Insert a user defined variable in an alias in Bash
Clash Royale CLAN TAG#URR8PPP
7
I would like to add an alias to a command in my .bashrc file, as in this example:
alias take_row="cat prova.csv_001.txt | awk -v nrow="$nrow" 'if (NR==nrow) print'"
But I would like to specify the nrow variable when I launch the command from the terminal. As an example:
take_row 1 --> the script takes the first row
take_row 20 --> the script takes the 20th row
How do I write inside the command something like <command> | <read nrow user input>, etc.?
bash variable alias
edited Jan 28 at 0:23
wjandrea
498413
asked Jan 27 at 16:49
ccc.nrcccc.nrc
414
add a comment |
7
I would like to add an alias to a command in my .bashrc file, as in this example:
alias take_row="cat prova.csv_001.txt | awk -v nrow="$nrow" 'if (NR==nrow) print'"
But I would like to specify the nrow variable when I launch the command from the terminal. As an example:
take_row 1 --> the script takes the first row
take_row 20 --> the script takes the 20th row
How do I write inside the command something like <command> | <read nrow user input>, etc.?
bash variable alias
edited Jan 28 at 0:23
wjandrea
498413
asked Jan 27 at 16:49
ccc.nrcccc.nrc
414
add a comment |
7
7
7
I would like to add an alias to a command in my .bashrc file, as in this example:
alias take_row="cat prova.csv_001.txt | awk -v nrow="$nrow" 'if (NR==nrow) print'"
But I would like to specify the nrow variable when I launch the command from the terminal. As an example:
take_row 1 --> the script takes the first row
take_row 20 --> the script takes the 20th row
How do I write inside the command something like <command> | <read nrow user input>, etc.?
bash variable alias
edited Jan 28 at 0:23
wjandrea
498413
asked Jan 27 at 16:49
ccc.nrcccc.nrc
414
I would like to add an alias to a command in my .bashrc file, as in this example:
alias take_row="cat prova.csv_001.txt | awk -v nrow="$nrow" 'if (NR==nrow) print'"
But I would like to specify the nrow variable when I launch the command from the terminal. As an example:
take_row 1 --> the script takes the first row
take_row 20 --> the script takes the 20th row
How do I write inside the command something like <command> | <read nrow user input>, etc.?
bash variable alias
bash variable alias
edited Jan 28 at 0:23
wjandrea
498413
asked Jan 27 at 16:49
ccc.nrcccc.nrc
414
edited Jan 28 at 0:23
wjandrea
498413
asked Jan 27 at 16:49
ccc.nrcccc.nrc
414
edited Jan 28 at 0:23
wjandrea
498413
edited Jan 28 at 0:23
wjandrea
498413
edited Jan 28 at 0:23
wjandrea
498413
498413
asked Jan 27 at 16:49
ccc.nrcccc.nrc
414
asked Jan 27 at 16:49
ccc.nrcccc.nrc
414
asked Jan 27 at 16:49
ccc.nrcccc.nrc
414
414
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
11
Use a function instead of an alias, in which case you can refer to positional parameters such as $1, etc.
take_row ()
local nrow=$1
cat prova.csv_001.txt
In this particular case, you can also address the useless use of cat:
take_row ()
local nrow=$1
awk -v nrow="$nrow" 'if (NR==nrow) print' prova.csv_001.txt
answered Jan 27 at 16:53
filbrandenfilbranden
8,27621139
1
Thanks a lot for the suggestions!
– ccc.nrc
Jan 27 at 17:04
Better yet, use an external shell script kept, for example, in ~/bin.
– jrw32982
Jan 30 at 19:28
1
@jrw32982 There are limitations to an external script, for example you can't change directories or modify the environment of the current shell, which an alias can do and so can a function... But if neither of these are needed, then yes an external script is a possibility.
– filbranden
Jan 30 at 21:41
1
My thesis is that unless there is a limitation which requires an alias or function due to some reason similar to those you suggest, then an external shell script should be preferred virtually always over an alias or a function defined by .bashrc. In the case at hand, there seems to be no reason which would prevent using an external script, and therefore, that is the preferred solution.
– jrw32982
Jan 31 at 1:17
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
11
Use a function instead of an alias, in which case you can refer to positional parameters such as $1, etc.
take_row ()
local nrow=$1
cat prova.csv_001.txt
In this particular case, you can also address the useless use of cat:
take_row ()
local nrow=$1
awk -v nrow="$nrow" 'if (NR==nrow) print' prova.csv_001.txt
answered Jan 27 at 16:53
filbrandenfilbranden
8,27621139
1
Thanks a lot for the suggestions!
– ccc.nrc
Jan 27 at 17:04
Better yet, use an external shell script kept, for example, in ~/bin.
– jrw32982
Jan 30 at 19:28
1
@jrw32982 There are limitations to an external script, for example you can't change directories or modify the environment of the current shell, which an alias can do and so can a function... But if neither of these are needed, then yes an external script is a possibility.
– filbranden
Jan 30 at 21:41
1
My thesis is that unless there is a limitation which requires an alias or function due to some reason similar to those you suggest, then an external shell script should be preferred virtually always over an alias or a function defined by .bashrc. In the case at hand, there seems to be no reason which would prevent using an external script, and therefore, that is the preferred solution.
