I want to save my files in specific path..
I have used like this
file_name = gets
F = open.(Dir.pwd, /data/folder /#{#file_name },w+)
I'm not sure whether the above line is correct or not! Where Dir.pwd tell the directory path followed by my folder path and the file name given.
It should get store the value on the specific path with the specific file name given. Can anyone tell me how to do that.
Your code has multiple errors. Have you ever tried to execute the script?
Your script ends with:
test.rb:7: unknown regexp options - fldr
test.rb:7: syntax error, unexpected end-of-input
F = open.(Dir.pwd, /data/folder /#{#file_name },w+)
First: You need to define the strings with ' or ":
file_name = gets
F = open.(Dir.pwd, "/data/folder/#{#file_name}","w+")
Some other errors:
You use file_name and later #file_name.
The open method belongs to File and needs two parameters.
The file is defined as a constant F. I would use a variable.
The path must be concatenated. I'd use File.join for it.
You don't close the file.
After all these changes you get:
file_name = gets
f = File.open(File.join(Dir.pwd, "/data/folder/#{file_name}"),"w+")
##
f.close
and the error:
test.rb:29:in `initialize': No such file or directory # rb_sysopen - C:/Temp/data/folder/sdssd (Errno::ENOENT)
The folder must exist, so you must create it first.
Now the script looks like:
require 'fileutils'
dirname = "data/folder"
file_name = gets.strip
FileUtils.mkdir_p(dirname) unless Dir.exists?(dirname)
f = File.open(File.join(Dir.pwd, dirname, file_name),"w+")
##fill the content
f.close
Related
Silly question, but I want to do some processing on a dataset and put them into different CSVs, like UDID1.csv, UDID2.csv, ..., UDID1000.csv. So this is my code:
for i in 1..1000
logfile = File.new('C:\Users\hp1\Desktop\Datasets\New File\UDID#{i}\.csv',"a")
#I'll do some processing here
end
But the program throws an error when running because of the UDID#{i} part. So, how to overcome this issue? Thanks.
Edit: This is the error:
in `initialize': No such file or directory # rb_sysopen - C:\Users\hp1\Desktop\Datasets\New File\udid#{1}\.csv (Errno::ENOENT)from C:/Ruby21/bin/hashedUDID.rb:38:in `new' from C:/Ruby21/bin/hashedUDID.rb:38:in '<main>'
The ' is one problem, another problem is the path.
In your posting the New File must exist as a directory. Inside this directory must exist another directories like UDID0001. This gets a .csv file.
Correct is (I don't use the non-rubyesk for-loop):
1.upto(1000) do |i|
logfile = File.new("C:\\Users\\hp1\\Desktop\\Datasets\\UDID#{i}.csv", "a")
#I'll do some processing here
logfile.close #Don't forget to close the file
end
Inside " the backslash must be masked (\\). Instead you may use /:
logfile = File.new("C:/Users/hp1/Desktop/Datasets/New File/UDID#{i}/.csv", "a")
Another possibility is the usage of %i to insert the number:
logfile = File.new("C:/Users/hp1/Desktop/Datasets/New File/UDID%02i/.csv" % i, "a")
I prefer to use open, then the file is closed with the end of the block:
File.open("C:/Users/hp1/Desktop/Datasets/New File/UDID%04i/.csv" % i, "a") do |logfile|
#I'll do some processing here
end #closes the file
Warning:
I'm not sure, if you really want to create 1000 log files (The File is opened inside the loop. so each step creates a file.).
If yes, then the %04i-version has the advantage, that the files get all the same number of digits (starting with 0001 and ending with 1000).
(1..10).each { |i| logfile = File.new("/base/path/UDID#{i}.csv") }
You must use double quote (") when you need string interpolation.
#{} can only be used in strings with double quotes ". So change your code to:
for i in 1..1000
logfile = File.new("C:\Users\hp1\Desktop\Datasets\New File\UDID#{i}\.csv","a")
# other stuff
end
I am trying to edit particular html files that I download in python. I am running into a problem where I run my code to edit the file and my python context locks up. I checked the file it's writing to and found that there are two files. The html file and a .bak file.
