How to treat a command output as text
I need to compare a command output with a string.
This is the scenario:
pvs_var=$(pvs | grep "sdb1")
so pvs var is: /dev/sdb1 vg_name lvm2 a-- 100.00g 0
if [[ $($pvs_var | awk '{ print $2 }') = vg_name ]]; then
do something
fi
The issue is that the output of the if statement is
-bash: /dev/sdb1: Permission denied
I don't understand this behavior.
Thank you
bash command string output
add a comment |
I need to compare a command output with a string.
This is the scenario:
pvs_var=$(pvs | grep "sdb1")
so pvs var is: /dev/sdb1 vg_name lvm2 a-- 100.00g 0
if [[ $($pvs_var | awk '{ print $2 }') = vg_name ]]; then
do something
fi
The issue is that the output of the if statement is
-bash: /dev/sdb1: Permission denied
I don't understand this behavior.
Thank you
bash command string output
add a comment |
I need to compare a command output with a string.
This is the scenario:
pvs_var=$(pvs | grep "sdb1")
so pvs var is: /dev/sdb1 vg_name lvm2 a-- 100.00g 0
if [[ $($pvs_var | awk '{ print $2 }') = vg_name ]]; then
do something
fi
The issue is that the output of the if statement is
-bash: /dev/sdb1: Permission denied
I don't understand this behavior.
Thank you
bash command string output
I need to compare a command output with a string.
This is the scenario:
pvs_var=$(pvs | grep "sdb1")
so pvs var is: /dev/sdb1 vg_name lvm2 a-- 100.00g 0
if [[ $($pvs_var | awk '{ print $2 }') = vg_name ]]; then
do something
fi
The issue is that the output of the if statement is
-bash: /dev/sdb1: Permission denied
I don't understand this behavior.
Thank you
bash command string output
bash command string output
asked Nov 30 at 15:42
intore
10619
10619
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
You are attempting to execute the contents of $pvs_var
as a command, rather than passing the string to awk.
To fix this, add an echo
or printf
in your if statement:
if [[ $(echo "$pvs_var" | awk '{ print $2 }') = vg_name ]]; then
do something
fi
Since the shell is bash, might as wellawk '{ print $2 }' <<<"$pvs_var"
– D. Ben Knoble
Nov 30 at 23:13
@D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
– Peschke
Nov 30 at 23:22
@Peschke With single brackets, you'd want to add double-quotes around the$( )
expression.
– Gordon Davisson
Dec 1 at 3:49
You definitely wantprintf
in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
– Kevin
Dec 1 at 17:08
add a comment |
Get the output in JSON format, and then you'll be able to extract information in a more reliable way:
pv_info=$(pvs -o pv_all,vg_all --unit b --nosuffix --reportformat json)
sdb1_vg=$(
printf '%sn' "$pv_info" |
jq -r '.report.pv|select(.pv_name == "/dev/sdb1").vg_name'
)
if [ "$sdb1_vg" = vg_name ]; then...
Or use a proper programming language with a JSON library instead of a shell (ksh93
does have JSON support though in its upcoming version).
If it's just that one query you want to do, pvs
can also do all the work for you:
sdb1_vg=$(
pvs -o vg_name -S pv_name=/dev/sdb1 --no-heading --config 'log{prefix=""}'
)
Also remember to quote your variables and avoid echo
.
add a comment |
If you want to compare the VG from the output, then it might be easier to pre-process that:
# if you still need it
pvs_var=$(pvs | grep "sdb1")
vg_name=$(pvs | grep "sdb1" | awk '{print $2}')
if [ "$vg_name" = "vg_name" ]; then
echo do something
fi
What you were doing with
$($pvs_var | awk '{ print $2 }')
was initiating a command substitution $( ... )
whose first command was $pvs_var
. Bash dutifully substituted the value of the variable and then attempted to execute it. That's not what you wanted.
Another alternative would be to send the variable as a here-string to the awk command:
# ...
if [ $(awk '{print $2}' <<< "$pvs_var") = "vg_name" ]; then
# ...
Here, the command substitution is calling awk and passing it input on stdin -- the contents of the $pvs_var
variable.
add a comment |
You can try only with awk :
pvs | awk -v search='vg_name' '/sdb1/&&$2==search{exit 1}' || echo "ok"
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are attempting to execute the contents of $pvs_var
as a command, rather than passing the string to awk.
To fix this, add an echo
or printf
in your if statement:
if [[ $(echo "$pvs_var" | awk '{ print $2 }') = vg_name ]]; then
do something
fi
Since the shell is bash, might as wellawk '{ print $2 }' <<<"$pvs_var"
– D. Ben Knoble
Nov 30 at 23:13
@D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
– Peschke
Nov 30 at 23:22
@Peschke With single brackets, you'd want to add double-quotes around the$( )
expression.
– Gordon Davisson
Dec 1 at 3:49
You definitely wantprintf
in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
– Kevin
Dec 1 at 17:08
add a comment |
You are attempting to execute the contents of $pvs_var
as a command, rather than passing the string to awk.
