.txt is in a different directory, give a path to it.
If you want to use the path search that require performs, then add a template for .txt in package.path, with the correct path and then do require"grid". Note the absence of suffix: require loads modules identified by names, not by paths.
If you want require("grid.txt") to work should someone try that then yes, you'll need to manually loadfile and run the script and put whatever it returns (or whatever require is documented to return when the module doesn't return anything) into package.loaded["grid.txt"].
Alternatively, you could write your own loader just for entries like this which you set into package.preload["grid.txt"] which finds and loads/runs the file or, more generically, you could write yourself a loader function, insert it into package.loaders, and then let it do its job whenever it sees a "*.txt" module come its way.
I found the following code snippet on the internet, and want to use it in my own .vimrc.
augroup CodeFormatters
autocmd!
autocmd BufReadPost,FileReadPost *.py :silent %!PythonTidy.py
augroup END
However, I'm not quite sure where to put the PythonTidy.py script, so that it is accessible from everywhere.
Furthermore I read that using BufReadPre is better than BufReadPost, respectively FileReadPre, is that true?
As it stands, PythonTidy.py must be accessible through your PATH. If you have a convenient place already contained in there, e.g. ~/bin, just place it there.
Alternatively, you can place it somewhere into your .vim directory, and use something like expand('<sfile>:p:h') to resolve the directory of your Vimscript, and build a relative path from there.
As you want to filter the read buffer contents with the :%! command, you have to use the BufReadPost event; with BufReadPre, the buffer hasn't yet been read and nothing would be sent to the filter.
PythonTidy is a command line executable: put it somewhere in your $PATH.
You can also put it anywhere and use an absolute path in the autocmd:
autocmd BufReadPost,FileReadPost *.py :silent %!/path/to/PythonTidy.py
I'm working on a script that processes a folder and there is always one file in it I need to rename. The new name should be the parent directory name. How do I get this in a batch file? The full path to the dir is known.
It is not very clear how the script is supposed to become acquainted with the path in question, but the following example should at least give you an idea of how to proceed:
FOR %%D IN ("%CD%") DO SET "DirName=%%~nxD"
ECHO %DirName%
This script gets the path from the CD variable and extracts the name only from it to DirName.
You can use basename command:
FULLPATH=/the/full/path/is/known
JUSTTHENAME=$(basename "$FULLPATH")
You can use built-in bash tricks:
FULLPATH=/the/full/path/is/known
JUSTTHENAME=${FULLPATH##*/}
Explanations:
first # means 'remove the pattern from the begining'
second # means 'remove the longer possible pattern'
*/ is the pattern
Using built-in bash avoid to call an external command (i.e. basename) therefore this optimises you script. However the script is less portable.