I need to extract a variable length sub string using Korn shell on Linux.
Sample String: "SID_LIST_ORADBPOC1LSN ="
Need to extract substring: "ORADBPOC1LSN"
Note: The sample substring is of variable length.
Thanks in advance.
FR
With pure bash's parameter expansion capability.
var="SID_LIST_ORADBPOC1LSN =" ##Creating shell variable here.
temp1="${var##*_}" ##Removing everything till _ ## for longest match here.
echo "${temp1/ =/}" ##Substituting space = with null here.
ORADBPOC1LSN
I am printing value of temp1's parameter expansion, you could save this into variable too as per your need.
OR if you want to do it in a single awk or so then try:
echo "$var" | awk -F'_| ' '{print $3}'
Related
I have this variable:
A="Some variable has value abc.123"
I need to extract this value i.e abc.123. Is this possible in bash?
Simplest is
echo "$A" | awk '{print $NF}'
Edit: explanation of how this works...
awk breaks the input into different fields, using whitespace as the separator by default. Hardcoding 5 in place of NF prints out the 5th field in the input:
echo "$A" | awk '{print $5}'
NF is a built-in awk variable that gives the total number of fields in the current record. The following returns the number 5 because there are 5 fields in the string "Some variable has value abc.123":
echo "$A" | awk '{print NF}'
Combining $ with NF outputs the last field in the string, no matter how many fields your string contains.
Yes; this:
A="Some variable has value abc.123"
echo "${A##* }"
will print this:
abc.123
(The ${parameter##word} notation is explained in §3.5.3 "Shell Parameter Expansion" of the Bash Reference Manual.)
Some examples using parameter expansion
A="Some variable has value abc.123"
echo "${A##* }"
abc.123
Longest match on " " space
echo "${A% *}"
Some variable has value
Longest match on . dot
echo "${A%.*}"
Some variable has value abc
Shortest match on " " space
echo "${A%% *}"
some
Read more Shell-Parameter-Expansion
The documentation is a bit painful to read, so I've summarised it in a simpler way.
Note that the '*' needs to swap places with the ' ' depending on whether you use # or %. (The * is just a wildcard, so you may need to take off your "regex hat" while reading.)
${A% *} - remove shortest trailing * (strip the last word)
${A%% *} - remove longest trailing * (strip the last words)
${A#* } - remove shortest leading * (strip the first word)
${A##* } - remove longest leading * (strip the first words)
Of course a "word" here may contain any character that isn't a literal space.
You might commonly use this syntax to trim filenames:
${A##*/} removes all containing folders, if any, from the start of the path, e.g.
/usr/bin/git -> git
/usr/bin/ -> (empty string)
${A%/*} removes the last file/folder/trailing slash, if any, from the end:
/usr/bin/git -> /usr/bin
/usr/bin/ -> /usr/bin
${A%.*} removes the last extension, if any (just be wary of things like my.path/noext):
archive.tar.gz -> archive.tar
How do you know where the value begins? If it's always the 5th and 6th words, you could use e.g.:
B=$(echo "$A" | cut -d ' ' -f 5-)
This uses the cut command to slice out part of the line, using a simple space as the word delimiter.
As pointed out by Zedfoxus here. A very clean method that works on all Unix-based systems. Besides, you don't need to know the exact position of the substring.
A="Some variable has value abc.123"
echo "$A" | rev | cut -d ' ' -f 1 | rev
# abc.123
More ways to do this:
(Run each of these commands in your terminal to test this live.)
For all answers below, start by typing this in your terminal:
A="Some variable has value abc.123"
The array example (#3 below) is a really useful pattern, and depending on what you are trying to do, sometimes the best.
