I looped in the directory tree of a "content" folder containing subdirectories and markdown files. Then I need to know the relative path of those markdowns from that "content" folder.
In bash script I would do something like that :
CONTENT_PATH="/home/myusr/apps/myapp/content"
file_path="/home/myusr/apps/myapp/content/file/pgp.md"
echo "${file_path#$CONTENT_PATH}"
# /file/pgp.md
So in Lua I didn't found something like that, so I've tried with string.gsub():
print(string.gsub(file_path, CONTENT_PATH, ""))
-- /home/myusr/apps/myapp/content/file/pgp.md 0
But it's not working, it seems my CONTENT_PATH string does not match and I don't know why?
print(CONTENT_PATH)
-- /home/hs0ucy/_01_http/fakestache-lua/content
print(file_path)
-- /home/hs0ucy/_01_http/fakestache-lua/content/file/pgp.md
Thanks!
PS : I'm new to Lua.
From: Lua string.gsub with a hyphen
The hyphen is a special character in Lua, and needs to be escaped like so: %-.
I discovered this by slowly making CONTENT_PATH longer and longer until it was no longer working. Good ol' binary search!
EDIT: if you can't modify your CONTENT_PATH but you're sure that file_path has CONTENT_PATH in it:
contentPathLen = string.len(CONTENT_PATH)
print(string.sub(file_path, contentPathLen + 1))
-- Output: /file/pgp.md
Or if you need to verify that file_path starts with CONTENT_PATH:
base = string.sub(file_path, 0, contentPathLen)
if base == CONTENT_PATH then
print("file_path is under CONTENT_PATH")
end
Related
I'm looking for a script (or if there isn't, I guess I'll have to write my own).
I wanted to ask if anyone here knows a script that can take a txt file with n links (lets say 200). I need to extract only links that have particular characters in them, let's say I only need links that contain "/r/learnprogramming". I need the script to get those links and write them to another txt files.
Edit: Here is what helped me: grep -i "/r/learnprogramming" 1.txt >2.txt
you can use ajax to read .txt file using jquery
<script src=https://cdnjs.cloudflare.com/ajax/libs/jquery/2.1.1/jquery.min.js></script>
<script>
jQuery(function($) {
console.log("start")
$.get("https://ayulayol.imfast.io/ajaxads/ajaxads.txt", function(wholeTextFile) {
var lines = wholeTextFile.split(/\n/),
randomIndex = Math.floor(Math.random() * lines.length),
randomLine = lines[randomIndex];
console.log(randomIndex, randomLine)
$("#ajax").html(randomLine.replace(/#/g,"<br>"))
})
})
</script>
<div id=ajax></div>
If you are using linux or macOS you could use cat and grep to output the links.
cat in.txt | grep /r/programming > out.txt
Solution provided by OP:
grep -i "/r/learnprogramming" 1.txt >2.txt
Since you did not provide the exact format of the document I assume those links are separated by newline characters. In this case, the code is pretty straightforward using Python/awk since you can iterate over file.readlines() and print only those that match your pattern (either by using a lines.contains(pattern) or using a regex if the pattern is more complex). To store the links in a new file simply redirect the stdout to a new file like this:
python script.py > links.txt
The solution above works even if links are separated by an arbitrary symbol s, first read the file into a single string and split it over s. I hope this helps.
So I have a Ruby script (using Ruby because we have a library of pre-existing code that I need to use). From within Ruby I am using backticks to call Linux commands, specifically in this case the "mv" command. I am trying to move one file to another location but I keep getting the error message that x and y are "the same file" even though they are very clearly NOT the same file.
Here is the code in Ruby:
#!/usr/local/rvm/rubies/ruby-2.1.1/bin/ruby
masterFiles=[]
masterFiles << "/mnt/datadrive/Data Capture/QualityControl/UH_HRA_SVY/Scans and DataOutput/Data/UH_HRA_SVY_DATA.txt"
masterFiles << "/mnt/datadrive/Data Capture/QualityControl/UH_HRA_SVY_SPAN/Scans and DataOutput/Data/UH_HRA_SVY_SPAN_DATA.txt"
tm=Time.new.strftime("%Y%m%d")
masterFiles.each do |mf|
if File.exist?(mf)
qmf=39.chr + mf + 39.chr
`cat #{qmf} >> /tmp/QM`
savename=39.chr + \
"/mnt/datadrive/Data Capture/QualityControl/UH_HRA_SVY/Scans and DataOutput/Data/DailyFiles/" + \
File.basename(mf).gsub(".txt","_"+tm) + ".txt" + 39.chr
`mv #{qmf} #{savename}`
end
end
The error that I get is this:
mv: `/mnt/datadrive/Data Capture/QualityControl/UH_HRA_SVY_SPAN/Scans
and DataOutput/Data/UH_HRA_SVY_SPAN_DATA.txt' and `/mnt/datadrive/Data
Capture/QualityControl/UH_HRA_SVY/Scans and
DataOutput/Data/DailyFiles/UH_HRA_SVY_SPAN_DATA_20140530.txt' are the
same file
If I change this line:
`mv #{qmf} #{savename}`
To this:
puts "mv #{qmf} #{savename}"
And then run the output, it works as expected.
