I have done some processing in perl, and got the result in perl's hash data structure. Usually in bash, when I try to retrieve result from other script like
output=$(perl -E '...')
I got the output in string. Is it possible to save the result in bash array?
Assuming a perl variable hash is an associative array, please try:
declare -A "output=($(perl -e '
$hash{"foo"} = "xx"; # just an example
$hash{"bar"} = "yy"; # ditto
for (keys %hash) {print "[\"$_\"]=\"$hash{$_}\"\n"}'))"
for i in "${!output[#]}"; do
echo "$i => ${output[$i]}" # see the result
done
The outermost double quotes around output=.. is required to tell declare
to evaluate the argument.
[Update]
Considering tripleee's comment, here is a robust version against special characters:
mapfile -d "" -t a < <(perl -e '
$hash{"baz"} = "boo"; # example
$hash{"foo"} = "x\"x"; # example with a double quote
$hash{"bar"} = "y\ny"; # example with a newline
print join("\0", %hash), "\0"') # use a nul byte as a delimiter
declare -A output # bash associative array
for ((i = 0; i < ${#a[#]}; i+=2 )); do
output[${a[i]}]=${a[i+1]} # key and value pair
done
for i in "${!output[#]}"; do
echo "$i => ${output[$i]}" # see the result
done
The conversion from perl variables to bash variables works only if they are free of null bytes (\n), as perl can store null bytes in strings, but bash cannot.
At least, we can use that limitation to print the hash in perl with null delimiters and safely parse it in bash again:
declare -A "array=($(
perl -e 'print join("\0", %hash), "\0"' |
xargs -0 printf '[%q]=%q '
))"
Please note that neither %q nor -0 are specified by posix. For a more portable solution see tshiono's answer.
If the hash is very big such that ARG_MAX might be exceeded you should ensure that xargs does not split a key value pair across two calls to printf. To do so, add the option -n2 (or any other number 2n where you are sure that n key value pairs never exceed ARG_MAX).
Related
this is probably a very simple question. I looked at other answers but couldn't come up with a solution. I have a 365 line date file. file as below,
01-01-2000
02-01-2000
I need to read this file line by line and assign each day to a separate variable. like this,
d001=01-01-2000
d002=02-01-2000
I tried while read commands but couldn't get them to work.It takes a lot of time to shoot one by one. How can I do it quickly?
Trying to create named variable out of an associative array, is time waste and not supported de-facto. Better use this, using an associative array:
#!/bin/bash
declare -A array
while read -r line; do
printf -v key 'd%03d' $((++c))
array[$key]=$line
done < file
Output
for i in "${!array[#]}"; do echo "key=$i value=${array[$i]}"; done
key=d001 value=01-01-2000
key=d002 value=02-01-2000
Assumptions:
an array is acceptable
array index should start with 1
Sample input:
$ cat sample.dat
01-01-2000
02-01-2000
03-01-2000
04-01-2000
05-01-2000
One bash/mapfile option:
unset d # make sure variable is not currently in use
mapfile -t -O1 d < sample.dat # load each line from file into separate array location
This generates:
$ typeset -p d
declare -a d=([1]="01-01-2000" [2]="02-01-2000" [3]="03-01-2000" [4]="04-01-2000" [5]="05-01-2000")
$ for i in "${!d[#]}"; do echo "d[$i] = ${d[i]}"; done
d[1] = 01-01-2000
d[2] = 02-01-2000
d[3] = 03-01-2000
d[4] = 04-01-2000
d[5] = 05-01-2000
In OP's code, references to $d001 now become ${d[1]}.
A quick one-liner would be:
eval $(awk 'BEGIN{cnt=0}{printf "d%3.3d=\"%s\"\n",cnt,$0; cnt++}' your_file)
eval makes the shell variables known inside your script or shell. Use echo $d000 to show the first one of the newly defined variables. There should be no shell special characters (like * and $) inside your_file. Remove eval $() to see the result of the awk command. The \" quoted %s is to allow spaces in the variable values. If you don't have any spaces in your_file you can remove the \" before and after %s.
Consider the below variables which are dynamic and might change each time. Sometimes there might even be 5 variables, But the length of all the variables will be the same every time.
var1='a b c d e... upto z'
var2='1 2 3 4 5... upto 26'
var3='I II III IV V... upto XXVI'
I am looking for a generalized approach to iterate the variables in a for loop & My desired output should be like below.
a,1,I
b,2,II
c,3,III
d,4,IV
e,5,V
.
.
goes on upto
z,26,XXVI
If I use nested loops, then I get all possible combinations which is not the expected outcome.
