I'm new to shell programming. I intend to get directory name after zip file was extracted. The print statement of it is
$test.sh helloworld.zip
helloworld
Let's take a look at test.sh:
#! /bin/sh
length=echo `expr index "$1" .zip`
a=$1
echo $(a:0:length}
However I got the Bad substitution error from the compiler.
And when I mention about 'shell'.I just talking about shell for I don't know the difference between bash or the others.I just using Ubuntu 10.04 and using the terminal. (I am using bash.)
If your shell is a sufficiently recent version of bash, that parameter expansion notation should work.
In many other shells, it will not work, and a bad substitution error is the way the shell says 'You asked for a parameter substitution but it does not make sense to me'.
Also, given the script:
#! /bin/sh
length=echo `expr index "$1" .zip`
a=$1
echo $(a:0:length}
The second line exports variable length with value echo for the command that is generated by running expr index "$1" .zip. It does not assign to length. That should be just:
length=$(expr index "${1:?}" .zip)
where the ${1:?} notation generates an error if $1 is not set (if the script is invoked with no arguments).
The last line should be:
echo ${a:0:$length}
Note that if $1 holds filename.zip, the output of expr index $1 .zip is 2, because the letter i appears at index 2 in filename.zip. If the intention is to get the base name of the file without the .zip extension, then the classic way to do it is:
base=$(basename $1 .zip)
and the more modern way is:
base=${1%.zip}
There is a difference; if the name is /path/to/filename.zip, the classic output is filename and the modern one is /path/to/filename. You can get the classic output with:
base=${1%.zip}
base=${base##*/}
Or, in the classic version, you can get the path with:
base=$(dirname $1)/$(basename $1 .zip)`.)
If the file names can contain spaces, you need to think about using double quotes, especially in the invocations of basename and dirname.
Try running it with bash.
bash test.sh helloworld.zip
-likewise-
"try changing the first line to #!/bin/bash" as comment-answered by – #shellter
Try that in bash :
echo $1
len=$(wc -c <<< "$1")
a="${1}.zip"
echo ${a:0:$len}
Adapt it to fit your needs.
Related
I have following images.
10.jpg
11.jpg
12.jpg
I want to remove above images. I used following shell script file.
for file in /home/scrapping/imgs/*
do
COUNT=$(expr $COUNT + 1)
STRING="/home/scrapping/imgs/""Img_"$COUNT".jpg"
echo $STRING
mv "$file" "$STRING"
done
So, replaced file name
Img_1.jpg
Img_2.jpg
Img_3.jpg
But, I want to replace the file name like this:
Img_10.jpg
Img_11.jpg
Img_12.jpg
So, How to set COUNT value 10 to get my own output?
The expr syntax is pretty outdated, POSIX shell allows you to do arithmetic evaluation with $(()) syntax. You can just do
#!/usr/bin/env bash
count=10
for file in /home/scrapping/imgs/*; do
[ -f "$file" ] || continue
mv "$file" "/home/scrapping/imgs/Img_$((count++)).jpg"
done
Also from the errors reported in the comments, you seem to be running it from the dash shell. It does not seem to have all the features complying to the standard POSIX shell. Run it with the sh or the bash shell.
And always use lowercase letters for user defined variables in your shell script. Upper case letters are primarily for the environment variables managed by the shell itself.
With rename command you can suffix your files with Img_:
rename 's/^/Img_/' *
The ^ means replace the start of the filename with Img_, i.e: adds a suffix.
I am having a problem where cmd1 works, but not cmd2 in my Bash script ending in .sh. I have made the Bash script executable.
Additionally, I can execute cmd2 just fine from my Bash terminal. I have tried to make a minimally reproducible example, but my larger goal is to run a complicated executable with command line arguments and pass output to a file that may or may not exist (rather than displaying the output in the terminal).
Replacing > with >> also gives the same error in the script, but not the terminal.
My Bash script:
#!/bin/bash
cmd1="cat test.txt"
cmd2="cat test.txt > a"
echo $cmd1
$cmd1
echo $cmd2
$cmd2
test.txt has the words "dog" and "cat" on two separate lines without quotes.
Short answer: see BashFAQ #50: I'm trying to put a command in a variable, but the complex cases always fail!.
Long answer: the shell expands variable references (like $cmd1) toward the end of the process of parsing a command line, after it's done parsing redirects (like > a is supposed to be) and quotes and escapes and... In fact, the only thing it does with the expanded value is word splitting (e.g. treating cat test.txt > a as "cat" followed by "test.txt", ">", and finally "a", rather than a single string) and wildcard expansion (e.g. if $cmd expanded to cat *.txt, it'd replace the *.txt part with a list of matching files). (And it skips word splitting and wildcard expansion if the variable is in double-quotes.)
Partly as a result of this, the best way to store commands in variables is: don't. That's not what they're for; variables are for data, not commands. What you should do instead, though, depends on why you were storing the command in a variable.
