How to copy an array to a new array with dynamic name?
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I have a complex data structure in Bash like this:
Test1_Name="test 1"
Test1_Files=(
file01.txt
file02.txt
)
Test2_Name="test 2"
Test2_Files=(
file11.txt
file12.txt
)
TestNames="Test1 Test2"
In my improved script, I would like read the files from disk. Each test resides in a directory.
So I have a Bash snippet reading directories and reading all the file names. The result is present in an array: $Files[*]
.
How can I copy that array to an array prefixed with the test's name. Let's assume $TestName
holds the test's name.
I know how to create a dynamically named array:
for $Name in $TestNames; do
declare -a "${TestName}_Files"
done
Will create e.g. Test1_Files
and Test2_Files
. The variables are of type array, because I used -a
in declare.
But how can I copy $Files[*]
to "${TestName}_Files"
in such a loop?
I tried this:
declare -a "${TestName}_Files"=${Files[*]}
But it gives an error that =file01.txt
is not a valid identifier.
arrays bash
add a comment |
I have a complex data structure in Bash like this:
Test1_Name="test 1"
Test1_Files=(
file01.txt
file02.txt
)
Test2_Name="test 2"
Test2_Files=(
file11.txt
file12.txt
)
TestNames="Test1 Test2"
In my improved script, I would like read the files from disk. Each test resides in a directory.
So I have a Bash snippet reading directories and reading all the file names. The result is present in an array: $Files[*]
.
How can I copy that array to an array prefixed with the test's name. Let's assume $TestName
holds the test's name.
I know how to create a dynamically named array:
for $Name in $TestNames; do
declare -a "${TestName}_Files"
done
Will create e.g. Test1_Files
and Test2_Files
. The variables are of type array, because I used -a
in declare.
But how can I copy $Files[*]
to "${TestName}_Files"
in such a loop?
I tried this:
declare -a "${TestName}_Files"=${Files[*]}
But it gives an error that =file01.txt
is not a valid identifier.
arrays bash
add a comment |
I have a complex data structure in Bash like this:
Test1_Name="test 1"
Test1_Files=(
file01.txt
file02.txt
)
Test2_Name="test 2"
Test2_Files=(
file11.txt
file12.txt
)
TestNames="Test1 Test2"
In my improved script, I would like read the files from disk. Each test resides in a directory.
So I have a Bash snippet reading directories and reading all the file names. The result is present in an array: $Files[*]
.
How can I copy that array to an array prefixed with the test's name. Let's assume $TestName
holds the test's name.
I know how to create a dynamically named array:
for $Name in $TestNames; do
declare -a "${TestName}_Files"
done
Will create e.g. Test1_Files
and Test2_Files
. The variables are of type array, because I used -a
in declare.
But how can I copy $Files[*]
to "${TestName}_Files"
in such a loop?
I tried this:
declare -a "${TestName}_Files"=${Files[*]}
But it gives an error that =file01.txt
is not a valid identifier.
arrays bash
I have a complex data structure in Bash like this:
Test1_Name="test 1"
Test1_Files=(
file01.txt
file02.txt
)
Test2_Name="test 2"
Test2_Files=(
file11.txt
file12.txt
)
TestNames="Test1 Test2"
In my improved script, I would like read the files from disk. Each test resides in a directory.
So I have a Bash snippet reading directories and reading all the file names. The result is present in an array: $Files[*]
.
How can I copy that array to an array prefixed with the test's name. Let's assume $TestName
holds the test's name.
I know how to create a dynamically named array:
for $Name in $TestNames; do
declare -a "${TestName}_Files"
done
Will create e.g. Test1_Files
and Test2_Files
. The variables are of type array, because I used -a
in declare.
But how can I copy $Files[*]
to "${TestName}_Files"
in such a loop?
I tried this:
declare -a "${TestName}_Files"=${Files[*]}
But it gives an error that =file01.txt
is not a valid identifier.
arrays bash
arrays bash
asked Dec 31 '18 at 12:05
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PaebbelsPaebbels
7,42983577
7,42983577
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You'll want to use newarray=( "${oldarray[@]}" )
to keep the array elements intact. ${oldarray[*]}
will involve word splitting, which will break at least with elements containing white space.
However, the obvious declare -a "$name"=("${oldarray[@]}")
doesn't work, with the parenthesis quoted or not. One way to do it seems to be to use a name ref. This would copy the contents of old
to new
, where the name of new
can be given dynamically:
#!/bin/bash
old=("abc" "white space")
name=new
declare -a "$name" # declare the new array, make 'ref' point to it
declare -n ref="$name"
ref=( "${old[@]}" ) # copy
#unset -n ref # unset the nameref, if required
declare -p "$name" # verify results
What is the purpose ofdeclare -p "$name"
?
