Please can you show me a shell script that takes a string input by the user and outputs the number of characters in it? I have tried numerous times but i can't get it right.
Use the wc function like that:
echo -n abc | wc -c
or
echo -n abc | wc -m
The -n supresses the final newline which would count as an extra character.
check manual for wc.
This should do the trick:
#!/usr/bin/env bash
echo -n "$1" | wc -c
wc is a tool to count characters, lines or bytes. -c is the option for characters.
The -n option for echo avoids the newline which would be an extra character.
Make sure to make the script executable by:
chmod +x wordcount.sh
So you will get:
user#host$ ./wordcount.sh "My String"
9
Related
conducting a word count of a directory.
ls | wc -l
if output is "17", I would like the output to display as "017".
I have played with | printf with little luck.
Any suggestions would be appreciated.
printf is the way to go to format numbers:
printf "There were %03d files\n" "$(ls | wc -l)"
ls | wc -l will tell you how many lines it encountered parsing the output of ls, which may not be the same as the number of (non-dot) filenames in the directory. What if a filename has a newline? One reliable way to get the number of files in a directory is
x=(*)
printf '%03d\n' "${#x[#]}"
But that will only work with a shell that supports arrays. If you want a POSIX compatible approach, use a shell function:
countargs() { printf '%03d\n' $#; }
countargs *
This works because when a glob expands the shell maintains the words in each member of the glob expansion, regardless of the characters in the filename. But when you pipe a filename the command on the other side of the pipe can't tell it's anything other than a normal string, so it can't do any special handling.
You coud use sed.
ls | wc -l | sed 's/^17$/017/'
And this applies to all the two digit numbers.
ls | wc -l | sed '/^[0-9][0-9]$/s/.*/0&/'
I would like to help out my girlfriend - she needs the specific count of certain characters in around 200 files (per file).
I already found How can I use the UNIX shell to count the number of times a letter appears in a text file?, but that only shows the complete number, not the number of occurrences per file. basically, what I want is the following:
$ ls
test1 test2
$ cat test1
ddddnnnn
ddnnddnnnn
$ cat test2
ddnnddnnnn
$ grep -o 'n' * | wc -w
16
$ <insert command here>
test1 10
test2 6
$
or something similar regarding the output. As this will be on her university machine, I cannot code anything in perl or so, just shell is allowed. My shell knowledge is a bit rusty, so I cannot come up with a better solution - maybe you could be of assistance.
grep -Ho n * | uniq -c
produces
10 test1:n
6 test2:n
If you want exactly your output:
grep -Ho n * | uniq -c | while read count file; do echo "${file%:n} $count"; done
It's not exactly elegant, but the most obvious solution is:
letter='n'
for file in *; do
count=`grep -o $letter "$file" | wc -w`
echo "$file contains $letter $count times"
done
Glen's answer is far better for the flavors of UNIX that support it. This will work on a UNIX that claims it is POSIX-compliant. This is meant for the poor folks for whom the other answer does not fly.
POSIX grep says nothing about grep -H -o See: http://pubs.opengroup.org/onlinepubs/009604499/utilities/grep.html
Get a list of the files you want call it list.txt. I chose the character ^ == shift 6 for no reason
while read fname
do
cnt=`tr -dc '^' < $fname | wc -c`
echo "$fname: $cnt"
done < list.txt
How can you limit the count of standard output characters that is redirected to file?
Other ways (external)
echo $out| head -c 20
echo $out | awk '{print substr($0,1,20) }'
echo $out | ruby -e 'print $_[0,19]'
echo $out | sed -r 's/(^.{20})(.*)/\1/'
You could use Command Substitution to wrap the output pre-redirection, then use the offset Parameter Expansion to limit the number of characters like so:
#!/bin/bash
limit=20
out=$(echo "this line has more than twenty characters in it")
echo ${out::limit} > /path/to/file
Proof of Concept
$ limit=20
$ out=$(echo "this line has more than twenty characters in it").
$ echo ${out::limit}
this line has more t
You can't do so directly into a file, but you can pipe through sed or head, etc. to pass on only part of the output. Or as #SiegeX says, capture the output in the shell (but I would be wary of that if the output mis likely to be large).
Suppose I have some output from a command (such as ls -1):
a
b
c
d
e
...
I want to apply a command (say echo) to each one, in turn. E.g.
echo a
echo b
echo c
echo d
echo e
...
What's the easiest way to do that in bash?
It's probably easiest to use xargs. In your case:
ls -1 | xargs -L1 echo
The -L flag ensures the input is read properly. From the man page of xargs:
-L number
Call utility for every number non-empty lines read.
A line ending with a space continues to the next non-empty line. [...]
You can use a basic prepend operation on each line:
ls -1 | while read line ; do echo $line ; done
Or you can pipe the output to sed for more complex operations:
ls -1 | sed 's/^\(.*\)$/echo \1/'
for s in `cmd`; do echo $s; done
If cmd has a large output:
cmd | xargs -L1 echo
You can use a for loop:
for file in * ; do
echo "$file"
done
Note that if the command in question accepts multiple arguments, then using xargs is almost always more efficient as it only has to spawn the utility in question once instead of multiple times.