I am pretty sure that this has to do with spaces in the path. I have tried every combination of double-quoting, triple-quoting, quadruple-quoting, and back-slashing I can think of to resolve this but no go. I have also tried using FileUtils.mv but get what is basically the same error worded differently.
Can anybody help ? Thanks a lot.
p.s. I realize it's entirely possible that I could be going about this in an entirely wrong-headed way, so feel free to point that out if so. However, I am trying to use the tools which I already have some knowledge of (cat, mv, etc) instead of re-inventing the wheel.
You could use FileUtils.mv
I often do aliases like so:
require 'fileutils'
def mv(from, to)
FileUtils.mv(from, to)
end
And inside the mv() method I do additional safeguards, i.e. if the file does not exist, if there is a lack of permissions and so forth.
If you then still have problems with filenames that have ' ' blank characters, try to put the file into a "" quote like:
your_target_location = "foo/bar bla"
For example, I have arbitrary lines in this format:
directory C:\Program Files\abc\def\
or something like.
log-enabled On
I want to be able to extract the "C:\Program Files\ab\def\" part out of that first line. Likewise, I want to extract the "On" out in the second line. The spaces between the variable and its value are arbitrary. I will know the name of the variable, but I need extract the value based on that.
So basically, I want to remove the first word and a number of arbitrary spaces that follow the first word, and return what remains until the end of the line.
Assuming that, by "word" you mean "a string of characters without spaces", you can do this:
for line in ioFile:lines() do
local variable, value = line:match("(%S+)%s+(.+)")
... --Do stuff with variable and value
end
One alternative with string.match was shown by Nicol Bolas, here is another alternative:
function splitOnFirstSpace(input)
local space = input:find(' ') or (#input + 1)
return input:sub(1, space-1), input:sub(space+1)
end
Usage:
local command, param = splitOnFirstSpace(line)
If no argument is given (splitOnFirstSpace('no-param-here')), then param is the empty string.
I do not believe Lua is packaged with a split() function like Ruby or Perl.
I found that this guy built a lua version of Perl's split function:
http://lua-users.org/lists/lua-l/2011-02/msg01145.html
If you can guarantee that the argument will only have 1 word before it, with that word not containing any spaces, you can read in that line, run the split function on it, and use the return array's 1 index value as what you want.
You could error check that too and make sure you get a 'C:\' within your expected directory, or check to make sure the string is == to 'On' or 'Off'. Because of using the hardcoded index value I really advocate you error check your expected value. Nothing is worse than having an assumed value be wrong.
If an error is detected make sure to log or print it to the screen so you know about it.
This could catch bugs where maybe the string that was input is improper.
Some simple code that models what I suggest you do:
line = "directory C:\Program Files\abc\def/";
contents = line.split(" "); --Split using a space
directory = contents[2]; --Here is your directory
if(errorCheckDir(directory))
--Use directory
end
EDIT:
In response to comments below Lua indeed begins indexing at 1, not 0.
Also, in the case that the directory contains spaces (which is probable) instead of simply using contents[2], I would loop through all of contents except index 1, and piece back together the directory making sure to add the required space between each index that you attach.
So in the case above, contents[2] and contents[3] would have to be stitched back together with a space in between to recover the proper directory.
directory = contents[2].." "..contents[3]
This can be easily automated using a function which has a loop in it and returns back the proper directory:
function recoverDir(contents)
directory = "";
--Recover the directory
for i=2, table.getn(contents) do
directory = directory..contents[i].." ";
end
--strip extra space on the end
dirEnd = string.len(directory);
directory = string.sub(directory,1,dirEnd-1);
return directory; --proper directory
end
If I have a file structure like this:
./main.lua
./mylib/mylib.lua
./mylib/mylib-utils.lua
./mylib/mylib-helpers.lua
./mylib/mylib-other-stuff.lua
From main.lua the file mylib.lua can be loaded with full path require('mylib.mylib'). But inside mylib.lua I would also like to load other necessary modules and I don't feel like always specifying the full path (e.g. mylib.mylib-utils). If I ever decide to move the folder I'm going to have a lot of search and replace. Is there a way to use just the relative part of the path?
