I written a Ruby tool, named foobar, having a default configuration into a
file (called .foobar).
When the tool is executed without any params, the configuration file's params
can be used: ~/.foobar.
But, if the current tool's path is ~/projects/foobar and if
~/projects/foobar/.foobar exists, then this file should be used instead of
~/.foobar.
That's why the way to look for this configuration file should start from the
current folder until the current user folder.
Is there a simple way to look for this file?
cfg_file = File.open(".foobar", "r") rescue File.open("~/.foobar", "r")
Although to be honest, I almost always do two things: provide an option for a config file path, and allow overriding of default config values. The problem with the latter is that unless you know config files are ordered/have precedence, it can be confusing.
I would do this:
if File.exists(".foobar")
# open .foobar
else
# open ~/..
end
Related
Oracle's sqlldr defaults to a .dat extension. That I want to override. I don't like to rename the file. When googled get to know few answers to use . like data='fileName.' which is not working. Share your ideas, please.
Error message is fileName.dat is not found.
Sqlloder has default extension for all input files data,log,control...
data= .dat
log= .log
control = .ctl
bad =.bad
PARFILE = .par
But you have to pass filename without apostrophe and dot
sqlloder pass/user#db control=control data=data
sqloader will add extension. control.ctl data.dat
Nevertheless i do not understand why you do not want to specify extension?
You can't, at least in Unix/Linux environments. In Windows you can use the trailing period trick, specifying either INFILE 'filename.' in the control file or DATA=filename. on the command line. WIndows file name handling allows that; you can for instance do DIR filename. at a command prompt and it will list the file with no extension (as will DIR filename). But you can't do that with *nix, from a shell prompt or anywhere else.
You said you don't want to copy or rename the file. Temporarily renaming it might be the simplest solution, but as you may have a reason not to do that even briefly you could instead create a hard or soft link to the file which does have an extension, and use that link as the target instead. You could wrap that in a shell script that takes the file name argument:
# set variable from correct positional parameter; if you pass in the control
# file name or other options, this might not be $1 so adjust as needed
# if the tmeproary file won't be int he same directory, need to be full path
filename=$1
# optionally check file exists, is readable, etc. but overkill for demo
# can also check temporary file does not already exist - stop or remove
# create soft link somewhere it won't impact any other processes
ln -s ${filename} /tmp/${filename##*/}.dat
# run SQL*Loader with soft link as target
sqlldr user/password#db control=file.ctl data=/tmp/${filename##*/}.dat
# clean up
rm -f /tmp/${filename##*/}.dat
You can then call that as:
./scriptfile.sh /path/to/filename
If you can create the link in the same directory then you only need to pass the file, but if it's somewhere else - which may be necessary depending on why renaming isn't an option, and desirable either way - then you need to pass the full path of the data file so the link works. (If the temporary file will be int he same filesystem you could use a hard link, and you wouldn't have to pass the full path then either, but it's still cleaner to do so).
As you haven't shown your current command line options you may have to adjust that to take into account anything else you currently specify there rather than in the control file, particularly which positional argument is actually the data file path.
I have the same issue. I get a monthly download of reference data used in medical application and the 485 downloaded files don't have file extensions (#2gb). Unless I can load without file extensions I have to copy the files with .dat and load from there.
I am trying to read a list of baby names from the year 1880 in CSV format. My program, when run in the terminal on OS X returns an error indicating yob1880.txt doesnt exist.
No such file or directory # rb_sysopen - /names/yob1880.txt (Errno::ENOENT)
from names.rb:2:in `<main>'
The location of both the script and the text file is /Users/*****/names.
lines = []
File.expand_path('../yob1880.txt', __FILE__)
IO.foreach('../yob1880.txt') do |line|
lines << line
if lines.size >= 1000
lines = FasterCSV.parse(lines.join) rescue next
store lines
lines = []
end
end
store lines
If you're running the script from the /Users/*****/names directory, and the files also exist there, you should simply remove the "../" from your pathnames to prevent looking in /Users/***** for the files.
Use this approach to referencing your files, instead:
File.expand_path('yob1880.txt', __FILE__)
IO.foreach('yob1880.txt') do |line|
Note that the File.expand_path is doing nothing at the moment, as the return value is not captured or used for any purpose; it simply consumes resources when it executes. Depending on your actual intent, it could realistically be removed.
Going deeper on this topic, it may be better for the script to be explicit about which directory in which it locates files. Consider these approaches:
Change to the directory in which the script exists, prior to opening files
Dir.chdir(File.dirname(File.expand_path(__FILE__)))
IO.foreach('yob1880.txt') do |line|
This explicitly requires that the script and the data be stored relative to one another; in this case, they would be stored in the same directory.
Provide a specific path to the files
# do not use Dir.chdir or File.expand_path
IO.foreach('/Users/****/yob1880.txt') do |line|
This can work if the script is used in a small, contained environment, such as your own machine, but will be brittle if it data is moved to another directory or to another machine. Generally, this approach is not useful, except for short-lived scripts for personal use.
Never put a script using this approach into production use.
Work only with files in the current directory
# do not use Dir.chdir or File.expand_path
IO.foreach('yob1880.txt') do |line|
This will work if you run the script from the directory in which the data exists, but will fail if run from another directory. This approach typically works better when the script detects the contents of the directory, rather than requiring certain files to already exist there.
Many Linux/Unix utilities, such as cat and grep use this approach, if the command-line options do not override such behavior.
