how do i print using echo in bash so the row wont "jump" abit to the right cause of the length of the Variable can u please help me with a command that do so
Try using the printf shell command:
$ printf "%5d %s\n" 1 test
1 test
$ printf "%5d %s\n" 123 another
123 another
To trim leading whitespace inside a variable you can use Bash parameter expansion:
var=" value"
echo "${var#"${var%%[![:space:]]*}"}"
Use tabs to separate your columns.
echo -e "$var1\t$var2"
or, better, use printf to do it:
printf "%s\t%s\n" $var1 $var2
Or, as Greg Hewgill showed, use field widths (even with strings - the hyphen makes them left-aligned):
printf "%-6s %-8s %10s\n" abcde fghij 12345
Related
I'm relatively new to bash scripting and I would like someone to explain this properly, thank you. Here is my code:
#! /bin/bash
echo "first arg: $1"
echo "first arg: $2"
var="$( grep -rnw $1 -e $2 | cut -d ":" -f1 )"
var2=$( grep -rnw $1 -e $2 | cut -d ":" -f1 | awk '{print substr($0,length,1)}')
echo "$var"
echo "$var2"
The problem I have is with the output, the script I'm trying to write is a c++ function searcher, so upon launching my script I have 2 arguments, one for the directory and the second one as the function name. This is how my output looks like:
first arg: Projekt
first arg: iseven
Projekt/AX/include/ax.h
Projekt/AX/src/ax.cpp
h
p
Now my question is: how do can I save the line by line output as a variable, so that later on I can use var as a path, or to use var2 as a character to compare. My plan was to use IF() statements to determine the type, idea: IF(last_char == p){echo:"something"}What I've tried was this question: Capturing multiple line output into a Bash variable and then giving it an array. So my code looked like: "${var[0]}". Please explain how can I use my line output later on, as variables.
I'd use readarray to populate an array variable just in case there's spaces in your command's output that shouldn't be used as field separators that would end up messing up foo=( ... ). And you can use shell parameter expansion substring syntax to get the last character of a variable; no need for that awk bit in your var2:
#!/usr/bin/env bash
readarray -t lines < <(printf "%s\n" "Projekt/AX/include/ax.h" "Projekt/AX/src/ax.cpp")
for line in "${lines[#]}"; do
printf "%s\n%s\n" "$line" "${line: -1}" # Note the space before the -1
done
will display
Projekt/AX/include/ax.h
h
Projekt/AX/src/ax.cpp
p
I often use printf inside shell scripts to make some nice aligned outputs
The problem is, everytime there is an accent (éèà) in the printed string, it shifts the following string 1 step back.
Example :
printf "%-10s %s\n" "toto" "test"
printf "%-10s %s\n" "titi" "test"
printf "%-10s %s\n" "tété" "test"
printf "%-10s %s\n" "toto" "test"
Expected :
toto test
titi test
tété test
toto test
Got :
toto test
titi test
tété test
toto test
Does someone have an explanation on this and what can I do to make printf doing it right with special characters?
Thank you for your help
Does someone have an explanation on this
é is character encoded with two bytes.
what can I do to make printf doing it right with special characters?
Design your own method of padding that would take into account utf-8s. Ideally I believe a tool like wprintf or making %Ls format specifier call wcwidth() to determine character width or something similar would be welcomed and usefull.
As of now at least my bash when calculating string length takes utf-8 chars into account. You could insert the padding yourself:
printf "%-10s %s\n" "titi" "test";
s="tété";
# (echo -n "$s" | wc -c) is 6 , but ${#s} is 4!
printf "%s%*s %s\n" "$s" "$((10-${#s}))" "" "test"
Adapted my answer from https://unix.stackexchange.com/a/592479/310674
#!/usr/bin/env bash
align_left(){ printf %s%\*s "${2:0:$1}" $(($1-${#2})) '';}
printf '%s %s\n' \
"$(align_left 10 "toto")" "test" \
"$(align_left 10 "titi")" "test" \
"$(align_left 10 "tété")" "test" \
"$(align_left 10 "têtu")" "test"
Output:
toto test
titi test
tété test
têtu test
But you can use other tool to print your report in that manner. Following example uses awk:
echo "toto" | awk '{printf "%-10s test\n", $1}'
echo "tété" | awk '{printf "%-10s test\n", $1}'
echo "titi" | awk '{printf "%-10s test\n", $1}'
EDIT:
The following statement was partially wrong:
printf might not be part of bash, but coreutils. Coreutils have a long history with multibyte characters - https://crashcourse.housegordon.org/coreutils-multibyte-support.html.