1. with awk, as the main answer shows
echo "$A" | awk '{print $NF}'
2. with grep:
echo "$A" | grep -o '[^ ]*$'
the -o says to only retain the matching portion of the string
the [^ ] part says "don't match spaces"; ie: "not the space char"
the * means: "match 0 or more instances of the preceding match pattern (which is [^ ]), and the $ means "match the end of the line." So, this matches the last word after the last space through to the end of the line; ie: abc.123 in this case.
3. via regular bash "indexed" arrays and array indexing
Convert A to an array, with elements being separated by the default IFS (Internal Field Separator) char, which is space:
Option 1 (will "break in mysterious ways", as #tripleee put it in a comment here, if the string stored in the A variable contains certain special shell characters, so Option 2 below is recommended instead!):
# Capture space-separated words as separate elements in array A_array
A_array=($A)
Option 2 [RECOMMENDED!]. Use the read command, as I explain in my answer here, and as is recommended by the bash shellcheck static code analyzer tool for shell scripts, in ShellCheck rule SC2206, here.
# Capture space-separated words as separate elements in array A_array, using
# a "herestring".
# See my answer here: https://stackoverflow.com/a/71575442/4561887
IFS=" " read -r -d '' -a A_array <<< "$A"
Then, print only the last elment in the array:
# Print only the last element via bash array right-hand-side indexing syntax
echo "${A_array[-1]}" # last element only
Output:
abc.123
Going further:
What makes this pattern so useful too is that it allows you to easily do the opposite too!: obtain all words except the last one, like this:
array_len="${#A_array[#]}"
array_len_minus_one=$((array_len - 1))
echo "${A_array[#]:0:$array_len_minus_one}"
Output:
Some variable has value
For more on the ${array[#]:start:length} array slicing syntax above, see my answer here: Unix & Linux: Bash: slice of positional parameters, and for more info. on the bash "Arithmetic Expansion" syntax, see here:
https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html#Arithmetic-Expansion
https://www.gnu.org/savannah-checkouts/gnu/bash/manual/bash.html#Shell-Arithmetic
You can use a Bash regex:
A="Some variable has value abc.123"
[[ $A =~ [[:blank:]]([^[:blank:]]+)$ ]] && echo "${BASH_REMATCH[1]}" || echo "no match"
Prints:
abc.123
That works with any [:blank:] delimiter in the current local (Usually [ \t]). If you want to be more specific:
A="Some variable has value abc.123"
pat='[ ]([^ ]+)$'
[[ $A =~ $pat ]] && echo "${BASH_REMATCH[1]}" || echo "no match"
echo "Some variable has value abc.123"| perl -nE'say $1 if /(\S+)$/'
I am newbie in shell script , may be stupid query to experts, I am using following code to remove leading and trailing spaces from value, how do I assign output of echo variable to StringVar variable again or to other Variable. I am using ksh shell.
StringVar= ' abc '
echo StringVar | awk '{$1=$1};1'
x=$(command)
where x is the variable to which you want to assign the output of the command.
In your case, do not give space before or after the assignment operator =
StringVar=' abc '
x=$(echo "$StringVar" | awk '{$1=$1};1')
I have tried few options but that not working on my case. My requirement is..
Suppose I have a parameter in a file and wanted to capture the details as below and run a shell script(ksh).
PARAMETR=aname1:7,aname2:5
The parameter contains 2 values delimited by a comma and each value separated by a colon.
So, wanted to process it as if the string matched as aname1 then print both in different variable $v1=aname1 and $v2=7. The same applies to the other value too if string searched as aname2 then $v1=aname2 and $v2=5.
Thank you in advance.
