How to iterate through a set of variables, then expand the variables in Bash
I'm trying to set up a "check if machines are online" script with Bash, but running into an issue of when and where to define the variables so they're expanded properly. Something like:
#!/bin/bash
rm01="c01 c02 c03"
rm02="d01 d02 d03"
rm10="e11 e22 e33"
for room in rm01 rm02 rm03; do
echo $room
for computer in $room; do
#run various nslookup/ping tests and report
done
done
exit 0
I'm running into issues because I can't find a way to expand $room for its corresponding set of computers (in $rm01, $rm02, $rm10) listed at the beginning.
What am I doing wrong?
bash variables iteration
edited Nov 21 '18 at 15:50
Benjamin W.
20.4k134656
asked Nov 21 '18 at 15:37
phonedog365phonedog365
133
add a comment |
I'm trying to set up a "check if machines are online" script with Bash, but running into an issue of when and where to define the variables so they're expanded properly. Something like:
#!/bin/bash
rm01="c01 c02 c03"
rm02="d01 d02 d03"
rm10="e11 e22 e33"
for room in rm01 rm02 rm03; do
echo $room
for computer in $room; do
#run various nslookup/ping tests and report
done
done
exit 0
I'm running into issues because I can't find a way to expand $room for its corresponding set of computers (in $rm01, $rm02, $rm10) listed at the beginning.
What am I doing wrong?
bash variables iteration
edited Nov 21 '18 at 15:50
Benjamin W.
20.4k134656
asked Nov 21 '18 at 15:37
phonedog365phonedog365
133
add a comment |
I'm trying to set up a "check if machines are online" script with Bash, but running into an issue of when and where to define the variables so they're expanded properly. Something like:
#!/bin/bash
rm01="c01 c02 c03"
rm02="d01 d02 d03"
rm10="e11 e22 e33"
for room in rm01 rm02 rm03; do
echo $room
for computer in $room; do
#run various nslookup/ping tests and report
done
done
exit 0
I'm running into issues because I can't find a way to expand $room for its corresponding set of computers (in $rm01, $rm02, $rm10) listed at the beginning.
What am I doing wrong?
bash variables iteration
edited Nov 21 '18 at 15:50
Benjamin W.
20.4k134656
asked Nov 21 '18 at 15:37
phonedog365phonedog365
133
I'm trying to set up a "check if machines are online" script with Bash, but running into an issue of when and where to define the variables so they're expanded properly. Something like:
#!/bin/bash
rm01="c01 c02 c03"
rm02="d01 d02 d03"
rm10="e11 e22 e33"
for room in rm01 rm02 rm03; do
echo $room
for computer in $room; do
#run various nslookup/ping tests and report
done
done
exit 0
I'm running into issues because I can't find a way to expand $room for its corresponding set of computers (in $rm01, $rm02, $rm10) listed at the beginning.
What am I doing wrong?
bash variables iteration
bash variables iteration
edited Nov 21 '18 at 15:50
Benjamin W.
20.4k134656
asked Nov 21 '18 at 15:37
phonedog365phonedog365
133
edited Nov 21 '18 at 15:50
Benjamin W.
20.4k134656
asked Nov 21 '18 at 15:37
phonedog365phonedog365
133
edited Nov 21 '18 at 15:50
Benjamin W.
20.4k134656
edited Nov 21 '18 at 15:50
Benjamin W.
20.4k134656
edited Nov 21 '18 at 15:50
Benjamin W.
20.4k134656
20.4k134656
asked Nov 21 '18 at 15:37
phonedog365phonedog365
133
asked Nov 21 '18 at 15:37
phonedog365phonedog365
133
asked Nov 21 '18 at 15:37
phonedog365phonedog365
133
133
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The quick fix is to use variable indirection:
for computer in ${!room}; do
Relying on word splitting is rarely the best idea, though. You could use arrays and namerefs instead (requires Bash 4.3 or newer):
#!/usr/bin/env bash
# Declare arrays
rm01=(c01 c02 c03)
rm02=(d01 d02 d03)
rm03=(e11 e22 e33)
# Declare room as nameref
declare -n room
# Using nameref as control variable sets room as reference to each variable in turn
for room in rm{01..03}; do
# Properly quoted array expansion
for computer in "${room[@]}"; do
echo "$computer" # or whatever needs to be done
done
done
exit 0
answered Nov 21 '18 at 15:49
Benjamin W.Benjamin W.
20.4k134656
Is there a way to do the arrays/nameref method with older bash? I'm stuck on 3.4 because Apple updates its BSD binaries infrequently. The first example is working just fine, thanks.
– phonedog365
Nov 22 '18 at 19:19
@phonedog365 You could install a more recent Bash with Homebrew; I think you're actually on Bash 3.2 with macOS. If you can't use namerefs, you can use indirection as in this Q&A. Another good read is BashFAQ/006.
– Benjamin W.
Nov 22 '18 at 19:31
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The quick fix is to use variable indirection:
for computer in ${!room}; do
Relying on word splitting is rarely the best idea, though. You could use arrays and namerefs instead (requires Bash 4.3 or newer):
#!/usr/bin/env bash
# Declare arrays
rm01=(c01 c02 c03)
rm02=(d01 d02 d03)
rm03=(e11 e22 e33)
# Declare room as nameref
declare -n room
# Using nameref as control variable sets room as reference to each variable in turn
for room in rm{01..03}; do
# Properly quoted array expansion
for computer in "${room[@]}"; do
echo "$computer" # or whatever needs to be done
done
done
exit 0
answered Nov 21 '18 at 15:49
Benjamin W.Benjamin W.
