I need a help, please
I have script
#!/bin/bash
NUMBER=$(echo "$1" | sed 's/[^0-9]*//g')
echo $NUMBER
./test.sh tas1 tefst2 thgst3 ynft4 jhuf5 hjh6 jhd7
1
returns only 1 but I need it 1234567
The reason you only get "1" is because $1 is only the first parameter. Use "$*" to get a single string containing all the parameters, separated by spaces (by default).
You can also do this with bash variable substitution:
params="$*"
echo "${params//[^[:digit:]]/}"
Related
I am learning bash. I would like to get the return value and matched line by grep at once.
if cat 'file' | grep 'match_word'; then
match_by_grep="$(cat 'file' | grep 'match_word')"
read a b <<< "${match_by_grep}"
fi
In the code above, I used grep twice. I cannot think of how to do it by grep once. I am not sure match_by_grep is always empty even when there is no matched words because cat may output error message.
match_by_grep="$(cat 'file' | grep 'match_word')"
if [[ -n ${match_by_grep} ]]; then
# match_by_grep may be an error message by cat.
# So following a and b may have wrong value.
read a b <<< "${match_by_grep}"
fi
Please tell me how to do it. Thank you very much.
You can avoid the double use of grep by storing the search output in a variable and seeing if it is not empty.
Your version of the script without double grep.
#!/bin/bash
grepOutput="$(grep 'match_word' file)"
if [ ! -z "$grepOutput" ]; then
read a b <<< "${grepOutput}"
fi
An optimization over the above script ( you can remove the temporary variable too)
#!/bin/bash
grepOutput="$(grep 'match_word' file)"
[[ ! -z "$grepOutput" ]] && (read a b <<< "${grepOutput}")
Using double-grep once for checking if-condition and once to parse the search result would be something like:-
#!/bin/bash
if grep -q 'match_word' file; then
grepOutput="$(grep 'match_word' file)"
read a b <<< "${grepOutput}"
fi
When assigning a variable with a string containing a command expansion, the return code is that of the (rightmost) command being expanded.
In other words, you can just use the assignment as the condition:
if grepOutput="$(cat 'file' | grep 'match_word')"
then
echo "There was a match"
read -r a b <<< "${grepOutput}"
(etc)
else
echo "No match"
fi
Is this what you want to achieve?
grep 'match_word' file ; echo $?
$? has a return value of the command run immediately before.
If you would like to keep track of the return value, it will be also useful to have PS1 set up with $?.
Ref: Bash Prompt with Last Exit Code
I've got a program that I want to call by passing parameters from a shell variable. Throughout this question, I am going to assume that it is given by
#!/bin/sh
echo $#
i.e. that it prints out the number of arguments that are passed to it. Let's call it count-args.
I call my program like this:
X="arg1 arg2"
count-args $X
This works quite well. But now one of my arguments has a whitespace in it and I can't find a way to escape it, e.g. the following things do not work:
X="Hello\ World"
X="Hello\\ World"
X="'Hello World'"
In all of the cases, my program count-args prints out 2. I want to find a way so I can pass the string Hello World and that it returns 1 instead. How?
Just for clarification: I do not want to pass all parameters as a single string, e.g.
X="Hello World"
count-args $X
should print out 2. I want a way to pass parameters that contain whitespaces.
Use an array to store multiple, space-containing arguments.
$ args=("first one" "second one")
$ count-args "${args[#]}"
2
This can be solved with xargs. By replacing
count-args $X
with
echo $X | xargs count-args
I can use backslashes to escape whitespaces in $X, e.g.
X="Hello\\ World"
echo $X | xargs count-args
prints out 1 and
X="Hello World"
echo $X | xargs count-args
prints out 2.
count-args "$X"
The quotes ensure in bash, that the whole content of variable X is passed as a single parameter.
Your Counting script:
$ cat ./params.sh
#!/bin/sh
echo $#
For completeness here is what happens with various arguments:
$ ./params.sh
0
$ ./params.sh 1 2
2
$ ./params.sh
0
$ ./params.sh 1
1
$ ./params.sh 1 2
2
$ ./params.sh "1 2"
1
And here is what you get with variables:
$ XYZ="1 2" sh -c './params.sh $XYZ'
2
$ XYZ="1 2" sh -c './params.sh "$XYZ"'
1
Taking this a bit further:
$ cat params-printer.sh
#!/bin/sh
echo "Count: $#"
echo "1 : '$1'"
echo "2 : '$2'"
We get:
$ XYZ="1 2" sh -c './params-printer.sh "$XYZ"'
Count: 1
1 : '1 2'
2 : ''
This looks like what you wanted to do.