Accept a command-line option to find data files
require 'optparse'
base_directory = "."
OptionParser.new do |opts|
opts.banner = "Usage: example.rb [options]"
opts.on('-d', '--dir NAME', 'Directory name') {|v| base_directory = Dir.chdir(File.dirname(File.expand_path(v))) }
end
IO.foreach(File.join(base_directory, 'yob1880.txt')) do |line|
# do lines
end
This will give your script a -d or --dir option in which to specify the directory in which to find files.
Use a configuration file to find data files
This code would allow you to use a YAML configuration file to define where the files are located:
require 'yaml'
config_filename = File.expand_path("~/yob/config.yml")
config = {}
name = nil
config = YAML.load_file(config_filename)
base_directory = config["base"]
IO.foreach(File.join(base_directory, 'yob1880.txt')) do |line|
# do lines
end
This doesn't include any error handling related to finding and loading the config file, but it gets the point across. For additional information on using a YAML config file with error handling, see my answer on Asking user for information, and never having to ask again.
Final thoughts
You have the tools to establish ways to locate your data files. You can even mix-and-match solutions for a more sophisticated solution. For instance, you could default to the current directory (or the script directory) when no config file exists, and allow the command-line option to manually override the directory, when necessary.
Here's a technique I always use when I want to normalize the current working directory for my scripts. This is a good idea because in most cases you code your script and place the supporting files in the same folder, or in a sub-folder of the main script.
This resets the current working directory to the same folder as where the script is situated in. After that it's much easier to figure out the paths to everything:
# Reset working directory to same folder as current script file
Dir.chdir(File.dirname(File.expand_path(__FILE__)))
After that you can open your data file with just:
IO.foreach('yob1880.txt')
I want to delete the line configuration (This word "configuration" is inside the angular brackets in the file "hdfs-site.xml"). I tried using this code but no luck.
ruby_block "delete_lines_in_hdfs_site" do
block do
file = Chef::Util::FileEdit.new("/opt/hadoop-2.4.1/etc/hadoop/hdfs-site.xml")
file.search_file_delete_line(/<configuration>/)
end
end
You need to end the ruby block with
file.write_file
Otherwise the changes are not written out.
While I agree that using a template is preferred and considered best practices, there are times you need to edit an existing file which may have variations and not fit the template model.
You do not add to or remove lines from files with chef. Instead you replace the configuration file with yours, which you put into your cookbook under files/ or templates/ forlder.
template "/opt/hadoop-2.4.1/etc/hadoop/hdfs-site.xml"
or
cookbook_file "/opt/hadoop-2.4.1/etc/hadoop/hdfs-site.xml"
When you just add/replace lines in config files, you cannot be sure, that after upgrade of installed software your config files are right.
I'm using Lua in interactive mode on a Mac (thanks to rudix.org).
When I want to load a file I do:
dofile("/my/long/path/to/my/directory/file.lua")
I want to do a different thing, that is:
put all my files in a desktop directory myDirectory;
then call the file from the terminal this way dofile("file.lua");
Is this possible? How?
If the path is fixed, you can just redefine dofile:
local _dofile=dofile
local path=("/my/long/path/to/my/directory/")
function dofile(x)
return _dofile(path..x)
end
You may put this (and other initializations) in a file and set the environment variable LUA_INIT to its location. After this, every invocation of lua will see the version of dofile redefined above and the users will be able to say simply dofile("foo.lua").
Alternatively, you can use require, which looks for modules in a list of paths in package.path or LUA_PATH.
So, I have a function that creates an object specifying user data. Then, using the Ruby YAML gem and some code, I put the object to a YAML file and save it. This saves the YAML file to the location where the Ruby script was run from. How can I tell it to save to a certain file directory? (A simplified version of) my code is this
print "Please tell me your name: "
$name=gets.chomp
$name.capitalize!
print "Please type in a four-digit PIN number: "
$pin=gets.chomp
I also have a function that enforces that the pin be a four-digit integer, but that is not important.
Then, I add this to an object
new_user=Hash.new (false)
new_user["name"]=$name
new_user["pin"]=$pin
and then add it to a YAML file and save it. If the YAML file doesn't exist, one is created. It creates it in the same file directory as the script is run in. Is there a way to change the save location?
The script fo save the object to a YAML file is this.
def put_to_yaml (new_user)
File.write("#{new_user["name"]}.yaml", new_user.to_yaml)
end
put_to_yaml(new_user)
Ultimately, the question is this: How can I change the save location of the file? And when I load it again, how can i tell it where to get the file from?
Thanks for any help
Currently when you use File.write it takes your current working directory, and appends the file name to that location. Try:
puts Dir.pwd # Will print the location you ran ruby script from.
You can specify the absolute path if you want to write it in a specific location everytime:
File.write("/home/chameleon/different_location/#{new_user["name"]}.yaml")
Or you can specify a relative path to your current working directory:
# write one level above your current working directory
File.write("../#{new_user["name"]}.yaml", new_user.to_yaml)
You can also specify relative to your current executing ruby file:
file_path = File.expand_path(File.dirname(__FILE__))
absolute_path = File.join(file_path, file_name)
File.write(absolute_path, new_user.to_yaml)
You are supplying a partial pathname (a mere file name), so we read and write from the current directory. Thus you have two choices:
Supply a full absolute pathname (personally, I like to use the Pathname class for this); or
Change the current directory first (with Dir.chdir)