To fix this, add an echo
or printf
in your if statement:
if [[ $(echo "$pvs_var" | awk '{ print $2 }') = vg_name ]]; then
do something
fi
Since the shell is bash, might as wellawk '{ print $2 }' <<<"$pvs_var"
– D. Ben Knoble
Nov 30 at 23:13
@D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
– Peschke
Nov 30 at 23:22
@Peschke With single brackets, you'd want to add double-quotes around the$( )
expression.
– Gordon Davisson
Dec 1 at 3:49
You definitely wantprintf
in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
– Kevin
Dec 1 at 17:08
add a comment |
You are attempting to execute the contents of $pvs_var
as a command, rather than passing the string to awk.
To fix this, add an echo
or printf
in your if statement:
if [[ $(echo "$pvs_var" | awk '{ print $2 }') = vg_name ]]; then
do something
fi
You are attempting to execute the contents of $pvs_var
as a command, rather than passing the string to awk.
To fix this, add an echo
or printf
in your if statement:
if [[ $(echo "$pvs_var" | awk '{ print $2 }') = vg_name ]]; then
do something
fi
edited Dec 1 at 0:50
answered Nov 30 at 15:58
Peschke
2,455924
2,455924
Since the shell is bash, might as wellawk '{ print $2 }' <<<"$pvs_var"
– D. Ben Knoble
Nov 30 at 23:13
@D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
– Peschke
Nov 30 at 23:22
@Peschke With single brackets, you'd want to add double-quotes around the$( )
expression.
– Gordon Davisson
Dec 1 at 3:49
You definitely wantprintf
in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
– Kevin
Dec 1 at 17:08
add a comment |
Since the shell is bash, might as wellawk '{ print $2 }' <<<"$pvs_var"
– D. Ben Knoble
Nov 30 at 23:13
@D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
– Peschke
Nov 30 at 23:22
@Peschke With single brackets, you'd want to add double-quotes around the$( )
expression.
– Gordon Davisson
Dec 1 at 3:49
You definitely wantprintf
in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.
– Kevin
Dec 1 at 17:08
Since the shell is bash, might as well
awk '{ print $2 }' <<<"$pvs_var"
– D. Ben Knoble
Nov 30 at 23:13
Since the shell is bash, might as well
awk '{ print $2 }' <<<"$pvs_var"
– D. Ben Knoble
Nov 30 at 23:13
@D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
– Peschke
Nov 30 at 23:22
@D.BenKnoble: Even better, we can replace the double brackets with single brackets and then it’s POSIX compliant.
– Peschke
Nov 30 at 23:22
@Peschke With single brackets, you'd want to add double-quotes around the
$( )
expression.– Gordon Davisson
Dec 1 at 3:49
@Peschke With single brackets, you'd want to add double-quotes around the
$( )
expression.– Gordon Davisson
Dec 1 at 3:49
You definitely want
printf
in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.– Kevin
Dec 1 at 17:08
You definitely want
printf
in the general case. Echo can do strange things if the variable contains backslashes or starts with a dash.– Kevin
Dec 1 at 17:08
add a comment |
Get the output in JSON format, and then you'll be able to extract information in a more reliable way:
pv_info=$(pvs -o pv_all,vg_all --unit b --nosuffix --reportformat json)
sdb1_vg=$(
printf '%sn' "$pv_info" |
jq -r '.report.pv|select(.pv_name == "/dev/sdb1").vg_name'
)
if [ "$sdb1_vg" = vg_name ]; then...
Or use a proper programming language with a JSON library instead of a shell (ksh93
does have JSON support though in its upcoming version).
If it's just that one query you want to do, pvs
can also do all the work for you:
sdb1_vg=$(
pvs -o vg_name -S pv_name=/dev/sdb1 --no-heading --config 'log{prefix=""}'
)
Also remember to quote your variables and avoid echo
.
add a comment |
Get the output in JSON format, and then you'll be able to extract information in a more reliable way:
pv_info=$(pvs -o pv_all,vg_all --unit b --nosuffix --reportformat json)
sdb1_vg=$(
printf '%sn' "$pv_info" |
jq -r '.report.pv|select(.pv_name == "/dev/sdb1").vg_name'
)
if [ "$sdb1_vg" = vg_name ]; then...
Or use a proper programming language with a JSON library instead of a shell (ksh93
does have JSON support though in its upcoming version).
If it's just that one query you want to do, pvs
can also do all the work for you:
sdb1_vg=$(
pvs -o vg_name -S pv_name=/dev/sdb1 --no-heading --config 'log{prefix=""}'
)
Also remember to quote your variables and avoid echo
.
add a comment |
Get the output in JSON format, and then you'll be able to extract information in a more reliable way:
pv_info=$(pvs -o pv_all,vg_all --unit b --nosuffix --reportformat json)
sdb1_vg=$(
printf '%sn' "$pv_info" |
jq -r '.report.pv|select(.pv_name == "/dev/sdb1").vg_name'
)
if [ "$sdb1_vg" = vg_name ]; then...
Or use a proper programming language with a JSON library instead of a shell (ksh93
does have JSON support though in its upcoming version).