Now: If you have a script you cannot control and neither can you control the way the script is invoked. Then there is very little you can do to prevent a variable with spaces turning into multiple arguments.
There are quite a few questions around this on StackOverflow which indicate that you need the ability to control how the command is invoked else there is little you can do.
Passing arguments with spaces between (bash) script
Passing a string with spaces as a function argument in bash
Passing arguments to a command in Bash script with spaces
And wow! this has been asked so many times before:
How to pass argument with spaces to a shell script function
The below code does not give any output:
$echo `cat time`
19991213100942
$a=$(echo `cat time`) | echo $a | echo ${a:0:4}
Please tell where I am making mistake.
a=$(echo `cat time`)
assigns the output of the command inside the brackets $(...) to the variable $a.
Later in the script, you can print the variable:
echo $a
That prints: 19991213100942
echo ${a:0:4}
That prints: 1999
You can reference the varibale by its name $a.
First, you don't need to echo the output of cat time: just cat time.
Second, as #Etan says (kind of), replace the pipes with semicolons or newlines
a=$(< time) # a bash builtin, equivalent to but faster than: a=$(cat time)
echo $a
echo ${a:0:4}
This question has 3 parts, and each alone is easy, but combined together is not trivial (at least for me) :)
Need write a script what should take as its arguments:
one name of another command
several arguments for the command
list of files
Examples:
./my_script head -100 a.txt b.txt ./xxx/*.txt
./my_script sed -n 's/xxx/aaa/' *.txt
and so on.
Inside my script for some reason I need distinguish
what is the command
what are the arguments for the command
what are the files
so probably the most standard way write the above examples is:
./my_script head -100 -- a.txt b.txt ./xxx/*.txt
./my_script sed -n 's/xxx/aaa/' -- *.txt
Question1: Is here any better solution?
Processing in ./my_script (first attempt):
command="$1";shift
args=`echo $* | sed 's/--.*//'`
filenames=`echo $* | sed 's/.*--//'`
#... some additional processing ...
"$command" "$args" $filenames #execute the command with args and files
This solution will fail when the filenames will contain spaces and/or '--', e.g.
/some--path/to/more/idiotic file name.txt
Question2: How properly get $command its $args and $filenames for the later execution?
Question3: - how to achieve the following style of execution?
echo $filenames | $command $args #but want one filename = one line (like ls -1)
Is here nice shell solution, or need to use for example perl?
First of all, it sounds like you're trying to write a script that takes a command and a list of filenames and runs the command on each filename in turn. This can be done in one line in bash:
$ for file in a.txt b.txt ./xxx/*.txt;do head -100 "$file";done
$ for file in *.txt; do sed -n 's/xxx/aaa/' "$file";done
However, maybe I've misinterpreted your intent so let me answer your questions individually.
Instead of using "--" (which already has a different meaning), the following syntax feels more natural to me:
./my_script -c "head -100" a.txt b.txt ./xxx/*.txt
./my_script -c "sed -n 's/xxx/aaa/'" *.txt
To extract the arguments in bash, use getopts:
SCRIPT=$0
while getopts "c:" opt; do
case $opt in
c)
command=$OPTARG
;;
esac
done
shift $((OPTIND-1))
if [ -z "$command" ] || [ -z "$*" ]; then
echo "Usage: $SCRIPT -c <command> file [file..]"
exit
fi
If you want to run a command for each of the remaining arguments, it would look like this:
for target in "$#";do
eval $command \"$target\"
done
If you want to read the filenames from STDIN, it would look more like this:
while read target; do
eval $command \"$target\"
done
The $# variable, when quoted will be able to group parameters as they should be:
for parameter in "$#"
do
echo "The parameter is '$parameter'"
done
If given:
head -100 test this "File name" out
Will print
the parameter is 'head'
the parameter is '-100'
the parameter is 'test'
the parameter is 'this'
the parameter is 'File name'
the parameter is 'out'
Now, all you have to do is parse the loop out. You can use some very simple rules:
The first parameter is always the file name
The parameters that follow that start with a dash are parameters
After the "--" or once one doesn't start with a "-", the rest are all file names.
You can check to see if the first character in the parameter is a dash by using this:
if [[ "x${parameter}" == "x${parameter#-}" ]]
If you haven't seen this syntax before, it's a left filter. The # divides the two parts of the variable name. The first part is the name of the variable, and the second is the glob filter (not regular expression) to cut off. In this case, it's a single dash. As long as this statement isn't true, you know you have a parameter. BTW, the x may or may not be needed in this case. When you run a test, and you have a string with a dash in it, the test might mistake it for a parameter of the test and not the value.
