On a Linux system I have a binary (bin.exe) which needs to read an input file (input.cfg), where the names of other data files (data.txt) are specified. Usually both binary, input file and data files were in the same directory. Now and for organization reasons I need binary file to be in $SOMEPATH/bin and input and data files in $SOMEPATH/input.
I do not know how to do this. If I try
$SOMEPATH/bin/bin.exe $SOMEPATH/input/input.cfg
I get
error, "data.txt" not found
One solution would be to include absolute of relative path of "data.txt" in input.cfg, but the binary does not accept this.
I thought about somehow fooling the binary so that it thinks it is in $SOMEPATH/input, so that I just do
$SOMEPATH/bin/bin.exe input.cfg
and it works, but I do not know whether this is possible. any hints?
(cd $SOMEPATH/input && $SOMEPATH/bin/bin.exe input.cfg)
This is assuming that the program is relying on the current working directory to find the files. If the program is trying hard to find them in the same location as the executable, by consulting /proc/<pid>/exe for example, then you may be out of luck.
Related
I'm trying to use Atom to run a Lua script. However, when I try to load files via the require() command, it always says it's unable to locate them. The files are all in the same folder. For example, to load utils.lua I have tried
require 'utils'
require 'utils.lua'
require 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua'
require 'D:\\Users\\Mike\\Dropbox\\Lua Modeling\\utils.lua'
require 'D:/Users/Mike/Dropbox/Lua Modeling/utils.lua'
I get errors like
Lua: D:\Users\Mike\Dropbox\Lua Modeling\main.lua:12: module 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua' not found:
no field package.preload['D:\Users\Mike\Dropbox\Lua Modeling\utils.lua']
no file '.\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
no file 'D:\Program Files (x86)\Lua\5.1\lua\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
no file 'D:\Program Files (x86)\Lua\5.1\lua\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua\init.lua'
no file 'D:\Program Files (x86)\Lua\5.1\D:\Users\Mike\Dropbox\Lua Modeling\utils\lua.lua'
The messages says on the first line that 'D:\Users\Mike\Dropbox\Lua Modeling\utils.lua' was not found, even though that is the full path of the file. What am I doing wrong?
Thanks.
The short answer
You should be able to load utils.lua by using the following code:
require("utils")
And by starting your program from the directory that utils.lua is in:
cd "D:\Users\Mike\Dropbox\Lua Modeling"
lua main.lua
The long answer
To understand what is going wrong here, it is helpful to know a little bit about how require works. The first thing that require does is to search for the module in the module path. From Programming in Lua chapter 8.1:
The path used by require is a little different from typical paths. Most programs use paths as a list of directories wherein to search for a given file. However, ANSI C (the abstract platform where Lua runs) does not have the concept of directories. Therefore, the path used by require is a list of patterns, each of them specifying an alternative way to transform a virtual file name (the argument to require) into a real file name. More specifically, each component in the path is a file name containing optional interrogation marks. For each component, require replaces each ? by the virtual file name and checks whether there is a file with that name; if not, it goes to the next component. The components in a path are separated by semicolons (a character seldom used for file names in most operating systems). For instance, if the path is
?;?.lua;c:\windows\?;/usr/local/lua/?/?.lua
then the call require"lili" will try to open the following files:
lili
lili.lua
c:\windows\lili
/usr/local/lua/lili/lili.lua
Judging from your error message, your Lua path seems to be the following:
.\?.lua;D:\Program Files (x86)\Lua\5.1\lua\?.lua;D:\Program Files (x86)\Lua\5.1\lua\?\init.lua;D:\Program Files (x86)\Lua\5.1\?.lua
To make that easier to read, here are each the patterns separated by line breaks:
.\?.lua
D:\Program Files (x86)\Lua\5.1\lua\?.lua
D:\Program Files (x86)\Lua\5.1\lua\?\init.lua
D:\Program Files (x86)\Lua\5.1\?.lua
From this list you can see that when calling require
Lua fills in the .lua extension for you
Lua fills in the rest of the file path for you
In other words, you should just specify the module name, like this:
require("utils")
Now, Lua also needs to know where the utils.lua file is. The easiest way is to run your program from the D:\Users\Mike\Dropbox\Lua Modeling folder. This means that when you run require("utils"), Lua will expand the first pattern .\?.lua into .\utils.lua, and when it checks that path it will find the utils.lua file in the current directory.
In other words, running your program like this should work:
cd "D:\Users\Mike\Dropbox\Lua Modeling"
lua main.lua
An alternative
If you can't (or don't want to) change your working directory to run the program, you can use the LUA_PATH environment variable to add new patterns to the path that require uses to search for modules.
set LUA_PATH=D:\Users\Mike\Dropbox\Lua Modeling\?.lua;%LUA_PATH%;
lua "D:\Users\Mike\Dropbox\Lua Modeling\main.lua"
There is a slight trick to this. If the LUA_PATH environment variable already exists, then this will add your project's folder to the start of it. If LUA_PATH doesn't exist, this will add ;; to the end, which Lua fills in with the default path.
I able to load libpcap.dylib which is confusing cause I can't figure out the actual file location. Doing find / -name libpcap.A.dylib or libpcap.dylib says no such file.
Also finder search with libpcap just results in libpcap.A.tbd and libpcap.rb.
libpcap.A.tbd shows "Install location /usr/lib/libpcap.A.dylib", but it does not actually exist there.