– jrw32982
Jan 31 at 1:17
add a comment |
11
Use a function instead of an alias, in which case you can refer to positional parameters such as $1, etc.
take_row ()
local nrow=$1
cat prova.csv_001.txt
In this particular case, you can also address the useless use of cat:
take_row ()
local nrow=$1
awk -v nrow="$nrow" 'if (NR==nrow) print' prova.csv_001.txt
answered Jan 27 at 16:53
filbrandenfilbranden
8,27621139
1
Thanks a lot for the suggestions!
– ccc.nrc
Jan 27 at 17:04
Better yet, use an external shell script kept, for example, in ~/bin.
– jrw32982
Jan 30 at 19:28
1
@jrw32982 There are limitations to an external script, for example you can't change directories or modify the environment of the current shell, which an alias can do and so can a function... But if neither of these are needed, then yes an external script is a possibility.
– filbranden
Jan 30 at 21:41
1
My thesis is that unless there is a limitation which requires an alias or function due to some reason similar to those you suggest, then an external shell script should be preferred virtually always over an alias or a function defined by .bashrc. In the case at hand, there seems to be no reason which would prevent using an external script, and therefore, that is the preferred solution.
– jrw32982
Jan 31 at 1:17
add a comment |
11
11
11
Use a function instead of an alias, in which case you can refer to positional parameters such as $1, etc.
take_row ()
local nrow=$1
cat prova.csv_001.txt
In this particular case, you can also address the useless use of cat:
take_row ()
local nrow=$1
awk -v nrow="$nrow" 'if (NR==nrow) print' prova.csv_001.txt
answered Jan 27 at 16:53
filbrandenfilbranden
8,27621139
Use a function instead of an alias, in which case you can refer to positional parameters such as $1, etc.
take_row ()
local nrow=$1
cat prova.csv_001.txt
In this particular case, you can also address the useless use of cat:
take_row ()
local nrow=$1
awk -v nrow="$nrow" 'if (NR==nrow) print' prova.csv_001.txt
answered Jan 27 at 16:53
filbrandenfilbranden
8,27621139
answered Jan 27 at 16:53
filbrandenfilbranden
8,27621139
answered Jan 27 at 16:53
filbrandenfilbranden
8,27621139
answered Jan 27 at 16:53
filbrandenfilbranden
8,27621139
8,27621139
1
Thanks a lot for the suggestions!
– ccc.nrc
Jan 27 at 17:04
Better yet, use an external shell script kept, for example, in ~/bin.
– jrw32982
Jan 30 at 19:28
1
@jrw32982 There are limitations to an external script, for example you can't change directories or modify the environment of the current shell, which an alias can do and so can a function... But if neither of these are needed, then yes an external script is a possibility.
– filbranden
Jan 30 at 21:41
1
My thesis is that unless there is a limitation which requires an alias or function due to some reason similar to those you suggest, then an external shell script should be preferred virtually always over an alias or a function defined by .bashrc. In the case at hand, there seems to be no reason which would prevent using an external script, and therefore, that is the preferred solution.
– jrw32982
Jan 31 at 1:17
add a comment |
1
Thanks a lot for the suggestions!
– ccc.nrc
Jan 27 at 17:04
Better yet, use an external shell script kept, for example, in ~/bin.
– jrw32982
Jan 30 at 19:28
1
@jrw32982 There are limitations to an external script, for example you can't change directories or modify the environment of the current shell, which an alias can do and so can a function... But if neither of these are needed, then yes an external script is a possibility.
– filbranden
Jan 30 at 21:41
1
My thesis is that unless there is a limitation which requires an alias or function due to some reason similar to those you suggest, then an external shell script should be preferred virtually always over an alias or a function defined by .bashrc. In the case at hand, there seems to be no reason which would prevent using an external script, and therefore, that is the preferred solution.
– jrw32982
Jan 31 at 1:17
1
1
Thanks a lot for the suggestions!
– ccc.nrc
Jan 27 at 17:04
Thanks a lot for the suggestions!
– ccc.nrc
Jan 27 at 17:04
Better yet, use an external shell script kept, for example, in ~/bin.
– jrw32982
Jan 30 at 19:28
Better yet, use an external shell script kept, for example, in ~/bin.
– jrw32982
Jan 30 at 19:28
1
1
@jrw32982 There are limitations to an external script, for example you can't change directories or modify the environment of the current shell, which an alias can do and so can a function... But if neither of these are needed, then yes an external script is a possibility.
– filbranden
Jan 30 at 21:41
@jrw32982 There are limitations to an external script, for example you can't change directories or modify the environment of the current shell, which an alias can do and so can a function... But if neither of these are needed, then yes an external script is a possibility.
– filbranden
Jan 30 at 21:41
1
1
My thesis is that unless there is a limitation which requires an alias or function due to some reason similar to those you suggest, then an external shell script should be preferred virtually always over an alias or a function defined by .bashrc. In the case at hand, there seems to be no reason which would prevent using an external script, and therefore, that is the preferred solution.
– jrw32982
Jan 31 at 1:17
My thesis is that unless there is a limitation which requires an alias or function due to some reason similar to those you suggest, then an external shell script should be preferred virtually always over an alias or a function defined by .bashrc. In the case at hand, there seems to be no reason which would prevent using an external script, and therefore, that is the preferred solution.
– jrw32982
Jan 31 at 1:17
add a comment |
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