20.4k134656
Is there a way to do the arrays/nameref method with older bash? I'm stuck on 3.4 because Apple updates its BSD binaries infrequently. The first example is working just fine, thanks.
– phonedog365
Nov 22 '18 at 19:19
@phonedog365 You could install a more recent Bash with Homebrew; I think you're actually on Bash 3.2 with macOS. If you can't use namerefs, you can use indirection as in this Q&A. Another good read is BashFAQ/006.
– Benjamin W.
Nov 22 '18 at 19:31
add a comment |
The quick fix is to use variable indirection:
for computer in ${!room}; do
Relying on word splitting is rarely the best idea, though. You could use arrays and namerefs instead (requires Bash 4.3 or newer):
#!/usr/bin/env bash
# Declare arrays
rm01=(c01 c02 c03)
rm02=(d01 d02 d03)
rm03=(e11 e22 e33)
# Declare room as nameref
declare -n room
# Using nameref as control variable sets room as reference to each variable in turn
for room in rm{01..03}; do
# Properly quoted array expansion
for computer in "${room[@]}"; do
echo "$computer" # or whatever needs to be done
done
done
exit 0
answered Nov 21 '18 at 15:49
Benjamin W.Benjamin W.
20.4k134656
Is there a way to do the arrays/nameref method with older bash? I'm stuck on 3.4 because Apple updates its BSD binaries infrequently. The first example is working just fine, thanks.
– phonedog365
Nov 22 '18 at 19:19
@phonedog365 You could install a more recent Bash with Homebrew; I think you're actually on Bash 3.2 with macOS. If you can't use namerefs, you can use indirection as in this Q&A. Another good read is BashFAQ/006.
– Benjamin W.
Nov 22 '18 at 19:31
add a comment |
The quick fix is to use variable indirection:
for computer in ${!room}; do
Relying on word splitting is rarely the best idea, though. You could use arrays and namerefs instead (requires Bash 4.3 or newer):
#!/usr/bin/env bash
# Declare arrays
rm01=(c01 c02 c03)
rm02=(d01 d02 d03)
rm03=(e11 e22 e33)
# Declare room as nameref
declare -n room
# Using nameref as control variable sets room as reference to each variable in turn
for room in rm{01..03}; do
# Properly quoted array expansion
for computer in "${room[@]}"; do
echo "$computer" # or whatever needs to be done
done
done
exit 0
answered Nov 21 '18 at 15:49
Benjamin W.Benjamin W.
20.4k134656
The quick fix is to use variable indirection:
for computer in ${!room}; do
Relying on word splitting is rarely the best idea, though. You could use arrays and namerefs instead (requires Bash 4.3 or newer):
#!/usr/bin/env bash
# Declare arrays
rm01=(c01 c02 c03)
rm02=(d01 d02 d03)
rm03=(e11 e22 e33)
# Declare room as nameref
declare -n room
# Using nameref as control variable sets room as reference to each variable in turn
for room in rm{01..03}; do
# Properly quoted array expansion
for computer in "${room[@]}"; do
echo "$computer" # or whatever needs to be done
done
done
exit 0
answered Nov 21 '18 at 15:49
Benjamin W.Benjamin W.
20.4k134656
edited Nov 21 '18 at 15:59
answered Nov 21 '18 at 15:49
Benjamin W.Benjamin W.
20.4k134656
answered Nov 21 '18 at 15:49
Benjamin W.Benjamin W.
20.4k134656
answered Nov 21 '18 at 15:49
Benjamin W.Benjamin W.
20.4k134656
20.4k134656
Is there a way to do the arrays/nameref method with older bash? I'm stuck on 3.4 because Apple updates its BSD binaries infrequently. The first example is working just fine, thanks.
– phonedog365
Nov 22 '18 at 19:19
@phonedog365 You could install a more recent Bash with Homebrew; I think you're actually on Bash 3.2 with macOS. If you can't use namerefs, you can use indirection as in this Q&A. Another good read is BashFAQ/006.
– Benjamin W.
Nov 22 '18 at 19:31
add a comment |
Is there a way to do the arrays/nameref method with older bash? I'm stuck on 3.4 because Apple updates its BSD binaries infrequently. The first example is working just fine, thanks.
– phonedog365
Nov 22 '18 at 19:19
@phonedog365 You could install a more recent Bash with Homebrew; I think you're actually on Bash 3.2 with macOS. If you can't use namerefs, you can use indirection as in this Q&A. Another good read is BashFAQ/006.
– Benjamin W.
Nov 22 '18 at 19:31
Is there a way to do the arrays/nameref method with older bash? I'm stuck on 3.4 because Apple updates its BSD binaries infrequently. The first example is working just fine, thanks.
– phonedog365
Nov 22 '18 at 19:19
Is there a way to do the arrays/nameref method with older bash? I'm stuck on 3.4 because Apple updates its BSD binaries infrequently. The first example is working just fine, thanks.
– phonedog365
Nov 22 '18 at 19:19
@phonedog365 You could install a more recent Bash with Homebrew; I think you're actually on Bash 3.2 with macOS. If you can't use namerefs, you can use indirection as in this Q&A. Another good read is BashFAQ/006.
– Benjamin W.
Nov 22 '18 at 19:31
@phonedog365 You could install a more recent Bash with Homebrew; I think you're actually on Bash 3.2 with macOS. If you can't use namerefs, you can use indirection as in this Q&A. Another good read is BashFAQ/006.
– Benjamin W.
Nov 22 '18 at 19:31
add a comment |
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