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How to convert CSV file of strings seperated by a new line to array
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}
I have a csv file with data like this:
UserName
UserName
UserName
Each piece of data is seperated by a new line.
I need that data to be converted to an array of strings and stored in a variable.
How do I achieve this with just bash or shell execution?
bash shell
asked 2 hours ago
DavidDavid
53 bronze badges
add a comment |
I have a csv file with data like this:
UserName
UserName
UserName
Each piece of data is seperated by a new line.
I need that data to be converted to an array of strings and stored in a variable.
How do I achieve this with just bash or shell execution?
bash shell
asked 2 hours ago
DavidDavid
53 bronze badges
Can there be multiple strings per line (separated by space or even tab or comma)? If yes, should they be stored as one or as multiple values?
– Freddy
2 hours ago
only one string per line, theyre each stored as a single value. each string is only seperated by a new line
– David
2 hours ago
add a comment |
I have a csv file with data like this:
UserName
UserName
UserName
Each piece of data is seperated by a new line.
I need that data to be converted to an array of strings and stored in a variable.
How do I achieve this with just bash or shell execution?
bash shell
asked 2 hours ago
DavidDavid
53 bronze badges
I have a csv file with data like this:
UserName
UserName
UserName
Each piece of data is seperated by a new line.
I need that data to be converted to an array of strings and stored in a variable.
How do I achieve this with just bash or shell execution?
bash shell
bash shell
asked 2 hours ago
DavidDavid
53 bronze badges
asked 2 hours ago
DavidDavid
53 bronze badges
asked 2 hours ago
DavidDavid
53 bronze badges
asked 2 hours ago
DavidDavid
53 bronze badges
asked 2 hours ago
DavidDavid
53 bronze badges
53 bronze badges
Can there be multiple strings per line (separated by space or even tab or comma)? If yes, should they be stored as one or as multiple values?
– Freddy
2 hours ago
only one string per line, theyre each stored as a single value. each string is only seperated by a new line
– David
2 hours ago
add a comment |
Can there be multiple strings per line (separated by space or even tab or comma)? If yes, should they be stored as one or as multiple values?
– Freddy
2 hours ago
only one string per line, theyre each stored as a single value. each string is only seperated by a new line
– David
2 hours ago
Can there be multiple strings per line (separated by space or even tab or comma)? If yes, should they be stored as one or as multiple values?
– Freddy
2 hours ago
Can there be multiple strings per line (separated by space or even tab or comma)? If yes, should they be stored as one or as multiple values?
– Freddy
2 hours ago
only one string per line, theyre each stored as a single value. each string is only seperated by a new line
– David
2 hours ago
only one string per line, theyre each stored as a single value. each string is only seperated by a new line
– David
2 hours ago
add a comment |
3 Answers
3
active
oldest
votes
With mapfile:
$ mapfile -t array < <(cat yourfile)
$ declare -p array # print array content
declare -a array=([0]="UserName" [1]="UserName" [2]="UserName")
answered 2 hours ago
guillermo chamorroguillermo chamorro
50112 bronze badges
mapfile -t array < yourfilewill do here. mapfile reads stdin, process substitution isn't needed to read an existing file (unless you need to transform it in some way first, e.g. with grep or sed or awk). this is a UUOC with a pretty bell and collar :-) butmapfileis a good answer, so +1.
– cas
1 hour ago
add a comment |
$ arr=($(<file.csv))
$ declare -p arr # print array content
declare -a arr=([0]="UserName" [1]="UserName" [2]="UserName")
arr=(…)create array
$(<file.csv)read filefile.csv(like$(cat file.csv))
The elements are split at newline, space or tab characters and trimmed.
answered 2 hours ago
FreddyFreddy
6,8481 gold badge6 silver badges24 bronze badges
add a comment |
i=0
while read line
do
var[$i]="$line"
i=$((i+1))
done < filename
Here var array variable stores all names. And can be accessed by
echo ${var[0]}
echo ${var[1]}
...
...
answered 2 hours ago
NaviNavi
1
New contributor
Navi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
With mapfile:
$ mapfile -t array < <(cat yourfile)
$ declare -p array # print array content
declare -a array=([0]="UserName" [1]="UserName" [2]="UserName")
answered 2 hours ago
guillermo chamorroguillermo chamorro
50112 bronze badges
mapfile -t array < yourfilewill do here. mapfile reads stdin, process substitution isn't needed to read an existing file (unless you need to transform it in some way first, e.g. with grep or sed or awk). this is a UUOC with a pretty bell and collar :-) butmapfileis a good answer, so +1.
– cas
1 hour ago
add a comment |
With mapfile:
$ mapfile -t array < <(cat yourfile)
$ declare -p array # print array content
declare -a array=([0]="UserName" [1]="UserName" [2]="UserName")
answered 2 hours ago
guillermo chamorroguillermo chamorro
50112 bronze badges
mapfile -t array < yourfilewill do here. mapfile reads stdin, process substitution isn't needed to read an existing file (unless you need to transform it in some way first, e.g. with grep or sed or awk). this is a UUOC with a pretty bell and collar :-) butmapfileis a good answer, so +1.
