can bash show one array item id and value using `declare -p`?
$ str="a'"b"
$ declare -p str
declare -- str="a'"b" # see " was escaped, possibly other chars will too
$ astr=("$str" "c")
$ declare -p astr
declare -ax astr='([0]="a'''"b" [1]="c")'
so, is there some way to do something like declare -p astr[0] and retrieve something like this: declare -- astr[0]="a'"b" ?
I could use sed or something else, but I would like to know if bash allow accessing astr[n] thru declare -p in some way I couldn't guess yet?
bash array
asked yesterday
Aquarius PowerAquarius Power
1,67232036
add a comment |
$ str="a'"b"
$ declare -p str
declare -- str="a'"b" # see " was escaped, possibly other chars will too
$ astr=("$str" "c")
$ declare -p astr
declare -ax astr='([0]="a'''"b" [1]="c")'
so, is there some way to do something like declare -p astr[0] and retrieve something like this: declare -- astr[0]="a'"b" ?
I could use sed or something else, but I would like to know if bash allow accessing astr[n] thru declare -p in some way I couldn't guess yet?
bash array
asked yesterday
Aquarius PowerAquarius Power
1,67232036
add a comment |
$ str="a'"b"
$ declare -p str
declare -- str="a'"b" # see " was escaped, possibly other chars will too
$ astr=("$str" "c")
$ declare -p astr
declare -ax astr='([0]="a'''"b" [1]="c")'
so, is there some way to do something like declare -p astr[0] and retrieve something like this: declare -- astr[0]="a'"b" ?
I could use sed or something else, but I would like to know if bash allow accessing astr[n] thru declare -p in some way I couldn't guess yet?
bash array
asked yesterday
Aquarius PowerAquarius Power
1,67232036
$ str="a'"b"
$ declare -p str
declare -- str="a'"b" # see " was escaped, possibly other chars will too
$ astr=("$str" "c")
$ declare -p astr
declare -ax astr='([0]="a'''"b" [1]="c")'
so, is there some way to do something like declare -p astr[0] and retrieve something like this: declare -- astr[0]="a'"b" ?
I could use sed or something else, but I would like to know if bash allow accessing astr[n] thru declare -p in some way I couldn't guess yet?
bash array
bash array
asked yesterday
Aquarius PowerAquarius Power
1,67232036
asked yesterday
Aquarius PowerAquarius Power
1,67232036
asked yesterday
Aquarius PowerAquarius Power
1,67232036
asked yesterday
Aquarius PowerAquarius Power
1,67232036
asked yesterday
Aquarius PowerAquarius Power
1,67232036
1,67232036
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
If you are just looking for a way of displaying the data with escaped special characters, then the %q format string of printf in bash would do that for you:
printf '%qn' "${astr[0]}"
To replicate the declare -p-like output that you suggest:
printf 'declare -- astr[0]="%q"n' "${astr[0]}"
This is from the bash manual, regarding the %q format string of printf:
%q
causes
printfto output the corresponding argument in a format that can be reused
as shell input.
answered yesterday
KusalanandaKusalananda
124k16233384
it is a lot simpler than using sed and does the same thing I was looking for, thx!
– Aquarius Power
yesterday
add a comment |
With bash v4.4 you can use the @A parameter expansion operator to get similar results, but not effectively for a single array element:
A
The expansion is a string in the form of an assignment statement or declare command that, if evaluated, will recreate parameter with its attributes and value.
$ str="a'"b"
$ astr=("$str" "c")
$ echo ${astr[0]@A}
declare -a astr='a'''"b'
$ echo ${astr[@]@A}
declare -a astr=([0]="a'"b" [1]="c")
Or similar to the %q printf format you can use the @Q operator:
$ echo ${astr[0]@Q}
'a'''"b'
$ echo "declare -- astr[0]=${astr[0]@Q}"
declare -- astr[0]='a'''"b'
answered yesterday
Jesse_bJesse_b
12k23064
1
a new feature, interesting! mine is v4.3.48, I am almost installing 18.04 (16.04 here), I will check then :)
– Aquarius Power
yesterday
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
oldest
votes
If you are just looking for a way of displaying the data with escaped special characters, then the %q format string of printf in bash would do that for you:
printf '%qn' "${astr[0]}"
To replicate the declare -p-like output that you suggest:
printf 'declare -- astr[0]="%q"n' "${astr[0]}"
This is from the bash manual, regarding the %q format string of printf:
%q
causes
printfto output the corresponding argument in a format that can be reused
as shell input.
answered yesterday
KusalanandaKusalananda
124k16233384
it is a lot simpler than using sed and does the same thing I was looking for, thx!
– Aquarius Power
yesterday
add a comment |
If you are just looking for a way of displaying the data with escaped special characters, then the %q format string of printf in bash would do that for you:
printf '%qn' "${astr[0]}"
To replicate the declare -p-like output that you suggest:
printf 'declare -- astr[0]="%q"n' "${astr[0]}"
This is from the bash manual, regarding the %q format string of printf:
%q
causes
printfto output the corresponding argument in a format that can be reused
as shell input.
answered yesterday
KusalanandaKusalananda
124k16233384
it is a lot simpler than using sed and does the same thing I was looking for, thx!