As noted in a comment by #charles-duffy - printf, in this case, is shell builtin. You can check it with:
[Alex#NormandySR2 ~]$ type printf
printf is a shell builtin
I also agree with the fact that most shell implements their own printf. I checked the following:
fish
bash
zsh
tcsh
ksh
dash
oil
All of them uses printf builtin that can differ in details. So my assumption about printf as part of coreutils, in this case, was wrong.
Suppose I have the following string:
some letters foo/substring/goo/some additional letters
I need to extract this substring supposing that foo/ and /goo are constant strings that are known in advance. How can I do that?
This sed one-liner does it.
sed 's#.*foo/##;s#/goo/.*##' file
Except for sed, awk, grep can do the job too. Or with zsh:
kent$ v="some letters foo/substring/goo/some additional letters"
kent$ echo ${${v##*foo/}%%/goo/*}
substring
Note that:
comment by #Nahuel Fouilleul
in ${var%%/goo/*} var must be a variable name, and can't be the result of expansion
The line should be divided into two statements, if work with bash.
$ echo $0
bash
$ v="some letters foo/substring/goo/some additional letters"
$ v=${v##*foo/}
$ v=${v%%/goo/*}
$ echo $v
substring
The line I executed in zsh, worked, but just I tested in bash, it didn't work.
$ echo $0
-zsh
$ v="some letters foo/substring/goo/some additional letters"
$ echo ${${v##*foo/}%%/goo/*}
substring
With variable expansion
line='some letters foo/substring/goo/some additional letters'
line=${line%%/goo*} # remove suffix /goo*
line=${line##*foo/} # remove prefix *ffo/
echo "$line"
or bash regular expression
line='some letters foo/substring/goo/some additional letters'
if [[ $line =~ foo/([^/]*)/goo ]]; then
echo "${BASH_REMATCH[1]}"
fi
If you know there are no other / in your "other letters", you can use cut :
> echo "some letters foo/substring/goo/some additional letters" | cut -d'/' -f2
In terms of readability I think awk is a good solution
echo "some letters foo/substring/goo/some additional letters" | awk -v FS="(foo/|/goo)" '{print $2}'
Suppose I have a Unix shell variable as below
variable=abc,def,ghij
I want to extract all the values (abc, def and ghij) using a for loop and pass each value into a procedure.
The script should allow extracting arbitrary number of comma-separated values from $variable.
Not messing with IFS
Not calling external command
variable=abc,def,ghij
for i in ${variable//,/ }
do
# call your procedure/other scripts here below
echo "$i"
done
Using bash string manipulation http://www.tldp.org/LDP/abs/html/string-manipulation.html
You can use the following script to dynamically traverse through your variable, no matter how many fields it has as long as it is only comma separated.
variable=abc,def,ghij
for i in $(echo $variable | sed "s/,/ /g")
do
# call your procedure/other scripts here below
echo "$i"
done
Instead of the echo "$i" call above, between the do and done inside the for loop, you can invoke your procedure proc "$i".
Update: The above snippet works if the value of variable does not contain spaces. If you have such a requirement, please use one of the solutions that can change IFS and then parse your variable.
If you set a different field separator, you can directly use a for loop:
IFS=","
for v in $variable
do
# things with "$v" ...
done
You can also store the values in an array and then loop through it as indicated in How do I split a string on a delimiter in Bash?:
IFS=, read -ra values <<< "$variable"
for v in "${values[#]}"
do
# things with "$v"
done
Test
$ variable="abc,def,ghij"
$ IFS=","
$ for v in $variable
> do
> echo "var is $v"
> done
var is abc
var is def
var is ghij
You can find a broader approach in this solution to How to iterate through a comma-separated list and execute a command for each entry.