– cas
1 hour ago
add a comment |
With mapfile:
$ mapfile -t array < <(cat yourfile)
$ declare -p array # print array content
declare -a array=([0]="UserName" [1]="UserName" [2]="UserName")
answered 2 hours ago
guillermo chamorroguillermo chamorro
50112 bronze badges
With mapfile:
$ mapfile -t array < <(cat yourfile)
$ declare -p array # print array content
declare -a array=([0]="UserName" [1]="UserName" [2]="UserName")
answered 2 hours ago
guillermo chamorroguillermo chamorro
50112 bronze badges
edited 1 hour ago
answered 2 hours ago
guillermo chamorroguillermo chamorro
50112 bronze badges
answered 2 hours ago
guillermo chamorroguillermo chamorro
50112 bronze badges
answered 2 hours ago
guillermo chamorroguillermo chamorro
50112 bronze badges
50112 bronze badges
mapfile -t array < yourfilewill do here. mapfile reads stdin, process substitution isn't needed to read an existing file (unless you need to transform it in some way first, e.g. with grep or sed or awk). this is a UUOC with a pretty bell and collar :-) butmapfileis a good answer, so +1.
– cas
1 hour ago
add a comment |
mapfile -t array < yourfilewill do here. mapfile reads stdin, process substitution isn't needed to read an existing file (unless you need to transform it in some way first, e.g. with grep or sed or awk). this is a UUOC with a pretty bell and collar :-) butmapfileis a good answer, so +1.
– cas
1 hour ago
mapfile -t array < yourfile will do here. mapfile reads stdin, process substitution isn't needed to read an existing file (unless you need to transform it in some way first, e.g. with grep or sed or awk). this is a UUOC with a pretty bell and collar :-) but mapfile is a good answer, so +1.– cas
1 hour ago
mapfile -t array < yourfile will do here. mapfile reads stdin, process substitution isn't needed to read an existing file (unless you need to transform it in some way first, e.g. with grep or sed or awk). this is a UUOC with a pretty bell and collar :-) but mapfile is a good answer, so +1.– cas
1 hour ago
add a comment |
$ arr=($(<file.csv))
$ declare -p arr # print array content
declare -a arr=([0]="UserName" [1]="UserName" [2]="UserName")
arr=(…)create array
$(<file.csv)read filefile.csv(like$(cat file.csv))
The elements are split at newline, space or tab characters and trimmed.
answered 2 hours ago
FreddyFreddy
6,8481 gold badge6 silver badges24 bronze badges
add a comment |
$ arr=($(<file.csv))
$ declare -p arr # print array content
declare -a arr=([0]="UserName" [1]="UserName" [2]="UserName")
arr=(…)create array
$(<file.csv)read filefile.csv(like$(cat file.csv))
The elements are split at newline, space or tab characters and trimmed.
answered 2 hours ago
FreddyFreddy
6,8481 gold badge6 silver badges24 bronze badges
add a comment |
$ arr=($(<file.csv))
$ declare -p arr # print array content
declare -a arr=([0]="UserName" [1]="UserName" [2]="UserName")
arr=(…)create array
$(<file.csv)read filefile.csv(like$(cat file.csv))
The elements are split at newline, space or tab characters and trimmed.
answered 2 hours ago
FreddyFreddy
6,8481 gold badge6 silver badges24 bronze badges
$ arr=($(<file.csv))
$ declare -p arr # print array content
declare -a arr=([0]="UserName" [1]="UserName" [2]="UserName")
arr=(…)create array
$(<file.csv)read filefile.csv(like$(cat file.csv))
The elements are split at newline, space or tab characters and trimmed.
answered 2 hours ago
FreddyFreddy
6,8481 gold badge6 silver badges24 bronze badges
edited 2 hours ago
answered 2 hours ago
FreddyFreddy
6,8481 gold badge6 silver badges24 bronze badges
answered 2 hours ago
FreddyFreddy
6,8481 gold badge6 silver badges24 bronze badges
answered 2 hours ago
FreddyFreddy
6,8481 gold badge6 silver badges24 bronze badges
6,8481 gold badge6 silver badges24 bronze badges
add a comment |
add a comment |
i=0
while read line
do
var[$i]="$line"
i=$((i+1))
done < filename
Here var array variable stores all names. And can be accessed by
echo ${var[0]}
echo ${var[1]}
...
...
answered 2 hours ago
NaviNavi
1
New contributor
Navi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
i=0
while read line
do
var[$i]="$line"
i=$((i+1))
done < filename
Here var array variable stores all names. And can be accessed by
echo ${var[0]}
echo ${var[1]}
...
...
answered 2 hours ago
NaviNavi
1
New contributor
Navi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
i=0
while read line
do
var[$i]="$line"
i=$((i+1))
done < filename
Here var array variable stores all names. And can be accessed by
echo ${var[0]}
echo ${var[1]}
...
...
answered 2 hours ago
NaviNavi
1
New contributor
Navi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
i=0
while read line
do
var[$i]="$line"
i=$((i+1))
done < filename
Here var array variable stores all names. And can be accessed by
echo ${var[0]}
echo ${var[1]}
...
...
answered 2 hours ago
NaviNavi
1
New contributor
Navi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
NaviNavi
1
New contributor
Navi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 2 hours ago
NaviNavi
1
answered 2 hours ago
NaviNavi
1
1
New contributor
Navi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Navi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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Can there be multiple strings per line (separated by space or even tab or comma)? If yes, should they be stored as one or as multiple values?
– Freddy
2 hours ago
only one string per line, theyre each stored as a single value. each string is only seperated by a new line
– David
2 hours ago