Let me first describe my situation, I am working on a Linux platform and have a collection of .bmp files that add one to the picture number from filename0022.bmp up to filename0680.bmp. So a total of 658 pictures. I want to be able to run each of these pictures through a .exe file that operates on the picture then kicks out the file to a file specified by the user, it also has some threshold arguments: lower, upper. So the typical call for the executable is:
./filter inputfile outputfile lower upper
Is there a way that I can loop this call over all the files just from the terminal or by creating some kind of bash script? My problem is similar to this: Execute a command over multiple files with a batch file but this time I am working in a Linux command line terminal.
You may be interested in looking into bash scripting.
You can execute commands in a for loop directly from the shell.
A simple loop to generate the numbers you specifically mentioned. For example, from the shell:
user#machine $ for i in {22..680} ; do
> echo "filename${i}.bmp"
> done
This will give you a list from filename22.bmp to filename680.bmp. That simply handles the iteration of the range you had mentioned. This doesn't cover zero padding numbers. To do this you can use printf. The printf syntax is printf format argument. We can use the $i variable from our previous loop as the argument and apply the %Wd format where W is the width. Prefixing the W placeholder will specify the character to use. Example:
user#machine $ for i in {22..680} ; do
> echo "filename$(printf '%04d' $i).bmp"
> done
In the above $() acts as a variable, executing commands to obtain the value opposed to a predefined value.
This should now give you the filenames you had specified. We can take that and apply it to the actual application:
user#machine $ for i in {22..680} ; do
> ./filter "filename$(printf '%04d' $i).bmp" lower upper
> done
This can be rewritten to form one line:
user#machine $ for i in {22..680} ; do ./filter "filename$(printf '%04d' $i).bmp" lower upper ; done
One thing to note from the question, .exe files are generally compiled in COFF format where linux expects an ELF format executable.
here is a simple example:
for i in {1..100}; do echo "Hello Linux Terminal"; done
to append to a file:(>> is used to append, you can also use > to overwrite)
for i in {1..100}; do echo "Hello Linux Terminal" >> file.txt; done
You can try something like this...
#! /bin/bash
for ((a=022; a <= 658 ; a++))
do
printf "./filter filename%04d.bmp outputfile lower upper" $a | "sh"
done
You can leverage xargs for iterating:
ls | xargs -i ./filter {} {}_out lower upper
Note:
{} corresponds to one line output from the pipe, here it's the inputfile name.
Output files wouldbe named with postfix '_out'.
You can test that AS-IS in your shell :
for i in *; do
echo "$i" | tr '[:lower:]' '[:upper:]'
done
If you have a special path, change * by your path + a glob : Ex :
for i in /home/me/*.exe; do ...
See http://mywiki.wooledge.org/glob
This while prepend the name of the output images like filtered_filename0055.bmp
for i in *; do
./filter $i filtered_$i lower upper
done
Related
I have a file called files.txt and I need to split it based on lines. The command is as follows -
split -l 1 files.txt file --numeric-suffixes=1 --suffix-length=4
The numeric suffixes here start from file0001 to file9000. But I want it to be from 1 to 9000.
I can't seem to change it when --suffix-length=1, as split exhausted output filenames. Any suggestions using the same split command?
I don't think split will do what you want it to do, though I'm on macOS, so the *nix I'm using is Darwin not Linux; however, a simple shell script would do the trick:
#!/bin/bash
N=1
cat $1 | while read line
do
echo "$line" > file$N
N=`expr $N + 1`
done
Assuming you save it as mysplit (don't forget chmod -x mysplit), then you run it:
./mysplit files.txt
I want to have a command in a variable that runs a program and specifies the output filename for it depending on the number of files exits (to work on a new file each time).
Here is what I have:
export MY_COMMAND="myprogram -o ./dir/outfile-0.txt"
However I would like to make this outfile number increases each time MY_COMMAND is being executed. You may suppose myprogram creates the file soon enough before the next call. So the number can be retrieved from the number of files exists in the directory ./dir/. I do not have access to change myprogram itself or the use of MY_COMMAND.
Thanks in advance.
Given that you can't change myprogram — its -o option will always write to the file given on the command line, and assuming that something also out of your control is running MY_COMMAND so you can't change the way that MY_COMMAND gets called, you still have control of MY_COMMAND
For the rest of this answer I'm going to change the name MY_COMMAND to callprog mostly because it's easier to type.
You can define callprog as a variable as in your example export callprog="myprogram -o ./dir/outfile-0.txt", but you could instead write a shell script and name that callprog, and a shell script can do pretty much anything you want.
So, you have a directory full of outfile-<num>.txt files and you want to output to the next non-colliding outfile-<num+1>.txt.
Your shell script can get the numbers by listing the files, cutting out only the numbers, sorting them, then take the highest number.