Also, I know how to make this work for 2 variables using for loop and shift using below link
https://unix.stackexchange.com/questions/390283/how-to-iterate-two-variables-in-a-sh-script
With paste
paste -d , <(tr ' ' '\n' <<<"$var1") <(tr ' ' '\n' <<<"$var2") <(tr ' ' '\n' <<<"$var3")
a,1,I
b,2,II
c,3,III
d,4,IV
e...z,5...26,V...XXVI
But clearly having to add other parameter substitutions for more varN's is not scalable.
You need to "zip" two variables at a time.
var1='a b c d e...z'
var2='1 2 3 4 5...26'
var3='I II III IV V...XXVI'
zip_var1_var2 () {
set $var1
for v2 in $var2; do
echo "$1,$v2"
shift
done
}
zip_var12_var3 () {
set $(zip_var1_var2)
for v3 in $var3; do
echo "$1,$v3"
shift
done
}
for x in $(zip_var12_var3); do
echo "$x"
done
If you are willing to use eval and are sure it is safe to do so, you can write a single function like
zip () {
if [ $# -eq 1 ]; then
eval echo \$$1
return
fi
a1=$1
shift
x=$*
set $(eval echo \$$a1)
for v in $(zip $x); do
printf '=== %s\n' "$1,$v" >&2
echo "$1,$v"
shift
done
}
zip var1 var2 var3 # Note the arguments are the *names* of the variables to zip
If you can use arrays, then (for example, in bash)
var1=(a b c d e)
var2=(1 2 3 4 5)
var3=(I II III IV V)
for i in "${!var1[#]}"; do
printf '%s,%s,%s\n' "${var1[i]}" "${var2[i]}" "${var3[i]}"
done
Use this Perl one-liner:
perl -le '#in = map { [split] } #ARGV; for $i ( 0..$#{ $in[0] } ) { print join ",", map { $in[$_][$i] } 0..$#in; }' "$var1" "$var2" "$var3"
Prints:
a,1,I
b,2,II
c,3,III
d,4,IV
e,5,V
z,26,XXVI
The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-l : Strip the input line separator ("\n" on *NIX by default) before executing the code in-line, and append it when printing.
The input variables must be quoted with double quotes "like so", to keep the blank-separated words from being treated as separate arguments.
#ARGV is an array of the command line arguments, here $var1, $var2, $var3.
#in is an array of 3 elements, each element being a reference to an array obtained as a result of splitting the corresponding element of #ARGV on whitespace. Note that split splits the string on whitespace by default, but you can specify a different delimiter, it accepts regexes.
The subsequent for loop prints #in elements separated by comma.
SEE ALSO:
perldoc perlrun: how to execute the Perl interpreter: command line switches
perldoc perlvar: Perl predefined variables
The following is (almost) a copy of this answer with a few tweaks that make it fit this question.
The Original Question
First let’s assign a few variables to play with, 26 tokens in each of them:
var1="$(echo {a..z})"
var2="$(echo {1..26})"
var3="$(echo I II III IV \
V{,I,II,III} IX \
X{,I,II,III} XIV \
XV{,I,II,III} XIX \
XX{,I,II,III} XXIV \
XXV XXVI)"
var4="$(echo {A..Z})"
var5="$(echo {010101..262626..10101})"
Now we want a “magic” function that zips an arbitrary number of variables, ideally in pure Bash:
zip_vars var1 # a trivial test
zip_vars var{1..2} # a slightly less trivial test
zip_vars var{1..3} # the original question
zip_vars var{1..4} # more vars, becasuse we can
zip_vars var{1..5} # more vars, because why not
What could zip_vars look like? Here’s one in pure Bash, without any external commands:
zip_vars() {
local var
for var in "$#"; do
local -a "array_${var}"
local -n array_ref="array_${var}"
array_ref=(${!var})
local -ar "array_${var}"
done
local -n array_ref="array_${1}"
local -ir size="${#array_ref[#]}"
local -i i
local output
for ((i = 0; i < size; ++i)); do
output=
for var in "$#"; do
local -n array_ref="array_${var}"
output+=",${array_ref[i]}"
done
printf '%s\n' "${output:1}"
done
}
How it works:
It splits all variables (passed by reference (by variable name)) into arrays. For each variable varX it creates a local array array_varX.
It would be actually way easier if the input variables were already Bash arrays to start with (see below), but … we stick with the original question initially.
It determines the size of the first array and then blindly expects all arrays to be of that size.
For each index i from 0 to size - 1 it concatenates the ith elements of all arrays, separated by ,.