– Paebbels
Dec 31 '18 at 12:29
@Paebbels, just to print out the new copy to verify the result is correct
– ilkkachu
Dec 31 '18 at 13:35
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
You'll want to use newarray=( "${oldarray[@]}" )
to keep the array elements intact. ${oldarray[*]}
will involve word splitting, which will break at least with elements containing white space.
However, the obvious declare -a "$name"=("${oldarray[@]}")
doesn't work, with the parenthesis quoted or not. One way to do it seems to be to use a name ref. This would copy the contents of old
to new
, where the name of new
can be given dynamically:
#!/bin/bash
old=("abc" "white space")
name=new
declare -a "$name" # declare the new array, make 'ref' point to it
declare -n ref="$name"
ref=( "${old[@]}" ) # copy
#unset -n ref # unset the nameref, if required
declare -p "$name" # verify results
What is the purpose ofdeclare -p "$name"
?
– Paebbels
Dec 31 '18 at 12:29
@Paebbels, just to print out the new copy to verify the result is correct
– ilkkachu
Dec 31 '18 at 13:35
add a comment |
You'll want to use newarray=( "${oldarray[@]}" )
to keep the array elements intact. ${oldarray[*]}
will involve word splitting, which will break at least with elements containing white space.
However, the obvious declare -a "$name"=("${oldarray[@]}")
doesn't work, with the parenthesis quoted or not. One way to do it seems to be to use a name ref. This would copy the contents of old
to new
, where the name of new
can be given dynamically:
#!/bin/bash
old=("abc" "white space")
name=new
declare -a "$name" # declare the new array, make 'ref' point to it
declare -n ref="$name"
ref=( "${old[@]}" ) # copy
#unset -n ref # unset the nameref, if required
declare -p "$name" # verify results
What is the purpose ofdeclare -p "$name"
?
– Paebbels
Dec 31 '18 at 12:29
@Paebbels, just to print out the new copy to verify the result is correct
– ilkkachu
Dec 31 '18 at 13:35
add a comment |
You'll want to use newarray=( "${oldarray[@]}" )
to keep the array elements intact. ${oldarray[*]}
will involve word splitting, which will break at least with elements containing white space.
However, the obvious declare -a "$name"=("${oldarray[@]}")
doesn't work, with the parenthesis quoted or not. One way to do it seems to be to use a name ref. This would copy the contents of old
to new
, where the name of new
can be given dynamically:
#!/bin/bash
old=("abc" "white space")
name=new
declare -a "$name" # declare the new array, make 'ref' point to it
declare -n ref="$name"
ref=( "${old[@]}" ) # copy
#unset -n ref # unset the nameref, if required
declare -p "$name" # verify results
You'll want to use newarray=( "${oldarray[@]}" )
to keep the array elements intact. ${oldarray[*]}
will involve word splitting, which will break at least with elements containing white space.
However, the obvious declare -a "$name"=("${oldarray[@]}")
doesn't work, with the parenthesis quoted or not. One way to do it seems to be to use a name ref. This would copy the contents of old
to new
, where the name of new
can be given dynamically:
#!/bin/bash
old=("abc" "white space")
name=new
declare -a "$name" # declare the new array, make 'ref' point to it
declare -n ref="$name"
ref=( "${old[@]}" ) # copy
#unset -n ref # unset the nameref, if required
declare -p "$name" # verify results
edited Dec 31 '18 at 13:34
answered Dec 31 '18 at 12:16
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ilkkachuilkkachu
3,509318
3,509318
What is the purpose ofdeclare -p "$name"
?
– Paebbels
Dec 31 '18 at 12:29
@Paebbels, just to print out the new copy to verify the result is correct
– ilkkachu
Dec 31 '18 at 13:35
add a comment |
What is the purpose ofdeclare -p "$name"
?
– Paebbels
Dec 31 '18 at 12:29
@Paebbels, just to print out the new copy to verify the result is correct
– ilkkachu
Dec 31 '18 at 13:35
What is the purpose of
declare -p "$name"
?– Paebbels
Dec 31 '18 at 12:29
What is the purpose of
declare -p "$name"
?– Paebbels
Dec 31 '18 at 12:29
@Paebbels, just to print out the new copy to verify the result is correct
– ilkkachu
Dec 31 '18 at 13:35
@Paebbels, just to print out the new copy to verify the result is correct
– ilkkachu
Dec 31 '18 at 13:35
add a comment |
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