If we have these files in dir:
outfile-0.txt
outfile-1.txt
outfile-5.txt
outfile-10.txt
ls -1 ./dir/outfile*.txt produces the list
./dir/outfile-0.txt
./dir/outfile-1.txt
./dir/outfile-10.txt
./dir/outfile-5.txt
(using outfile and .txt means this will work even if there are other files not name outfile)
Scrape out the number by piping it through the stream editor sed … capture the number and keep only that part:
ls -1 ./dir/outfile*.txt | sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:'
(I'm using colon : instead of the standard slash / so I don't have to escape the directory separator in dir/outfile)
Now you just need to pick the highest number. Sort the numbers and take the top
| sort -rn | head -1
Sorting with -n is numeric, not lexigraphic sorting, -r reverses so the highest number will be first, not last.
Putting it all together, this will list the files, edit the names keeping only the numeric part, sort, and get just the first entry. You want to assign that to a variable to work with it, so it is:
high=$(ls -1 ./dir/outfile*.txt | sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:' | sort -rn | head -1)
In the shell (I'm using bash) you can do math on that, $[high + 1] so if high is 10, the expression produces 11
You would use that as the numeric part of your filename.
The whole shell script then just needs to use that number in the filename. Here it is, with lines broken for better readability:
#!/bin/sh
high=$(ls -1 ./dir/outfile*.txt \
| sed -e 's:^.*dir/outfile-\([0-9][0-9]*\)\.txt$:\1:' \
| sort -rn | head -1)
echo "myprogram -o ./dir/outfile-$[high + 1].txt"
Of course you wouldn't echo myprogram, you'd just run it.
you could do this in a bash function under your .bashrc by using wc to get the number of files in the dir and then adding 1 to the result
yourfunction () {
dir=/path/to/dir
filenum=$(expr $(ls $dir | wc -w) + 1)
myprogram -o $dir/outfile-${filenum}.txt
}
this should get the number of files in $dir and append 1 to that number to get the number you need for the filename. if you place it in your .bashrc or under .bash_aliases and source .bashrc then it should work like any other shell command
You can try exporting a function for MY_COMMAND to run.
next_outfile () {
my_program -o ./dir/outfile-${_next_number}.txt
((_next_number ++ ))
}
export -f next_outfile
export MY_COMMAND="next_outfile" _next_number=0
This relies on a "private" global variable _next_number being initialized to 0 and not otherwise modified.
#! /bin/bash
for i in $(ls);
do
j=1
echo "$i"
not expected Output:-
autodeploy
bin
config
console-ext
edit.lok
need Output like below if give input 2 it should print "bin" based on below condition, but I want out put like Directory list
1.)autodeploy
2.)bin
3.)config
4.)console-ext
5.)edit.lok
and if i like as input:- 2 then it should print "bin"
Per BashFAQ #1, a while read loop is the correct way to read content line-by-line:
#!/usr/bin/env bash
enumerate() {
local line i
i=0
while IFS= read -r line; do
((++i))
printf '%d.) %s\n' "$i" "$line"
done
}
ls | enumerate
However, ls is not an appropriate tool for programmatic use; the above is acceptable if the results of ls are only for human consumption, but not if they're going to be parsed by a machine -- see Why you shouldn't parse the output of ls(1).
If you want to list files and let the user choose among them by number, pass the results of a glob expression to select:
select filename in *; do
echo "$filename" && break
done
I don't understand what you mean in your question by like Directory list, but following your example, you do not need to write a loop:
ls|nl -s '.)' -w 1
If you want to avoid ls, you can do the following (but be careful - this only works if the directory entries do not contain white spaces (because this would make fmt to break them into two lines):
echo *|fmt -w 1 |nl -s '.)' -w 1
I am trying to cat a file.txt and loop it twice through the whole content and copy it to a new file file_new.txt. The bash command I am using is as follows:
for i in {1..3}; do cat file.txt > file_new.txt; done
The above command is just giving me the same file contents as file.txt. Hence file_new.txt is also of the same size (1 GB).
Basically, if file.txt is a 1GB file, then I want file_new.txt to be a 2GB file, double the contents of file.txt. Please, can someone help here? Thank you.
Simply apply the redirection to the for loop as a whole:
for i in {1..3}; do cat file.txt; done > file_new.txt
The advantage of this over using >> (aside from not having to open and close the file multiple times) is that you needn't ensure that a preexisting output file is truncated first.
Note that the generalization of this approach is to use a group command ({ ...; ...; }) to apply redirections to multiple commands; e.g.:
$ { echo hi; echo there; } > out.txt; cat out.txt
hi
there
Given that whole files are being output, the cost of invoking cat for each repetition will probably not matter that much, but here's a robust way to invoke cat only once:[1]
# Create an array of repetitions of filename 'file' as needed.
files=(); for ((i=0; i<3; ++i)); do files[i]='file'; done
# Pass all repetitions *at once* as arguments to `cat`.
cat "${files[#]}" > file_new.txt
[1] Note that, hypothetically, you could run into your platform's command-line length limit, as reported by getconf ARG_MAX - given that on Linux that limit is 2,097,152 bytes (2MB) that's not likely, though.
You could use the append operator, >>, instead of >. Then adjust your loop count as needed to get the output size desired.