Put it together would be something like this:
parameterFlag=""
for parameter in "$#" #Quotes are important!
do
if [[ "x${parameter}" == "x${parameter#-}" ]]
then
parameterFlag="Tripped!"
fi
if [[ "x${parameter}" == "x--" ]]
then
print "Parameter \"$parameter\" ends the parameter list"
parameterFlag="TRIPPED!"
fi
if [ -n $parameterFlag ]
then
print "\"$parameter\" is a file"
else
echo "The parameter \"$parameter\" is a parameter"
fi
done
Question 1
I don't think so, at least not if you need to do this for arbitrary commands.
Question 3
command=$1
shift
while [ $1 != '--' ]; do
args="$args $1"
shift
done
shift
while [ -n "$1" ]; do
echo "$1"
shift
done | $command $args
Question 2
How does that differ from question 3?
I have a Perl script that gives me a defined list of random numbers that correspond to the lines of a file. Next I want to extract those lines from the file using sed.
#!/bin/bash
count=$(cat last_queries.txt | wc -l)
var=$(perl test.pl test2 $count)
The variable var returns an output like: cat last_queries.txt | sed -n '12p;500p;700p'. The problem is that I can't run this last command. I tried with $var, but the output is not correct (if I run manually the command it works fine, so no problem there). What is the correct way to do this?
P.S: Sure I could do all the work in Perl, but I'm trying to learn this way, because it could help me in other situations.
You just need to do:
#!/bin/bash
count=$(cat last_queries.txt | wc -l)
$(perl test.pl test2 $count)
However, if you want to call your Perl command later, and that's why you want to assign it to a variable, then:
#!/bin/bash
count=$(cat last_queries.txt | wc -l)
var="perl test.pl test2 $count" # You need double quotes to get your $count value substituted.
...stuff...
eval $var
As per Bash's help:
~$ help eval
eval: eval [arg ...]
Execute arguments as a shell command.
Combine ARGs into a single string, use the result as input to the shell,
and execute the resulting commands.
Exit Status:
Returns exit status of command or success if command is null.
You're are probably looking for eval $var.
line=$((${RANDOM} % $(wc -l < /etc/passwd)))
sed -n "${line}p" /etc/passwd
just with your file instead.
In this example I used the file /etc/password, using the special variable ${RANDOM} (about which I learned here), and the sed expression you had, only difference is that I am using double quotes instead of single to allow the variable expansion.
There're 2 basic ways of executing a string command in a shell script whether it's given as parameter or not here's.
COMMAND="ls -lah"
$(echo $COMMAND)
or
COMMAND="ls -lah"
bash -c $COMMAND
In the case where you have multiple variables containing the arguments for a command you're running, and not just a single string, you should not use eval directly, as it will fail in the following case:
function echo_arguments() {
echo "Argument 1: $1"
echo "Argument 2: $2"
echo "Argument 3: $3"
echo "Argument 4: $4"
}
# Note we are passing 3 arguments to `echo_arguments`, not 4
eval echo_arguments arg1 arg2 "Some arg"
Result:
Argument 1: arg1
Argument 2: arg2
Argument 3: Some
Argument 4: arg
Note that even though "Some arg" was passed as a single argument, eval read it as two.
Instead, you can just use the string as the command itself:
# The regular bash eval works by jamming all its arguments into a string then
# evaluating the string. This function treats its arguments as individual
# arguments to be passed to the command being run.
function eval_command() {
"$#";
}
Note the difference between the output of eval and the new eval_command function:
eval_command echo_arguments arg1 arg2 "Some arg"
Result:
Argument 1: arg1
Argument 2: arg2
Argument 3: Some arg
Argument 4:
Better ways to do it
Using a function:
# define it
myls() {
ls -l "/tmp/test/my dir"
}
# run it
myls
Using an array:
# define the array
mycmd=(ls -l "/tmp/test/my dir")
# run the command
"${mycmd[#]}"
cmd="ls -atr ${HOME} | tail -1" <br/>
echo "$cmd" <br/>
VAR_FIRST_FILE=$( eval "${cmd}" ) <br/>
or
cmd=("ls -atr ${HOME} | tail -1") <br/>
echo "$cmd" <br/>
VAR_FIRST_FILE=$( eval "${cmd[#]}" )