What does this command with two filenames in it mean?
./getProductAccess.sh access-2014-09-04.log
Normally there is only 1 file behind ./. Now there are two. It seems the second one is to provide input, but what should write in the first one to get the input? Because read doesn't work.
linux shell-script shell
asked 13 mins ago
VictorVictor
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./getProductAccess.sh access-2014-09-04.log
Normally there is only 1 file behind ./. Now there are two. It seems the second one is to provide input, but what should write in the first one to get the input? Because read doesn't work.
linux shell-script shell
asked 13 mins ago
VictorVictor
1
New contributor
Victor 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 |
./getProductAccess.sh access-2014-09-04.log
Normally there is only 1 file behind ./. Now there are two. It seems the second one is to provide input, but what should write in the first one to get the input? Because read doesn't work.
linux shell-script shell
asked 13 mins ago
VictorVictor
1
New contributor
Victor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
./getProductAccess.sh access-2014-09-04.log
Normally there is only 1 file behind ./. Now there are two. It seems the second one is to provide input, but what should write in the first one to get the input? Because read doesn't work.
linux shell-script shell
linux shell-script shell
asked 13 mins ago
VictorVictor
1
New contributor
Victor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 13 mins ago
VictorVictor
1
New contributor
Victor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 13 mins ago
VictorVictor
1
New contributor
Victor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 13 mins ago
VictorVictor
1
asked 13 mins ago
VictorVictor
1
1
New contributor
Victor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Victor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Victor is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
2 Answers
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votes
This really is beginners 101.
./getProductAccess.sh is the program to run (in the current directory) and access-2014-09-04.log is an argument supplied to that program.
If you're writing the program you could reference that first argument as "$1".
answered 8 mins ago
roaimaroaima
44.2k555119
add a comment |
There is no need to use read you can use $1.
So in your script if you will use echo "$1", then it will give output as access-2014-09-04.log.
$0 is the path(relative or absolute) of the file which you are executing. $1 is the first argument passed to file, $2 second and so on...
answered 11 mins ago
P_YadavP_Yadav
1,86631024
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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votes
active
oldest
votes
This really is beginners 101.
./getProductAccess.sh is the program to run (in the current directory) and access-2014-09-04.log is an argument supplied to that program.
If you're writing the program you could reference that first argument as "$1".
answered 8 mins ago
roaimaroaima
44.2k555119
add a comment |
This really is beginners 101.
./getProductAccess.sh is the program to run (in the current directory) and access-2014-09-04.log is an argument supplied to that program.
If you're writing the program you could reference that first argument as "$1".
answered 8 mins ago
roaimaroaima
44.2k555119
add a comment |
This really is beginners 101.
./getProductAccess.sh is the program to run (in the current directory) and access-2014-09-04.log is an argument supplied to that program.
If you're writing the program you could reference that first argument as "$1".
answered 8 mins ago
roaimaroaima
44.2k555119
This really is beginners 101.
./getProductAccess.sh is the program to run (in the current directory) and access-2014-09-04.log is an argument supplied to that program.
If you're writing the program you could reference that first argument as "$1".
answered 8 mins ago
roaimaroaima
44.2k555119
answered 8 mins ago
roaimaroaima
44.2k555119
answered 8 mins ago
roaimaroaima
44.2k555119
answered 8 mins ago
roaimaroaima
44.2k555119
44.2k555119
add a comment |
add a comment |
There is no need to use read you can use $1.
So in your script if you will use echo "$1", then it will give output as access-2014-09-04.log.
$0 is the path(relative or absolute) of the file which you are executing. $1 is the first argument passed to file, $2 second and so on...
answered 11 mins ago
P_YadavP_Yadav
1,86631024
add a comment |
There is no need to use read you can use $1.
So in your script if you will use echo "$1", then it will give output as access-2014-09-04.log.
$0 is the path(relative or absolute) of the file which you are executing. $1 is the first argument passed to file, $2 second and so on...
answered 11 mins ago
P_YadavP_Yadav
1,86631024
add a comment |
There is no need to use read you can use $1.
So in your script if you will use echo "$1", then it will give output as access-2014-09-04.log.
$0 is the path(relative or absolute) of the file which you are executing. $1 is the first argument passed to file, $2 second and so on...
answered 11 mins ago
P_YadavP_Yadav
1,86631024
There is no need to use read you can use $1.
So in your script if you will use echo "$1", then it will give output as access-2014-09-04.log.
$0 is the path(relative or absolute) of the file which you are executing. $1 is the first argument passed to file, $2 second and so on...
answered 11 mins ago
P_YadavP_Yadav
1,86631024
edited 3 mins ago
answered 11 mins ago
P_YadavP_Yadav
1,86631024
answered 11 mins ago
P_YadavP_Yadav
1,86631024
answered 11 mins ago
P_YadavP_Yadav
1,86631024
1,86631024
add a comment |
add a comment |
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