The html file starts out at 0kb and the .bak file constantly grows to a point, maybe 12 mb or so, then the .html file will grow to a larger size, then the .bak file will grow again. This seems to cycle endlessly. The html file I am editing is 22kb. I watched the output file grow to a gig once just to see if it would stop... It doesn't.
Here is the function I am using to edit the file:
def replace(self, search_str, replace_str):
f = open(self.path,'r+')
content = f.readlines()
for i, line in enumerate(content):
content[i] = line.replace(search_str, replace_str)
f.writelines(content)
f.close()
The issue, I imagine relates to the fact that the html file, as downloaded, is mostly in a single line with ~ 21,000 characters in it. Any ideas?
edit:
I have also tried another function, but get the same result:
def replace(self, search_str, replace_str):
assert self.path != None, 'No file path provided.'
fi = fileinput.FileInput(self.path,inplace=1)
for line in fi:
if search_str in line:
line=line.replace(search_str,replace_str)
print line
fi.close()
Try using generator. Thats the way to go if you need to read a large file
for line in open(self.path,'r+'):
# do stuff with line
I re-wrote the function to write everything out to a new file and it works.
def replace(self, search_str, replace_str):
f = open(self.path,'r+')
new_path = self.path.split('.')[0]+'.TEMP'
new_f = open(new_path,'w')
new_lines = [x.replace(search_str, replace_str) for x in f]
new_f.writelines(new_lines)
f.close()
new_f.close()
os.remove(self.path)
os.rename(new_path, self.path)
I am trying to read the contents of the file from a local disk as follows :
content = File.read("C:\abc.rb","r")
when I execute the rb file I get an exception as Error: No such file or directory .What am I missing in this?
In a double quoted string, "\a" is a non-printable bel character. Similar to how "\n" is a newline. (I think these originate from C)
You don't have a file with name "C:<BEL>bc.rb" which is why you get the error.
To fix, use single quotes, where these interpolations don't happen:
content = File.read('C:\abc.rb')
content = File.read("C:\/abc.rb","r")
First of all:
Try using:
Dir.glob(".")
To see what's in the directory (and therefore what directory it's looking at).
open("C:/abc.rb", "rb") { |io| a = a + io.read }
EDIT: Unless you're concatenating files together, you could write it as:
data = File.open("C:/abc.rb", "rb") { |io| io.read }
I would like to make a Ruby program which sorts the images in the current directory into different subfolders, for example:
tree001.jpg, ... tree131.jpg -> to folder "tree"
apple01, ... apple20.jpg -> to folder "apple"
plum1.jpg, plum2.jpg, ... plum33.jpg -> to folder "plum"
and so on, the program should automagically recognize which files belong together by their names. I have no clue how to achive this. Till now I make a small program which collect the files with command "Dir" into an array and sort it alphabetically to help finding the appropriate classes by the file names. Does anybody have a good idea?
Check out Find:
http://www.ruby-doc.org/stdlib-2.0/libdoc/find/rdoc/Find.html
Or Dir.glob:
http://ruby-doc.org/core-2.0/Dir.html#method-c-glob
For instance:
Dir.glob("*.jpg")
will return an array that you can iterate with each.
I'd go about it something like this:
files = %w[
tree001.jpg tree03.jpg tree9.jpg
apple1.jpg apple002.jpg
plum3.jpg plum300.jpg
].shuffle
# => ["tree001.jpg", "apple1.jpg", "tree9.jpg", "plum300.jpg", "apple002.jpg", "plum3.jpg", "tree03.jpg"]
grouped_files = files.group_by{ |fn| fn[/^[a-z]+/i] }
# => {"tree"=>["tree001.jpg", "tree9.jpg", "tree03.jpg"], "apple"=>["apple1.jpg", "apple002.jpg"], "plum"=>["plum300.jpg", "plum3.jpg"]}
grouped_files.each do |grp, files|
Dir.mkdir(grp) unless Dir.exist(grp)
files.each { |f| FileUtils.mv(f, "#{grp}/#{f}") }
end
I can't test that because I don't have all the files, nor am I willing to generate them.