I wanted to locate the actual dylib file cause I running into issue with being able to import function, So I wanted to check file to make sure I have function names correct.
So I wanted to check file to make sure I have function names correct.
The first thing to check is the pcap man page - from the command line, it'd be
man pcap
It's a bit long, but it should mention all the functions available in libpcap; it may be easier than
nm /usr/lib/libpcap.dylib | egrep ' T '
(and doesn't require you to remember that the leading underscores in the output of that command are NOT part of the name of the function, they're a leftover from ancient UNIX history).
Where is libpcap.dylib?
/usr/lib/libpcap.A.dylib. /usr/lib/libpcap.dylib is a symbolic link to it.
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.
How do . and .., as paths (vs. ranges, e.g., {1..10}, which I'm not concerned with), really work? I know what they do, and use them all the time, but don't fully grasp how/where they're interpreted. Does the shell handle them? The interpreting process? The OS?
The reason why I'm asking is that I'd like to be able to use ... to refer to ../.., .... to refer to ../../.., etc. (up to some small finite number; I don't need bash to process an arbitrarily large number of dots). I.e., if my current directory is /tmp/let/me/out, and I call cd ..., my resulting current directory should be /tmp/let. I don't particularly care if ... etc. show up in ls -a output like . and .. do, but I would like to be able to call cat /tmp/let/me/out/..../phew.txt to print the contents of /tmp/phew.txt.
Pointers to relevant documentation appreciated as well as direct answers. This kind of syntax question is very hard to Google.
I'm using bash 4.3.42, by the way, with the autocd and globstar shell options.
. and .. are genuine directory names. They are not "sort-cuts", aliases, or anything fake.
They happen to point to the same inode as the other name you use. A file or directory can have several names pointing to the same inode, these are usually known as hard links, to distinguish them from symbolic (or soft) links.
If you are on Linux or OS X you can use stat to look at most of the inode metadata - it is what ls looks at. You will see there is an inode number. If you stat . and stat current-directory-name you will see that number is the same.
The one thing that is not held in the inode is the filename - that is held in the directory.
So . and .. reside in the directory on the file system, they are not a figment of the shell's imagination. So, for example, I can use . and .. quite happily from C.
I doubt you can change them - personally I have never tried and I never will. You would have to change what these filenames linked to by editing the directory. If you managed it you would probably do irreparable damage to your file system.
I write this to clarify what has already been written before.
In many file systems a DIRECTORY is a file; a special type of file that the file system identifies as being distinctly a directly.
A directory file contains a list of names that map to files on the disk
A file, including a directly does not have an intrinsic name associated with it (not true in all file systems). The name of a file exists only in a directory.
The same file can have an entry in multiple directories (hard link). The same file can then have multiple names and multiple paths.
The file system maintains in every directory entries for "." and ".."
In such file systems there are always directory ENTRIES for the NAMES "." and "..". These entries are maintained by the file system.
The name "." links to its own directory.
The name ".." links to the parent directory EXCEPT for the top level directory where it links to itself (. and .. thus link to the same directory file).
So when you use "." and ".." as in /dir1/dir2/../dir3/./dir4/whatever,
"." and ".." are processed in the exact same way as "dir1" and "dir2".
This translation is done by the file system; not the shell.
cd ...
Does not work because there is no entry for "..." (at least not normally).
You can create a directory called "..." if you want.
You can actually achieve something like this, though this is an ugly hack:
You can run a command before every command entered to bash, and after every command. For that you trap the DEBUG pseudo signal and set a command to PROMPT_COMMAND, respectively.
trap 'ln -s ../.. ... &>/dev/null | true' DEBUG
PROMPT_COMMAND='rm ...'
With this, it seems like there's an additional entry in the current directory:
pwd
# /tmp/crazy-stuff
ls -a
# . .. ... foo
ls -a .../tmp/crazy-stuff
# . .. ... foo
Though this only works in the current directory, because the symbolic links is deleted after each command invokation. Thus ls foo/bar/... won't work this way.
Another ugly hack would be to "override" mkdir such that it populates every new directory with these symbolic links.
See also the comments on the second answer here, particularly Eliah's: https://askubuntu.com/questions/327126/what-is-a-dot-only-named-folder
Much in the same way that when you cd into some directory subdir, you're actually following a pointer that points to that directory, .. is a pointer added by the OS that points to the parent directory, and I'd imagine . works the same way.
Is it possible to tell ASDF that it should produce only one fas(l) file for entire system? This file should be concatenation (in right order) of all compiled files of the system, including all files of systems on which target system depends.
Yes, with compile-bundle-op (ASDF 3.1): http://common-lisp.net/project/asdf/asdf/Predefined-operations-of-ASDF.html
edit: Actually, monolithic-compile-bundle-op seemes to be asked for (as shown in other answers).
If you have to predict the extension, use uiop:compile-file-type.
And/or you can just call (asdf:output-files 'asdf:monolithic-compile-bundle-op :my-system) to figure out what is actually used.
Option monolithic-compile-bundle-op will create single compiled file which includes all dependencies, while compile-bundle-op creates a file for every system.
Example of use:
(asdf:operate 'asdf:monolithic-compile-bundle-op :my-system)
This command will create file my-system--all-systems.fas(l) in output directory of target project, as well as "bundle" files for every system, named like my-system--system.fas(l).