You should adjust your code so it is as follows:
for i in {1..3}; do cat file.txt >> file_new.txt; done
The >> operator appends data to a file rather than writing over it (>)
if file.txt is a 1GB file,
cat file.txt > file_new.txt
cat file.txt >> file_new.txt
The > operator will create file_new.txt(1GB),
The >> operator will append file_new.txt(2GB).
for i in {1..3}; do cat file.txt >> file_new.txt; done
This command will make file_new.txt(3GB),because for i in {1..3} will run three times.
As others have mentioned, you can use >> to append. But, you could also just invoke cat once and have it read the file 3 times. For instance:
n=3; cat $( yes file.txt | sed ${n}q ) > file_new.txt
Note that this solution exhibits a common anti-pattern and fails to properly quote the arguments, which will cause issues if the filename contains whitespace. See mklement's solution for a more robust solution.
Hi… Need a little help here…
I tried to emulate the DOS' dir command in Linux using bash script. Basically it's just a wrapped ls command with some parameters plus summary info. Here's the script:
#!/bin/bash
# default to current folder
if [ -z "$1" ]; then var=.;
else var="$1"; fi
# check file existence
if [ -a "$var" ]; then
# list contents with color, folder first
CMD="ls -lgG $var --color --group-directories-first"; $CMD;
# sum all files size
size=$(ls -lgGp "$var" | grep -v / | awk '{ sum += $3 }; END { print sum }')
if [ "$size" == "" ]; then size="0"; fi
# create summary
if [ -d "$var" ]; then
folder=$(find $var/* -maxdepth 0 -type d | wc -l)
file=$(find $var/* -maxdepth 0 -type f | wc -l)
echo "Found: $folder folders "
echo " $file files $size bytes"
fi
# error message
else
echo "dir: Error \"$var\": No such file or directory"
fi
The problem is when the argument contains an asterisk (*), the ls within the script acts differently compare to the direct ls command given at the prompt. Instead of return the whole files list, the script only returns the first file. See the video below to see the comparation in action. I don't know why it behaves like that.
Anyone knows how to fix it? Thank you.
Video: problem in action
UPDATE:
The problem has been solved. Thank you all for the answers. Now my script works as expected. See the video here: http://i.giphy.com/3o8dp1YLz4fIyCbOAU.gif
The asterisk * is expanded by the shell when it parses the command line. In other words, your script doesn't get a parameter containing an asterisk, it gets a list of files as arguments. Your script only works with $1, the first argument. It should work with "$#" instead.
This is because when you retrieve $1 you assume the shell does NOT expand *.
In fact, when * (or other glob) matches, it is expanded, and broken into segments by $IFS, and then passed as $1, $2, etc.
You're lucky if you simply retrieved the first file. When your first file's path contains spaces, you'll get an error because you only get the first segment before the space.
Seriously, read this and especially this. Really.
And please don't do things like
CMD=whatever you get from user input; $CMD;
You are begging for trouble. Don't execute arbitrary string from the user.
Both above answers already answered your question. So, i'm going a bit more verbose.
In your terminal is running the bash interpreter (probably). This is the program which parses your input line(s) and doing "things" based on your input.
When you enter some line the bash start doing the following workflow:
parsing and lexical analysis
expansion
brace expansion
tidle expansion
variable expansion
artithmetic and other substitutions
command substitution
word splitting
filename generation (globbing)
removing quotes
Only after all above the bash
will execute some external commands, like ls or dir.sh... etc.,
or will do so some "internal" actions for the known keywords and builtins like echo, for, if etc...
As you can see, the second last is the filename generation (globbing). So, in your case - if the test* matches some files, your bash expands the willcard characters (aka does the globbing).
So,
when you enter dir.sh test*,
and the test* matches some files
the bash does the expansion first
and after will execute the command dir.sh with already expanded filenames
e.g. the script get executed (in your case) as: dir.sh test.pas test.swift
BTW, it acts exactly with the same way for your ls example:
the bash expands the ls test* to ls test.pas test.swift
then executes the ls with the above two arguments
and the ls will print the result for the got two arguments.
with other words, the ls don't even see the test* argument - if it is possible - the bash expands the wilcard characters. (* and ?).
Now back to your script: add after the shebang the following line:
echo "the $0 got this arguments: $#"
and you will immediatelly see, the real argumemts how your script got executed.
also, in such cases is a good practice trying to execute the script in debug-mode, e.g.
bash -x dir.sh test*
and you will see, what the script does exactly.
Also, you can do the same for your current interpreter, e.g. just enter into the terminal
set -x
and try run the dir.sh test* = and you will see, how the bash will execute the dir.sh command. (to stop the debug mode, just enter set +x)
Everbody is giving you valuable advice which you should definitely should follow!
But here is the real answer to your question.
To pass unexpanded arguments to any executable you need to single quote them:
./your_script '*'
The best solution I have is to use the eval command, in this way:
#!/bin/bash
cmd="some command \"with_quetes_and_asterisk_in_it*\""
echo "$cmd"
eval $cmd
The eval command takes its arguments and evaluates them into the command as the shell does.
This solves my problem when I need to call a command with asterisk '*' in it from a script.