– Aquarius Power
yesterday
add a comment |
If you are just looking for a way of displaying the data with escaped special characters, then the %q format string of printf in bash would do that for you:
printf '%qn' "${astr[0]}"
To replicate the declare -p-like output that you suggest:
printf 'declare -- astr[0]="%q"n' "${astr[0]}"
This is from the bash manual, regarding the %q format string of printf:
%q
causes
printfto output the corresponding argument in a format that can be reused
as shell input.
answered yesterday
KusalanandaKusalananda
124k16233384
If you are just looking for a way of displaying the data with escaped special characters, then the %q format string of printf in bash would do that for you:
printf '%qn' "${astr[0]}"
To replicate the declare -p-like output that you suggest:
printf 'declare -- astr[0]="%q"n' "${astr[0]}"
This is from the bash manual, regarding the %q format string of printf:
%q
causes
printfto output the corresponding argument in a format that can be reused
as shell input.
answered yesterday
KusalanandaKusalananda
124k16233384
edited yesterday
answered yesterday
KusalanandaKusalananda
124k16233384
answered yesterday
KusalanandaKusalananda
124k16233384
answered yesterday
KusalanandaKusalananda
124k16233384
124k16233384
it is a lot simpler than using sed and does the same thing I was looking for, thx!
– Aquarius Power
yesterday
add a comment |
it is a lot simpler than using sed and does the same thing I was looking for, thx!
– Aquarius Power
yesterday
it is a lot simpler than using sed and does the same thing I was looking for, thx!
– Aquarius Power
yesterday
it is a lot simpler than using sed and does the same thing I was looking for, thx!
– Aquarius Power
yesterday
add a comment |
With bash v4.4 you can use the @A parameter expansion operator to get similar results, but not effectively for a single array element:
A
The expansion is a string in the form of an assignment statement or declare command that, if evaluated, will recreate parameter with its attributes and value.
$ str="a'"b"
$ astr=("$str" "c")
$ echo ${astr[0]@A}
declare -a astr='a'''"b'
$ echo ${astr[@]@A}
declare -a astr=([0]="a'"b" [1]="c")
Or similar to the %q printf format you can use the @Q operator:
$ echo ${astr[0]@Q}
'a'''"b'
$ echo "declare -- astr[0]=${astr[0]@Q}"
declare -- astr[0]='a'''"b'
answered yesterday
Jesse_bJesse_b
12k23064
1
a new feature, interesting! mine is v4.3.48, I am almost installing 18.04 (16.04 here), I will check then :)
– Aquarius Power
yesterday
add a comment |
With bash v4.4 you can use the @A parameter expansion operator to get similar results, but not effectively for a single array element:
A
The expansion is a string in the form of an assignment statement or declare command that, if evaluated, will recreate parameter with its attributes and value.
$ str="a'"b"
$ astr=("$str" "c")
$ echo ${astr[0]@A}
declare -a astr='a'''"b'
$ echo ${astr[@]@A}
declare -a astr=([0]="a'"b" [1]="c")
Or similar to the %q printf format you can use the @Q operator:
$ echo ${astr[0]@Q}
'a'''"b'
$ echo "declare -- astr[0]=${astr[0]@Q}"
declare -- astr[0]='a'''"b'
answered yesterday
Jesse_bJesse_b
12k23064
1
a new feature, interesting! mine is v4.3.48, I am almost installing 18.04 (16.04 here), I will check then :)
– Aquarius Power
yesterday
add a comment |
With bash v4.4 you can use the @A parameter expansion operator to get similar results, but not effectively for a single array element:
A
The expansion is a string in the form of an assignment statement or declare command that, if evaluated, will recreate parameter with its attributes and value.
$ str="a'"b"
$ astr=("$str" "c")
$ echo ${astr[0]@A}
declare -a astr='a'''"b'
$ echo ${astr[@]@A}
declare -a astr=([0]="a'"b" [1]="c")
Or similar to the %q printf format you can use the @Q operator:
$ echo ${astr[0]@Q}
'a'''"b'
$ echo "declare -- astr[0]=${astr[0]@Q}"
declare -- astr[0]='a'''"b'
answered yesterday
Jesse_bJesse_b
12k23064
With bash v4.4 you can use the @A parameter expansion operator to get similar results, but not effectively for a single array element:
A
The expansion is a string in the form of an assignment statement or declare command that, if evaluated, will recreate parameter with its attributes and value.
$ str="a'"b"
$ astr=("$str" "c")
$ echo ${astr[0]@A}
declare -a astr='a'''"b'
$ echo ${astr[@]@A}
declare -a astr=([0]="a'"b" [1]="c")
Or similar to the %q printf format you can use the @Q operator:
$ echo ${astr[0]@Q}
'a'''"b'
$ echo "declare -- astr[0]=${astr[0]@Q}"
declare -- astr[0]='a'''"b'
answered yesterday
Jesse_bJesse_b
12k23064
edited yesterday
answered yesterday
Jesse_bJesse_b
12k23064
answered yesterday
Jesse_bJesse_b
12k23064
answered yesterday
Jesse_bJesse_b
12k23064
12k23064
1
a new feature, interesting! mine is v4.3.48, I am almost installing 18.04 (16.04 here), I will check then :)
– Aquarius Power
yesterday
add a comment |
1
a new feature, interesting! mine is v4.3.48, I am almost installing 18.04 (16.04 here), I will check then :)
– Aquarius Power
yesterday
1
1
a new feature, interesting! mine is v4.3.48, I am almost installing 18.04 (16.04 here), I will check then :)
– Aquarius Power
yesterday
a new feature, interesting! mine is v4.3.48, I am almost installing 18.04 (16.04 here), I will check then :)
– Aquarius Power
yesterday
add a comment |
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