Arrays Make Things Easier
If you use Bash arrays from the very start, the script will be shorter and look simpler and there won’t be any string-to-array conversions.
zip_arrays() {
local -n array_ref="$1"
local -ir size="${#array_ref[#]}"
local -i i
local output
for ((i = 0; i < size; ++i)); do
output=
for arr in "$#"; do
local -n array_ref="$arr"
output+=",${array_ref[i]}"
done
printf '%s\n' "${output:1}"
done
}
arr1=({a..z})
arr2=({1..26})
arr3=( I II III IV
V{,I,II,III} IX
X{,I,II,III} XIV
XV{,I,II,III} XIX
XX{,I,II,III} XXIV
XXV
XXVI)
arr4=({A..Z})
arr5=({010101..262626..10101})
zip_arrays arr1 # a trivial test
zip_arrays arr{1..2} # a slightly less trivial test
zip_arrays arr{1..3} # (almost) the original question
zip_arrays arr{1..4} # more arrays, becasuse we can
zip_arrays arr{1..5} # more arrays, because why not
I wrote a bash script to read multiple inputs from the user
Here is the command:
read -a choice
In this way, I can put all the inputs in the choice variable as an array so that I can extract them using an index.
The problem is that when one of the inputs, which is a string has space in it, like
user1 google.com "login: myLogin\npassword: myPassword"
the read command will split the quoted string into 3 words. How can I stop this from happening?
bash doesn't process quotes in user input. The only thing I can think of is to use eval to execute an array assignment.
IFS= read -r input
eval "choice=($input)"
Unfortunately this is dangerous -- if the input contains executable code, it will be executed by eval.
You can use a tab instead of space as a field delimiter. For instance :
$ IFS=$'\t' read -a choice
value1 value2 a value with many words ## This is typed
$ echo ${choice[2]}
a value with many words
Regards!
Given risk of using eval, and the fact the input seems to have only two types of tokens: unquoted, and quoted, consider using scripting engine that will put all text into proper format that will be easy to read.
It's not clear from the example what other quoting rules are used. Example assume 'standard' escaped that can be processed with bash #E processor.
The following uses Perl one liner to generate TAB delimited tokens (hopefully, raw tabs can not be part of the input, but other character can be used instead).
input='user1 google.com "login: myLogin\npassword: myPassword"'
tsv_input=$(perl -e '$_ = " $ARGV[0]" ; print $2 // $3, "\t" while ( /\s+("([^"]*)"|(\S*))/g) ;' "$input")
IFS=$'\t' read -d '' id domain values <<< $(echo -e "${tsv_input#E}")
Or using a function to get more readable code
function data_to_tsv {
# Translate to TSV
local tsv_input=$(perl -e '$_ = " $ARGV[0]" ; print $2 // $3, "\t" while ( /\s+("([^"]*)"|(\S*))/g) ;' "$1")
# Process escapes
echo -n "${tsv_input#E}"
}
input='user1 google.com "login: myLogin\npassword: myPassword"'
IFS=$'\t' read -d '' id domain values <<< $(data_to_tsv "$input")
I have a file links.txt:
1 a.sh
3 b.sh
6 c.sh
4 d.sh
So, if i pass 1,4 as parameters to another file(master.sh), a.sh and d.sh should be stored in a variable.
sed '3!d' would print the 3rd line, but not the line that starts with 3. For that, you need sed '/^3 /!d'. The problem is you can't combine them for more lines, as this means "Delete everything that doesn't start with a 3", which means all other lines will be missed. So, use sed -n '/^3 /p' instead, i.e. don't print by default and tell sed what lines to print, not what lines to delete.
You can loop over the argument and create a sed script from them that prints the lines, then run sed using this output:
#!/bin/bash
file=$1
shift
for id in "$#" ; do
echo "/^$id /p"
done | sed -nf- "$file"
Run as script.sh filename 3 4.
If you want to remove the id from the output, you can either use
cut -f2 -d' '
or you can modify the generated sed script to do the work
echo "/^$id /s/.* //p"
i.e. only print if the substitution was successful.
This loops through each argument and greps for it in the links file. The result is piped into cut where we specify the delimiter as a space with -d flag and the field number as 2 with -f flag. Finally this is appended to the array called files.
links="links.txt"
files=()
for arg in $#; do
files=("${files[#]}" `grep "^$arg" "$links" | cut -d" " -f2`)
done;
echo ${files[#]}
Usage:
$ ./master.sh 1 4
a.sh d.sh
Edit:
As pointed out by mklement0, the solution above reads the file once per arg. The following first builds the pattern then reads the file just once.