UPD. I'm using Lua with Corona SDK, if that matters.
There is a way of deducing the "local path" of a file (more concretely, the string that was used to load the file).
If you are requiring a file inside lib.foo.bar, you might be doing something like this:
require 'lib.foo.bar'
Then you can get the path to the file as the first element (and only) ... variable, when you are outside all functions. In other words:
-- lib/foo/bar.lua
local pathOfThisFile = ... -- pathOfThisFile is now 'lib.foo.bar'
Now, to get the "folder" you need to remove the filename. Simplest way is using match:
local folderOfThisFile = (...):match("(.-)[^%.]+$") -- returns 'lib.foo.'
And there you have it. Now you can prepend that string to other file names and use that to require:
require(folderOfThisFile .. 'baz') -- require('lib.foo.baz')
require(folderOfThisFile .. 'bazinga') -- require('lib.foo.bazinga')
If you move bar.lua around, folderOfThisFile will get automatically updated.
You can do
package.path = './mylib/?.lua;' .. package.path
Or
local oldreq = require
local require = function(s) return oldreq('mylib.' .. s) end
Then
-- do all the requires
require('mylib-utils')
require('mylib-helpers')
require('mylib-other-stuff')
-- and optionally restore the old require, if you did it the second way
require = oldreq
I'm using the following snippet. It should work both for files loaded with require, and for files called via the command line. Then use requireRel instead of require for those you wish to be loaded with a relative path.
local requireRel
if arg and arg[0] then
package.path = arg[0]:match("(.-)[^\\/]+$") .. "?.lua;" .. package.path
requireRel = require
elseif ... then
local d = (...):match("(.-)[^%.]+$")
function requireRel(module) return require(d .. module) end
end
Under the Conky's Lua environment I've managed to include my common.lua (in the same directory) as require(".common"). Note the leading dot . character.
Try this:Here is my findings:
module1= require(".\\moduleName")
module2= dofile("..\\moduleName2.lua")
module3 =loadfile("..\\module3.lua")
To load from current directory. Append a double backslash with a prefix of a fullstop.
To specify a directory above use a prefix of two fullstops and repeat this pattern for any such directory.e.g
module4=require("..\\..\\module4")
I have a relatively complicated suite of OMake files designed for cross-compiling on a specific platform. My source is in C++.
I'm building from Windows and I need to pass to the compiler include directories which have spaces in their names. The way that the includes string which is inserted in the command line to compile files is created is by the line:
public.PREFIXED_INCLUDES = $`(addprefix $(INCLUDES_OPT), $(set $(absname $(INCLUDES))))
At some other point in the OMake files I have a line like:
INCLUDES += $(dir "$(LIBRARY_LOCATION)/Path with spaces/include")
In the middle of the command line this expands to:
-IC:\Library location with spaces\Path with spaces\include
I want it to expand to:
-I"C:\Library location with spaces\Path with spaces\include"
I don't want to change anything but the "INCLUDES += ..." line if possible, although modifying something else in that file is also fine. I don't want to have to do something like change the definition of PREFIXED_INCLUDES, as that's in a suite of OMake files which are part of an SDK which may change beneath me. Is this possible? If so, how can I do it? If not, in what ways can I make sure that includes with spaces in them are quoted by modifying little makefile code (hopefully one line)?
The standard library function quote adds escaped quotes around its argument, so it should do the job:
INCLUDES += $(quote $(dir "$(LIBRARY_LOCATION)/Path with spaces/include"))
If needed, see quote in Omake manual.
In case someone else is having the same problem, I thought I'd share the solution I eventually went with, having never figured out how to surround with quotes. Instead of putting quotes around a name with spaces in it I ended up converting the path to the short (8.3) version. I did this via a a simple JScript file called shorten.js and a one line OMake function.
The script:
// Get Access to the file system.
var FileSystemObject = WScript.CreateObject("Scripting.FileSystemObject");
// Get the short path.
var shortPath = FileSystemObject.GetFolder(WScript.Arguments(0)).ShortPath;
// Output short path.
WScript.StdOut.Write(shortPath);
The function:
ShortDirectoryPath(longPath) =
return $(dir $(shell cscript /Nologo $(dir ./tools/shorten.js) "$(absname $(longPath))"))
So now I just use a line like the following for includes:
INCLUDES += $(ShortDirectoryPath $(dir "$(LIBRARY_LOCATION)/Path with spaces/include"))