Perl substitution output drops characters from Bash script input

Question

I have a variable in a bash script with a length of 64 characters

authkey=$(LC_ALL=C tr -cd 'a-zA-Z0-9,;.:_#*+~!@$%&()=?{[]}|><-' < /dev/random | head -c 64)

if i parse the variable to perl to do a string substitution

perl -pi -e "s/'AUTH_KEY',         'put your unique phrase here'/'AUTH_KEY',         '$authkey'/" test.txt

depending on the selected random characters the length of the output differs.The output looks the following (The first string is the output in the resulting text file, the second string is the echo'ed output of the variable in the bash script)

q=dB7oUz59.IDBXI:i>ckW4oy3smX&k:-C.[rIf*9w}H=(N93yiB&nk{fP:y0_
q=dB7oUz59.IDBXI:i>ckW4oy3smX&k$s:-C.[rIf*9w}H=(N93yiB&nk{fP:y0_

5A+BwP~l3~<evp.ciTkMYtvmPjyMrL=):Qj1VaMI(,TSS,ZGMcd.m,4W
5A+BwP~l3~<evp.ciTkMYtvmPjyMrL=):Qj1VaMI(@Dk7UNgs,TSS,ZGMcd.m,4W

dX73}i5G1d;L*J=60WHHe<!61Ji_KJ)T5B~b2bCfaNDjBQr_N]}3HS=;GzAaX<gB
dX73}i5G1d;L*J=60WHHe<!61Ji_KJ)T5B~b2bCfaNDjBQr_N]}3HS=;GzAaX<gB

6Ndn(9+:>(6>*rh?B.m),3POp)>sfm8c1rh9vXr~fzZj;]!)kf3#60=M
6Ndn(9+:>(6>*rh?B.m),3POp)>sfm8c1rh9vXr~fzZj;]@YH!)kf3#$=$$ckt=M

FYMI,K|6WutC&dr-3]6)f(>QU-~{vBX>n!J-zq:kK84T|fZ7UW:{1&qU[nwYZLmC
FYMI,K|6WutC&dr-3]6)f(>QU-~{vBX>n!J-zq:kK84T|fZ7UW:{1&qU[nwYZLmC

5A+BwP~l3~<evp.ciTkMYtvmPjyMrL=):Qj1VaMI(,TSS,ZGMcd.m,4W
5A+BwP~l3~<evp.ciTkMYtvmPjyMrL=):Qj1VaMI(@Dk7UNgs,TSS,ZGMcd.m,4W

v1FR8c8}dZD(QGwOrr%M{FSUw*?h.JGI?Ay4tgRVp~l7C5eAxW<w<;c}emeX#S
v1FR8c8}dZD(QGwOrr%M{FSUw*?h.JGI?Ay4tgRVp@s~l7C5eAxW<w<;c}emeX#S

+MGg0=*NrhJ}.qPkk6v[lc)J.uiW1o?LL5t<HTC#Q-hSeqn%-ke!cwL5tk[e
+MGg$|=*NrhJ}.qPkk6v[lc)J.uiW1o?L$55L5t<HTC#Q-hSeqn%-ke!cwL5tk[e

each character dropout was caused by either a $ or @ at the beginning of the group of characters. Is there a way to prevent that behaviour? Best regards Ralf

Perl performs variable interpolation in the replacement string, and $ and @ are variable prefixes. See stackoverflow.com/questions/3694851/… for a related issue. — Barmar
– Barmar, Commented May 23, 2014 at 23:58

Borodin · Accepted Answer · 2014-05-24 01:20:12Z

1

Using a single quote ' as the delimiter instead of slash / for the substitution suppresses variable interpolation

$ foobar=\$bar; perl -p -e "s'foo'$foobar'"
xx
xx
foo
$bar
^C
$

Unfortunately the single quotes that are matched in the substitution now need escaping

foobar=\$bar; perl -p -e "s'\'foo'$foobar\''"
x
x
'foo
$bar'
^C

But that seems to get passed through to Perl OK, without munging the authkey contents with sed

edited May 24, 2014 at 1:20

Borodin

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answered May 24, 2014 at 1:17

Rob11311

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Comments

Tiago Lopo · Accepted Answer · 2014-05-24 00:05:20Z

0

You can escape $ and @ before calling perl:

 authey=$(echo -n "$authkey" | sed -re 's/(\$|\@)/\\\1/g')

answered May 24, 2014 at 0:05

Tiago Lopo

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