links="links.txt"
pattern="^$1\s"
for arg in ${#:2}; do
pattern+="|^$arg\s"
done
files=$(grep -E "$pattern" "$links" | cut -d" " -f2)
echo ${files[#]}
Usage:
$ ./master.sh 1 4
a.sh d.sh
Here is another example with grep and cut:
#!/bin/bash
for line in $(grep "$1\|$2" links.txt|cut -d' ' -f2)
do
echo $line
done
Example of usage:
./master.sh 1 4
a.sh
d.sh
Why not just stores the values and call them at will:
items=()
while read -r num file
do
items[num]="$file"
done<links.txt
for arg
do
echo "${items[arg]}"
done
Now you can use the items array any time you like :)
The following awk solution:
preserves the argument order; that is, the results reflect the order in which the lookup values were specified (as opposed to the order in which the lookup values happen to occur in the file).
If that is not important (i.e., if outputting the results in file order is acceptable), the readarray technique below can be combined with this one-liner, which is a generalized variant of Panta's answer:
grep -f <(printf "^%s\n" "$#") links.txt | cut -d' ' -f2-
performs well, because the input file is only read once; the only requirement is that all key-value pairs fit into memory as a whole (as a single associative Awk array (dictionary)).
works with any lookup values that don't have embedded whitespace.
Similarly, the assumption is that the output column values (containing values such as a.sh in the sample input) have no embedded whitespace. awk doesn't handle quoted fields well, so more work would be needed.
#!/bin/bash
readarray -t files < <(
awk -v idList="$*" '
BEGIN { count=split(idList, idArr); for (i in idArr) idDict[idArr[i]]++ }
$1 in idDict { idDict[$1] = $2 }
END { for (i=1; i<=count; ++i) print idDict[idArr[i]] }
' links.txt
)
# Print results.
printf '%s\n' "${files[#]}"
readarray -t files reads stdin input (<) line by line into array variable files.
Note: readarray requires Bash v4+; on Bash 3.x, such as on macOS, replace this part with
IFS=$'\n' read -d '' -ra files
<(...) is a Bash process substitution that, loosely speaking, presents the output from the enclosed command as if it were (self-deleting) temporary file.
This technique allows readarray to run in the current shell (as opposed to a subshell if a pipeline had been used), which is necessary for the files variable to remain defined in the remainder of the script.
The awk command breaks down as follows:
-v idList="$*" passes the space-separated list of all command-line arguments as a single string to Awk variable idList.
Note that this assumes that the arguments have no embedded spaces, which is indeed the case here and also generally the case with identifiers.
BEGIN { ... } is only executed once, before the individual lines are processed:
split(idList, idArr) splits the input ID list into an array by whitespace and stores the result in idArr.
for (i in idArr) idDict[idArr[i]]++ } then converts the (conceptually regular) array into associative array idDict (dictionary), whose keys are the input IDs - this enables efficient lookup by ID later, and also allows storing the lookup result for each ID.
$1 in idDict { idDict[$1] = $2 } is processed for every input line:
Pattern $1 in idDict returns true if the line's first whitespace-separated field ($1) - e.g., 6 - is among the keys (in) of associative array idDict, and, if so, executes the associated action ({...}).
Action { idDict[$1] = $2 } then assigns the second field ($2) - e.g., c.sh - to the iDict entry for key $1.
END { ... } is executed once, after all input lines have been processed:
for (i=1; i<=count; ++i) print idDict[idArr[i]] loops over all input IDs in order and prints each ID's lookup result, which is the value of the dictionary entry with that ID.
I have a variable equal to a string, which is a series of key/value pairs separated by newlines.
I want to then replace these newline characters with spaces, and set a new variable equal to the result
From various answers on the internet I've arrived at the following:
#test.txt has the content:
#test=example
#what=s0omething
vars="$(cat ./test.txt)"
formattedVars= $("$vars" | tr '\n' ' ')
echo "$taliskerEnvVars"
Problem is when I try to set formattedVars it tries to execute the second line:
script.sh: line 7: test=example
what=s0omething: command not found
I just want formattedVars to equal test=example what=s0omething
What trick am I missing?
Change your line to:
formattedVars=$(tr '\n' ' ' <<< "$secretsContent")
Notice the space of = in your code, which is not permitted in assignment statements.
I see that you are not setting secretsContent in your code, you are setting vars instead.
If possible, use an array to hold contents of the file:
readarray -t vars < ./test.txt # bash 4
or
# bash 3.x
declare -a vars
while IFS= read -r line; do
vars+=( "$line" )
done < ./test.txt
Then you can do what you need with the array. You can make your space-separated list with
formattedVars="${vars[*]}"
, but consider whether you need to. If the goal is to use them as a pre-command modifier, use, for instance,
"${vars[#]}" my_command arg1 arg2