You actually can use sed to do it, provided it is GNU sed.
... | sed 's/match/command \0/e'
How it works:
Substitute match with command match
On substitution execute command
Replace substituted line with command output.
A solution that works with filenames that have spaces in them, is:
ls -1 | xargs -I %s echo %s
The following is equivalent, but has a clearer divide between the precursor and what you actually want to do:
ls -1 | xargs -I %s -- echo %s
Where echo is whatever it is you want to run, and the subsequent %s is the filename.
Thanks to Chris Jester-Young's answer on a duplicate question.
xargs fails with with backslashes, quotes. It needs to be something like
ls -1 |tr \\n \\0 |xargs -0 -iTHIS echo "THIS is a file."
xargs -0 option:
-0, --null
Input items are terminated by a null character instead of by whitespace, and the quotes and backslash are
not special (every character is taken literally). Disables the end of file string, which is treated like
any other argument. Useful when input items might contain white space, quote marks, or backslashes. The
GNU find -print0 option produces input suitable for this mode.
ls -1 terminates the items with newline characters, so tr translates them into null characters.
This approach is about 50 times slower than iterating manually with for ... (see Michael Aaron Safyans answer) (3.55s vs. 0.066s). But for other input commands like locate, find, reading from a file (tr \\n \\0 <file) or similar, you have to work with xargs like this.
i like to use gawk for running multiple commands on a list, for instance
ls -l | gawk '{system("/path/to/cmd.sh "$1)}'
however the escaping of the escapable characters can get a little hairy.
Better result for me:
ls -1 | xargs -L1 -d "\n" CMD
I bumped into the following problem: I'm writing a Linux bash script which does the following:
Read line from file
Strip the \n character from the end of the line just read
Execute the command that's in there
Example:
commands.txt
ls
ls -l
ls -ltra
ps as
The execution of the bash file should get the first line, and execute it, but while the \n present, the shell just outputs "command not found: ls"
That part of the script looks like this
read line
if [ -n "$line" ]; then #if not empty line
#myline=`echo -n $line | tr -d '\n'`
#myline=`echo -e $line | sed ':start /^.*$/N;s/\n//g; t start'`
myline=`echo -n $line | tr -d "\n"`
$myline #execute it
cat $fname | tail -n+2 > $fname.txt
mv $fname.txt $fname
fi
Commented you have the things I tried before asking SO. Any solutions? I'm smashing my brains for the last couple of hours over this...
I always like perl -ne 'chomp and print' , for trimming newlines. Nice and easy to remember.
e.g. ls -l | perl -ne 'chomp and print'
However
I don't think that is your problem here though. Although I'm not sure I understand how you're passing the commands in the file through to the 'read' in your shell script.
With a test script of my own like this (test.sh)
read line
if [ -n "$line" ]; then
$line
fi
and a sample input file like this (test.cmds)
ls
ls -l
ls -ltra
If I run it like this ./test.sh < test.cmds, I see the expected result, which is to run the first command 'ls' on the current working directory.
Perhaps your input file has additional non-printable characters in it ?
mine looks like this
od -c test.cmds
0000000 l s \n l s - l \n l s - l t
0000020 r a \n
0000023
From your comments below, I suspect you may have carriage returns ( "\r" ) in your input file, which is not the same thing as a newline. Is the input file originally in DOS format ? If so, then you need to convert the 2 byte DOS line ending "\r\n" to the single byte UNIX one, "\n" to achieve the expected results.
You should be able to do this by swapping the "\n" for "\r" in any of your commented out lines.
Someone already wrote a program which executes shell commands: sh file
If you really only want to execute the first line of a file: head -n 1 file |sh
If your problem is carriage-returns: tr -d '\r' <file |sh
I tried this:
read line
echo -n $line | od -x
For the input 'xxxx', I get:
0000000 7878 7878
As you can see, there is no \n at the end of the contents of the variable. I suggest to run the script with the option -x (bash -x script). This will print all commands as they are executed.
[EDIT] Your problem is that you edited commands.txt on Windows. Now, the file contains CRLF (0d0a) as line delimiters which confuses read (and ls\r is not a known command). Use dos2unix or similar to turn it into a Unix file.
You may also try to replace carriage returns with newlines only using Bash builtins:
line=$'a line\r'
line="${line//$'\r'/$'\n'}"
#line="${line/%$'\r'/$'\n'}" # replace only at line end
printf "%s" "$line" | ruby -0777 -n -e 'p $_.to_s'
you need eval command
#!/bin/bash -x
while read cmd
do
if [ "$cmd" ]
then
eval "$cmd"
fi
done
I ran it as
./script.sh < file.txt
And file.txt was:
ls
ls -l
ls -ltra
ps as
though not working for ls, I recommend having a look at find’s -print0 option
The following script works (at least for me):
#!/bin/bash
while read I ; do if [ "$I" ] ; then $I ; fi ; done ;