If there's no real reason to store the command in a variable, then just use the command directly. For conditional redirects, just use a standard if statement:
if [ -f a ]; then
cat test.txt > a
else
cat test.txt
fi
If you need to define the command at one point, and use it later; or want to use the same command over and over without having to write it out in full each time, use a function:
cmd2() {
cat test.txt > a
}
cmd2
It sounds like you may need to be able to define the command differently depending on some condition, you can actually do that with a function as well:
if [ -f a ]; then
cmd() {
cat test.txt > a
}
else
cmd() {
cat test.txt
}
fi
cmd
Alternately, you can wrap the command (without redirect) in a function, then use a conditional to control whether it redirects:
cmd() {
cat test.txt
}
if [ -f a ]; then
cmd > a
else
cmd
fi
It's also possible to wrap a conditional redirect into a function itself, then pipe output to it:
maybe_redirect_to() {
if [ -f "$1" ]; then
cat > "$1"
else
cat
fi
}
cat test.txt | maybe_redirect_to a
(This creates an extra cat process that isn't really doing anything useful, but if it makes the script cleaner, I'd consider that worth it. In this particular case, you could minimize the stray cats by using maybe_redirect_to a < test.txt.)
As a last resort, you can store the command string in a variable, and use eval to parse it. eval basically re-runs the shell parsing process from the beginning, meaning that it'll recognize things like redirects in the string. But eval has a well-deserved reputation as a bug magnet, because it's easy for it to treat parts of the string you thought were just data as command syntax, which can cause some really weird (& dangerous) bugs.
If you must use eval, at least double-quote the variable reference, so it runs through the parsing process just once, rather than sort-of-once-and-a-half as it would unquoted. Here's an example of what I mean:
cmd3="echo '5 * 3 = 15'"
eval "$cmd3"
# prints: 5 * 3 = 15
eval $cmd3
# prints: 5 [list of files in the current directory] 3 = 15
# ...unless there are any files with shell metacharacters in their names, in
# which case something more complicated might happen.
BashFAQ #50 discusses some other possible reasons and solutions. Note that the array approach will not work here, since arrays also get expanded after redirects are parsed.
If you pop an 'eval' in front of $cmd2 it should work as expected:
#!/bin/bash
cmd2="cat test.txt > a"
eval $cmd2
If you're not sure about the operation of a script you could always use the debug mode to see if you can determine the error.
bash -x scriptname
This will run the command and display the output of variable evaluations. Hopefully this will reveal any issues with syntax.
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.
I can't understand the following code in a bash.
set `pwd` ; mfix=$1
It actually get the run directory name.But I don't how does it work.
What is the set command mean?
From the doc for the set:
This builtin is so complicated that it deserves its own section. set
allows you to change the values of shell options and set the
positional parameters, or to display the names and values of shell
variables.
e.g.
set v1 v2 v3 ; echo $1
will print
v1
The comand inside backticks is called as "command substitution". From the docs:
Bash performs the expansion by executing command and replacing the
command substitution with the standard output of the command, with any
trailing newlines deleted.
In your example, it sets the 1st positional argument $1 to the value of the result of execution of command inside the backticks. (called as command substitution). The command is pwd what shows the current working directory.
Anyway, if the path to the directory contains an space, the $1 will get only the first part of the path., e.g.
$ pwd
/some/path with/space
$ set `pwd`
$ echo $1
/some/path
$echo $2
with/space
Finally the all above is strange design, because you can simply:
mfix=$(pwd) #old school: mfix=`pwd`
It is better to use the $(command) instead of the backticks.
This code in bash put the result of the command pwd in the variable mfix.
You can print the result of the mfix variable by running
echo $mfix
I have two folders full of trainings and corresponding testfiles and I'd like to run the fitting pairs against each other using a shell script.
This is what I have so far:
for x in SpanishLS.train/*.train
do
timbl -f $x -t SpanishLS.test/$x.test
done
This is supposed to take file1(-n).train in one directory, look for file1(-n).test in the other, and run them trough a tool called timbl.
What it does instead is look for a file called SpanishLS.train/file1(-n).train.test which of course doesn't exist.
What I tried to do, to no avail, is truncate $x in a way that lets the script find the correct file, but whenever I do this, $x is truncated way too early, resulting in the script not even finding the .train file.
How should I code this?
If I got you right, this will do the job:
for x in SpanishLS.train/*.train
do
y=${x##*/} # strip basepath
y=${y%.*} # strip extention
timbl -f $x -t SpanishLS.test/$y.test
done
Use basename:
for x in SpanishLS.train/*.train
do
timbl -f $x -t SpanishLS.test/$(basename "$x" .train).test
done
That removes the directory prefix and the .train suffix from $x, and builds up the name you want.
In bash (and other POSIX-compliant shells), you can do the basename operation with two shell parameter expansions without invoking an external program. (I don't think there's a way to combine the two expansions into one.)
for x in SpanishLS.train/*.train
do
y=${x##*/} # Remove path prefix
timbl -f $x -t SpanishLS.test/${y%.train}.test # Remove .train suffix
done
Beware: bash supports quite a number of (useful) expansions that are not defined by POSIX. For example, ${y//.train/.test} is a bash-only notation (or bash and compatible shells notation).
Replace all occurences of .train in the filename to .text:
timbl -f $x -t $(echo $x | sed 's/\.train/.text/g')