The important thing is group_by. It makes it easy to group the similarly named files, making it easy to walk through them.
For your case, you'll want to replace the assignment to files with Dir.glob(...) or Dir.entries(...) to get your list of files.
If you want to separate the file path from the file name, look at File.split or File.dirname and File.basename:
File.split('/path/to/foo')
=> ["/path/to", "foo"]
File.dirname('/path/to/foo')
=> "/path/to"
File.basename('/path/to/foo')
=> "foo"
Assuming every file name starts with non-digit characters followed by at least one digit character, and the initial non-digit characters define the directory you want the file moved to:
require 'fileutils'
Dir.glob("*").select{|f| File.file? f}.each do |file| # For each regular file
dir = file.match(/[^\d]*/).to_s # Determine destination directory
FileUtils.mkdir_p(dir) # Make directory if necessary
FileUtils.mv(file, dir) # Move file
end
The directories are created if necessary. You can run it again after adding files. For example, if you added the file tree1.txt later and re-ran this, it would be moved to tree/ where tree001.jpg through tree131.jpg already are.
Update: In the comments, you added the requirement that you only want to do this for files which form groups of at least 10. Here's one way to do that:
require 'fileutils'
MIN_GROUP_SIZE = 10
reg_files = Dir.glob("*").select{|f| File.file? f}
reg_files.group_by{|f| f.match(/[^\d]*/).to_s}.each do |dir, files|
next if files.size < MIN_GROUP_SIZE
FileUtils.mkdir_p(dir)
files.each do |file|
FileUtils.mv(file, dir)
end
end
I'm writing my first Rakefile. The first things that I see in the doc is "there is no special format for a Rakefile" and "there is no special syntax in a Rakefile".
Ok, so I had to come up with something on my own, but I can see at least two problems with my creature:
1) I need to create a number of folders, five of them. The sequence of 6 directory tasks looks a bit weird. The list of 5 dependencies in deploy task looks even more weird. Can I shrink it down to one line somehow?
2) I need to repeat my directory name literals two times - when I define their deployment paths and when I copy the contents. Can I avoid that without introducing 5 more variables?
In Java Ant I would create a properties file with all name literals - can I do that with Rake?
This is what I've got:
WEBAPPSDIR = '/var/webapps/'
WEBAPPNAME = 'foo.local'
WEBAPPDIR = File.join(WEBAPPSDIR, WEBAPPNAME)
VIEWSDIR = File.join(WEBAPPDIR, 'views')
PUBLICDIR = File.join(WEBAPPDIR, 'public')
CSSDIR = File.join(PUBLICDIR, 'css')
IMAGESDIR = File.join(PUBLICDIR, 'images')
TMPDIR = File.join(WEBAPPDIR, 'tmp')
HTMLDIR = File.join(PUBLICDIR, 'html')
directory VIEWSDIR
directory CSSDIR
directory HTMLDIR
directory IMAGESDIR
directory TMPDIR
desc 'Deploy to webapps dir'
task :deploy => [VIEWSDIR, CSSDIR, IMAGESDIR, TMPDIR, HTMLDIR] do
cp 'config.ru', WEBAPPDIR
Dir.glob('*.rb') {|f| cp f, WEBAPPDIR}
Dir.glob('views/*.{mab,str}') {|f| cp f, VIEWSDIR}
Dir.glob('css/*.css') {|f| cp f, CSSDIR}
Dir.glob('images/*.{png,jpg,gif}') {|f| cp f, IMAGESDIR}
Dir.glob('html/*.html') {|f| cp f, VIEWSDIR}
end
desc 'Cleans webapp dir'
task :clean do
rm_r WEBAPPDIR, {force: true}
end
Other thoughts/links/examples are welcome too.
This does not really answer your question - but why don't you use capistrano ? If you don't know it already, it's a ruby tool frequently used to handle deployments smoothly