Examples on the second approach:
$ IFS=, read -ra vals <<< "abc,def,ghij"
$ printf "%s\n" "${vals[#]}"
abc
def
ghij
$ for v in "${vals[#]}"; do echo "$v --"; done
abc --
def --
ghij --
I think syntactically this is cleaner and also passes shell-check linting
variable=abc,def,ghij
for i in ${variable//,/ }
do
# call your procedure/other scripts here below
echo "$i"
done
#/bin/bash
TESTSTR="abc,def,ghij"
for i in $(echo $TESTSTR | tr ',' '\n')
do
echo $i
done
I prefer to use tr instead of sed, becouse sed have problems with special chars like \r \n in some cases.
other solution is to set IFS to certain separator
Another solution not using IFS and still preserving the spaces:
$ var="a bc,def,ghij"
$ while read line; do echo line="$line"; done < <(echo "$var" | tr ',' '\n')
line=a bc
line=def
line=ghij
Here is an alternative tr based solution that doesn't use echo, expressed as a one-liner.
for v in $(tr ',' '\n' <<< "$var") ; do something_with "$v" ; done
It feels tidier without echo but that is just my personal preference.
The following solution:
doesn't need to mess with IFS
doesn't need helper variables (like i in a for-loop)
should be easily extensible to work for multiple separators (with a bracket expression like [:,] in the patterns)
really splits only on the specified separator(s) and not - like some other solutions presented here on e.g. spaces too.
is POSIX compatible
doesn't suffer from any subtle issues that might arise when bash’s nocasematch is on and a separator that has lower/upper case versions is used in a match like with ${parameter/pattern/string} or case
beware that:
it does however work on the variable itself and pop each element from it - if that is not desired, a helper variable is needed
it assumes var to be set and would fail if it's not and set -u is in effect
while true; do
x="${var%%,*}"
echo $x
#x is not really needed here, one can of course directly use "${var%%:*}"
if [ -z "${var##*,*}" ] && [ -n "${var}" ]; then
var="${var#*,}"
else
break
fi
done
Beware that separators that would be special characters in patterns (e.g. a literal *) would need to be quoted accordingly.
Here's my pure bash solution that doesn't change IFS, and can take in a custom regex delimiter.
loop_custom_delimited() {
local list=$1
local delimiter=$2
local item
if [[ $delimiter != ' ' ]]; then
list=$(echo $list | sed 's/ /'`echo -e "\010"`'/g' | sed -E "s/$delimiter/ /g")
fi
for item in $list; do
item=$(echo $item | sed 's/'`echo -e "\010"`'/ /g')
echo "$item"
done
}
Try this one.
#/bin/bash
testpid="abc,def,ghij"
count=`echo $testpid | grep -o ',' | wc -l` # this is not a good way
count=`expr $count + 1`
while [ $count -gt 0 ] ; do
echo $testpid | cut -d ',' -f $i
count=`expr $count - 1 `
done
Working with printf in a bash script, adding no spaces after "\n" does not create a newline, whereas adding a space creates a newline, e. g.:
No space after "\n"
NewLine=`printf "\n"`
echo -e "Firstline${NewLine}Lastline"
Result:
FirstlineLastline
Space after "\n "
NewLine=`printf "\n "`
echo -e "Firstline${NewLine}Lastline"
Result:
Firstline
Lastline
Question: Why doesn't 1. create the following result:
Firstline
Lastline
I know that this specific issue could have been worked around using other techniques, but I want to focus on why 1. does not work.
Edited:
When using echo instead of printf, I get the expected result, but why does printf work differently?
NewLine=`echo "\n"`
echo -e "Firstline${NewLine}Lastline"
Result:
Firstline
Lastline
The backtick operator removes trailing new lines. See 3.4.5. Command substitution at http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_03_04.html
Note on edited question
Compare:
[alvaro#localhost ~]$ printf "\n"
[alvaro#localhost ~]$ echo "\n"
\n
[alvaro#localhost ~]$ echo -e "\n"
[alvaro#localhost ~]$
The echo command doesn't treat \n as a newline unless you tell him to do so:
NAME
echo - display a line of text
[...]