That will do what you're asking for
#!/bin/ksh
typeset -A valueArray
PARAMETR=aname1:7,aname2:5
paramArray=(${PARAMETR//,/ })
for ((i=0;i<=${#paramArray[#]};i++)); do
valueArray[${paramArray[$i]%:*}]=${paramArray[$i]#*:}
done
for j in ${!valueArray[#]}; do
print "$j = ${valueArray[$j]}"
done
Hope it can help
First split the line in two sets and than process each set.
echo "${PARAMETR}" | tr "," "\n" | while IFS=: read -r v1 v2; do
echo "v1=$v1 and v2=$v2"
done
Result:
v1=aname1 and v2=7
v1=aname2 and v2=5
Say I have a string like
s1="sxfn://xfn.oxbr.ac.uk:8843/xfn/mech2?XFN=/castor/
xf.oxbr.ac.uk/prod/oxbr.ac.uk/disk/xf20.m.ac.uk/prod/v1.8/pienug_ib-2/reco_c21_dr3809_r35057.dst"
or
s2="sxfn://xfn.gla.ac.uk:8841/xfn/mech2?XFN=/castor/
xf.gla.ac.uk/space/disk1/prod/v1.8/pienug_ib-2/reco_c21_dr3809_r35057.dst"
and I want in my script to extract the last part starting from prod/ i.e. "prod/v1.8/pienug_ib-2/reco_c21_dr3809_r35057.dst". Note that $s1 contains two occurrences of "prod/".
What is the most elegant way to do this in bash?
Using BASH string manipulations you can do:
echo "prod/${s1##*prod/}"
prod/v1.8/pienug_ib-2/reco_c21_dr3809_r35057.dst
echo "prod/${s2##*prod/}"
prod/v1.8/pienug_ib-2/reco_c21_dr3809_r35057.dst
With awk (which is a little overpowered for this, but it may be helpful if you have a file full of these strings you need to parse:
echo "sxfn://xfn.gla.ac.uk:8841/xfn/mech2?XFN=/castor/xf.gla.ac.uk/space/disk1/prod/v1.8/pienug_ib-2/reco_c21_dr3809_r35057.dst" | awk -F"\/prod" '{print "/prod"$NF}'
That's splitting the string by '/prod' then printing out the '/prod' delimiter and the last token in the string ($NF)
sed can do it nicely:
s1="sxfn://xfn.oxbr.ac.uk:8843/xfn/mech2?XFN=/castor/xf.oxbr.ac.uk/prod/oxbr.ac.uk/disk/xf20.m.ac.uk/prod/v1.8/pienug_ib-2/reco_c21_dr3809_r35057.dst"
echo "$s1" | sed 's/.*\/prod/\/prod/'
this relies on the earger matching of the .* part up front.
Can you please help me solve this puzzle? I am trying to print the location of a string (i.e., line #) in a file, first to the std output, and then capture that value in a variable to be used later. The string is “my string”, the file name is “myFile” which is defined as follows:
this is first line
this is second line
this is my string on the third line
this is fourth line
the end
Now, when I use this command directly at the command prompt:
% awk ‘s=index($0, “my string”) { print “line=” NR, “position= ” s}’ myFile
I get exactly the result I want:
% line= 3, position= 9
My question is: if I define a variable VAR=”my string”, why can’t I get the same result when I do this:
% awk ‘s=index($0, $VAR) { print “line=” NR, “position= ” s}’ myFile
It just won’t work!! I even tried putting the $VAR in quotation marks, to no avail? I tried using VAR (without the $ sign), no luck. I tried everything I could possibly think of ... Am I missing something?
awk variables are not the same as shell variables. You need to define them with the -v flag
For example:
$ awk -v var="..." '$0~var{print NR}' file
will print the line number(s) of pattern matches. Or for your case with the index
$ awk -v var="$Var" 'p=index($0,var){print NR,p}' file
using all uppercase may not be good convention since you may accidentally overwrite other variables.
to capture the output into a shell variable
$ info=$(awk ...)
for multi line output assignment to shell array, you can do
$ values=( $(awk ...) ); echo ${values[0]}
however, if the output contains more than one field, it will be assigned it's own array index. You can change it with setting the IFS variable, such as
$ IFS=$(echo -en "\n\b"); values=( $(awk ...) )
which will capture the complete lines as the array values.