Split String with a word delimiter into array in bash

Question

I have a string like below:

dn: uid=svc_ddvportal2ssh,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_ddvportal2ssh svc_ddvportal2ssh krbpasswordexpiration: 20180607182429Z dn: uid=svc_ddvrundeckdeploy,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_ddvrundeckdeploy svc_ddvrundeckdeploy krbpasswordexpiration: 20180607182430Z dn: uid=svc_bo2ansible_mon,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_bo2ansible_mon svc_bo2ansible_mon krbpasswordexpiration: 20990101200000Z

wanted to extract like below into array a

a[0] = dn: uid=svc_bo2icinga2ipa,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_bo2icinga2ipa svc_bo2icinga2ipa krbpasswordexpiration: 20180119194104Z
a[1] = dn: uid=svc_ddvrundeckdeploy,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_ddvrundeckdeploy svc_ddvrundeckdeploy krbpasswordexpiration: 20180607182430Z
a[2] = uid=svc_bo2ansible_mon,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_bo2ansible_mon svc_bo2ansible_mon krbpasswordexpiration: 20990101200000Z

I tried using IFS delimitor but not succeeded. Please someone help me into this

Which word, specifically, is the delimiter? Can you build shorter / easier to read sample data that makes the logic you want more obvious? (The minimal reproducible example rules don't just apply to code). — Charles Duffy
– Charles Duffy, Commented Sep 24, 2019 at 19:45
I have this suspicion that the line feeds aren't suppose to exist in the input string. — JNevill
– JNevill, Commented Sep 24, 2019 at 19:47
@JNevill, ...what line feeds? I don't see any in the question's text; it's just in single backticks to allow line wrapping to be done by the browser, so folks can actually read the input data without scrolling left-to-right. — Charles Duffy
– Charles Duffy, Commented Sep 24, 2019 at 19:48

Cyrus · Accepted Answer · 2019-09-24 19:56:34Z

2

Build your array with GNU sed:

mapfile -t a < <(sed -r 's/ .{14}Z /&\n/g' file)

See: The Stack Overflow Regular Expressions FAQ

edited Sep 24, 2019 at 19:56

answered Sep 24, 2019 at 19:53

Cyrus

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2 Comments

Charles Duffy Over a year ago

Maybe add a readarray or mapfile consuming the output of that sed command, since the OP wants to populate an array?

Charles Duffy Over a year ago

BTW, I'm curious which of us is right wrt. guessing the OP's intended delimiter -- you're stopping after Zs, I'm starting at a dn:. (BTW, maybe change it from .{14} to [[:digit:]]{14}?)

Charles Duffy · Accepted Answer · 2019-09-24 19:56:47Z

This is doable, yes. If you don't mind reversing the array's order, the easiest approach would look something like:

s='dn: uid=svc_ddvportal2ssh,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_ddvportal2ssh svc_ddvportal2ssh krbpasswordexpiration: 20180607182429Z dn: uid=svc_ddvrundeckdeploy,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_ddvrundeckdeploy svc_ddvrundeckdeploy krbpasswordexpiration: 20180607182430Z dn: uid=svc_bo2ansible_mon,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_bo2ansible_mon svc_bo2ansible_mon krbpasswordexpiration: 20990101200000Z'

while [[ $s =~ (.*)(dn: .*) ]]; do
  results+=( "${BASH_REMATCH[2]}" )
  s=${BASH_REMATCH[1]}
done
declare -p results # print the resulting array

...which emits as output (with line breaks added for readability):

declare -a results='(
  [0]="dn: uid=svc_bo2ansible_mon,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_bo2ansible_mon svc_bo2ansible_mon krbpasswordexpiration: 20990101200000Z"
  [1]="dn: uid=svc_ddvrundeckdeploy,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_ddvrundeckdeploy svc_ddvrundeckdeploy krbpasswordexpiration: 20180607182430Z "
  [2]="dn: uid=svc_ddvportal2ssh,cn=users,cn=accounts,dc=tenant,dc=ycsdev,dc=io cn: svc_ddvportal2ssh svc_ddvportal2ssh krbpasswordexpiration: 20180607182429Z "
)'

If by contrast you're willing to eat a performance penalty to retain the original ordering, change results+=( "${BASH_REMATCH[2]}" ) to results=( "${BASH_REMATCH[2]}" "${results[@]}" ).

What we're doing here is matching the sequences from the back to the front (since bash doesn't have non-greedy regular expressions, the first .* matches everything to the last dn:).

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