-e enable interpretation of backslash escapes
POSIX 7 specifies this behaviour here:
[...] with the standard output of the command, removing sequences of one or more characters at the end of the substitution
Maybe people will come here with the same problem I had:
echoing \n inside a code wrapped in backsticks. A little tip:
printf "astring\n"
# and
printf "%s\n" "astring"
# both have the same effect.
# So... I prefer the less typing one
The short answer is:
# Escape \n correctly !
# Using just: printf "$myvar\n" causes this effect inside the backsticks:
printf "banana
"
# So... you must try \\n that will give you the desired
printf "banana\n"
# Or even \\\\n if this string is being send to another place
# before echoing,
buffer="${buffer}\\\\n printf \"$othervar\\\\n\""
One common problem is that if you do inside the code:
echo 'Tomato is nice'
when surrounded with backsticks will produce the error
command Tomato not found.
The workaround is to add another echo -e or printf
printed=0
function mecho(){
#First time you need an "echo" in order bash relaxes.
if [[ $printed == 0 ]]; then
printf "echo -e $1\\\\n"
printed=1
else
echo -e "\r\n\r$1\\\\n"
fi
}
Now you can debug your code doing in prompt just:
(prompt)$ `mySuperFunction "arg1" "etc"`
The output will be nicely
mydebug: a value
otherdebug: whathever appended using myecho
a third string
and debuging internally with
mecho "a string to be hacktyped"
$ printf -v NewLine "\n"
$ echo -e "Firstline${NewLine}Lastline"
Firstline
Lastline
$ echo "Firstline${NewLine}Lastline"
Firstline
Lastline
It looks like BASH is removing trailing newlines.
e.g.
NewLine=`printf " \n\n\n"`
echo -e "Firstline${NewLine}Lastline"
Firstline Lastline
NewLine=`printf " \n\n\n "`
echo -e "Firstline${NewLine}Lastline"
Firstline
Lastline
Your edited echo version is putting a literal backslash-n into the variable $NewLine which then gets interpreted by your echo -e. If you did this instead:
NewLine=$(echo -e "\n")
echo -e "Firstline${NewLine}Lastline"
your result would be the same as in case #1. To make that one work that way, you'd have to escape the backslash and put the whole thing in single quotes:
NewLine=$(printf '\\n')
echo -e "Firstline${NewLine}Lastline"
or double escape it:
NewLine=$(printf "\\\n")
Of course, you could just use printf directly or you can set your NewLine value like this:
printf "Firstline\nLastline\n"
or
NewLine=$'\n'
echo "Firstline${NewLine}Lastline" # no need for -e
For people coming here wondering how to use newlines in arguments to printf, use %b instead of %s:
$> printf "a%sa" "\n"
a\na
$> printf "a%ba" "\n"
a
a
From the manual:
%b expand backslash escape sequences in the corresponding argument
We do not need "echo" or "printf" for creating the NewLine variable:
NewLine="
"
printf "%q\n" "${NewLine}"
echo "Firstline${NewLine}Lastline"
Bash delete all trailing newlines in commands substitution.
To save trailing newlines, assign printf output to the variable with printf -v VAR
instead of
NewLine=`printf "\n"`
echo -e "Firstline${NewLine}Lastline"
#FirstlineLastline
use
printf -v NewLine '\n'
echo -e "Firstline${NewLine}Lastline"
#Firstline
#Lastline
Explanation
According to bash man
3.5.4 Command Substitution
$(command)
or
`command`
Bash performs the expansion by executing command and replacing the command substitution with the standard output of the command, with any trailing newlines deleted. Embedded newlines are not deleted, but they may be removed during word splitting.
So, after adding any trailing newlines, bash will delete them.
var=$(printf '%s\n%s\n\n\n' 'foo' 'bar')
echo "$var"
output:
foo
bar
According to help printf
printf [-v var] format [arguments]
If the -v option is supplied, the output is placed into the value of the shell variable VAR rather than being sent to the standard output.
In this case, for safe copying of formatted text to the variable, use the [-v var] option:
printf -v var '%s\n%s\n\n\n' 'foo' 'bar'
echo "$var"
output:
foo
bar
Works ok if you add "\r"
$ nl=`printf "\n\r"` && echo "1${nl}2"
1
2