If it's just that one query you want to do, pvs
can also do all the work for you:
sdb1_vg=$(
pvs -o vg_name -S pv_name=/dev/sdb1 --no-heading --config 'log{prefix=""}'
)
Also remember to quote your variables and avoid echo
.
Get the output in JSON format, and then you'll be able to extract information in a more reliable way:
pv_info=$(pvs -o pv_all,vg_all --unit b --nosuffix --reportformat json)
sdb1_vg=$(
printf '%sn' "$pv_info" |
jq -r '.report.pv|select(.pv_name == "/dev/sdb1").vg_name'
)
if [ "$sdb1_vg" = vg_name ]; then...
Or use a proper programming language with a JSON library instead of a shell (ksh93
does have JSON support though in its upcoming version).
If it's just that one query you want to do, pvs
can also do all the work for you:
sdb1_vg=$(
pvs -o vg_name -S pv_name=/dev/sdb1 --no-heading --config 'log{prefix=""}'
)
Also remember to quote your variables and avoid echo
.
answered Nov 30 at 17:40
Stéphane Chazelas
298k54563910
298k54563910
add a comment |
add a comment |
If you want to compare the VG from the output, then it might be easier to pre-process that:
# if you still need it
pvs_var=$(pvs | grep "sdb1")
vg_name=$(pvs | grep "sdb1" | awk '{print $2}')
if [ "$vg_name" = "vg_name" ]; then
echo do something
fi
What you were doing with
$($pvs_var | awk '{ print $2 }')
was initiating a command substitution $( ... )
whose first command was $pvs_var
. Bash dutifully substituted the value of the variable and then attempted to execute it. That's not what you wanted.
Another alternative would be to send the variable as a here-string to the awk command:
# ...
if [ $(awk '{print $2}' <<< "$pvs_var") = "vg_name" ]; then
# ...
Here, the command substitution is calling awk and passing it input on stdin -- the contents of the $pvs_var
variable.
add a comment |
If you want to compare the VG from the output, then it might be easier to pre-process that:
# if you still need it
pvs_var=$(pvs | grep "sdb1")
vg_name=$(pvs | grep "sdb1" | awk '{print $2}')
if [ "$vg_name" = "vg_name" ]; then
echo do something
fi
What you were doing with
$($pvs_var | awk '{ print $2 }')
was initiating a command substitution $( ... )
whose first command was $pvs_var
. Bash dutifully substituted the value of the variable and then attempted to execute it. That's not what you wanted.
Another alternative would be to send the variable as a here-string to the awk command:
# ...
if [ $(awk '{print $2}' <<< "$pvs_var") = "vg_name" ]; then
# ...
Here, the command substitution is calling awk and passing it input on stdin -- the contents of the $pvs_var
variable.
add a comment |
If you want to compare the VG from the output, then it might be easier to pre-process that:
# if you still need it
pvs_var=$(pvs | grep "sdb1")
vg_name=$(pvs | grep "sdb1" | awk '{print $2}')
if [ "$vg_name" = "vg_name" ]; then
echo do something
fi
What you were doing with
$($pvs_var | awk '{ print $2 }')
was initiating a command substitution $( ... )
whose first command was $pvs_var
. Bash dutifully substituted the value of the variable and then attempted to execute it. That's not what you wanted.
Another alternative would be to send the variable as a here-string to the awk command:
# ...
if [ $(awk '{print $2}' <<< "$pvs_var") = "vg_name" ]; then
# ...
Here, the command substitution is calling awk and passing it input on stdin -- the contents of the $pvs_var
variable.
If you want to compare the VG from the output, then it might be easier to pre-process that:
# if you still need it
pvs_var=$(pvs | grep "sdb1")
vg_name=$(pvs | grep "sdb1" | awk '{print $2}')
if [ "$vg_name" = "vg_name" ]; then
echo do something
fi
What you were doing with
$($pvs_var | awk '{ print $2 }')
was initiating a command substitution $( ... )
whose first command was $pvs_var
. Bash dutifully substituted the value of the variable and then attempted to execute it. That's not what you wanted.
Another alternative would be to send the variable as a here-string to the awk command:
# ...
if [ $(awk '{print $2}' <<< "$pvs_var") = "vg_name" ]; then
# ...
Here, the command substitution is calling awk and passing it input on stdin -- the contents of the $pvs_var
variable.
answered Nov 30 at 15:55
Jeff Schaller
38.4k1053125
38.4k1053125
add a comment |
add a comment |
You can try only with awk :
pvs | awk -v search='vg_name' '/sdb1/&&$2==search{exit 1}' || echo "ok"
add a comment |
You can try only with awk :
pvs | awk -v search='vg_name' '/sdb1/&&$2==search{exit 1}' || echo "ok"
add a comment |
You can try only with awk :
pvs | awk -v search='vg_name' '/sdb1/&&$2==search{exit 1}' || echo "ok"
You can try only with awk :
pvs | awk -v search='vg_name' '/sdb1/&&$2==search{exit 1}' || echo "ok"
answered Nov 30 at 16:23
ctac_
1,344117
1,344117
add a comment |
add a comment |
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