Evaluate another awk expression inside awk
In the dataset I am processing, the 5th ($5
) column of each line contains a date in '%d/%m/%y'
format. I want to compare this with a user provided one.
I know that in order to easily compare dates with relational binary operators they need to be in %y-%m-%d
so I need to inline-convert the 5th columns value to that in order for the expression to evaluate to true:
Here's what I've come up with till now:
awk -F "$columnsep" -v dateA="$bornsince" ' awk -F '/' 'print $3"-"$2"-"$1'") ) print' $file
As you can see, the ' awk -F '/' 'print $3"-"$2"-"$1'") ) print'
is the expression I am trying to evaluate but I have trouble getting the system call to parse. I really just want to call awk -F '/' 'print $3"-"$2"-"$1'"
, piping $5
into it in order to convert the date properly.
- Needs to work on macos and linux (ubuntu)
Minimal example dataset:
1099511629352 Nunez Jorge female 17/11/1986
bash awk
add a comment |
In the dataset I am processing, the 5th ($5
) column of each line contains a date in '%d/%m/%y'
format. I want to compare this with a user provided one.
I know that in order to easily compare dates with relational binary operators they need to be in %y-%m-%d
so I need to inline-convert the 5th columns value to that in order for the expression to evaluate to true:
Here's what I've come up with till now:
awk -F "$columnsep" -v dateA="$bornsince" ' awk -F '/' 'print $3"-"$2"-"$1'") ) print' $file
As you can see, the ' awk -F '/' 'print $3"-"$2"-"$1'") ) print'
is the expression I am trying to evaluate but I have trouble getting the system call to parse. I really just want to call awk -F '/' 'print $3"-"$2"-"$1'"
, piping $5
into it in order to convert the date properly.
- Needs to work on macos and linux (ubuntu)
Minimal example dataset:
1099511629352 Nunez Jorge female 17/11/1986
bash awk
2
You could provide a minimal example of your dataset
– miimote
Nov 10 '18 at 16:44
1
awk version 20121220 on a macbook pro late 2017 up-to-date
– Zarkopafilis
Nov 10 '18 at 16:44
add a comment |
In the dataset I am processing, the 5th ($5
) column of each line contains a date in '%d/%m/%y'
format. I want to compare this with a user provided one.
I know that in order to easily compare dates with relational binary operators they need to be in %y-%m-%d
so I need to inline-convert the 5th columns value to that in order for the expression to evaluate to true:
Here's what I've come up with till now:
awk -F "$columnsep" -v dateA="$bornsince" ' awk -F '/' 'print $3"-"$2"-"$1'") ) print' $file
As you can see, the ' awk -F '/' 'print $3"-"$2"-"$1'") ) print'
is the expression I am trying to evaluate but I have trouble getting the system call to parse. I really just want to call awk -F '/' 'print $3"-"$2"-"$1'"
, piping $5
into it in order to convert the date properly.
- Needs to work on macos and linux (ubuntu)
Minimal example dataset:
1099511629352 Nunez Jorge female 17/11/1986
bash awk
In the dataset I am processing, the 5th ($5
) column of each line contains a date in '%d/%m/%y'
format. I want to compare this with a user provided one.
I know that in order to easily compare dates with relational binary operators they need to be in %y-%m-%d
so I need to inline-convert the 5th columns value to that in order for the expression to evaluate to true:
Here's what I've come up with till now:
awk -F "$columnsep" -v dateA="$bornsince" ' awk -F '/' 'print $3"-"$2"-"$1'") ) print' $file
As you can see, the ' awk -F '/' 'print $3"-"$2"-"$1'") ) print'
is the expression I am trying to evaluate but I have trouble getting the system call to parse. I really just want to call awk -F '/' 'print $3"-"$2"-"$1'"
, piping $5
into it in order to convert the date properly.
- Needs to work on macos and linux (ubuntu)
Minimal example dataset:
1099511629352 Nunez Jorge female 17/11/1986
bash awk
bash awk
edited Nov 10 '18 at 16:45
Zarkopafilis
asked Nov 10 '18 at 16:18
ZarkopafilisZarkopafilis
3471516
3471516
2
You could provide a minimal example of your dataset
– miimote
Nov 10 '18 at 16:44
1
awk version 20121220 on a macbook pro late 2017 up-to-date
– Zarkopafilis
Nov 10 '18 at 16:44
add a comment |
2
You could provide a minimal example of your dataset
– miimote
Nov 10 '18 at 16:44
1
awk version 20121220 on a macbook pro late 2017 up-to-date
– Zarkopafilis
Nov 10 '18 at 16:44
2
2
You could provide a minimal example of your dataset
– miimote
Nov 10 '18 at 16:44
You could provide a minimal example of your dataset
– miimote
Nov 10 '18 at 16:44
1
1
awk version 20121220 on a macbook pro late 2017 up-to-date
– Zarkopafilis
Nov 10 '18 at 16:44
awk version 20121220 on a macbook pro late 2017 up-to-date
– Zarkopafilis
Nov 10 '18 at 16:44
add a comment |
1 Answer
1
active
oldest
votes
Here is a gawk solution
awk -F "$columnsep" -v dateA="$bornsince" '
(FNR > 1 && dateA <= cnv_date($5))
function cnv_date(date, a)
split(date, a, "/")
return a[3] "-" a[2] "-" a[1]
' $file
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here is a gawk solution
awk -F "$columnsep" -v dateA="$bornsince" '
(FNR > 1 && dateA <= cnv_date($5))
function cnv_date(date, a)
split(date, a, "/")
return a[3] "-" a[2] "-" a[1]
' $file
add a comment |
Here is a gawk solution
awk -F "$columnsep" -v dateA="$bornsince" '
(FNR > 1 && dateA <= cnv_date($5))
function cnv_date(date, a)
split(date, a, "/")
return a[3] "-" a[2] "-" a[1]
' $file
add a comment |
Here is a gawk solution
awk -F "$columnsep" -v dateA="$bornsince" '
(FNR > 1 && dateA <= cnv_date($5))
function cnv_date(date, a)
split(date, a, "/")
return a[3] "-" a[2] "-" a[1]
' $file
Here is a gawk solution
awk -F "$columnsep" -v dateA="$bornsince" '
(FNR > 1 && dateA <= cnv_date($5))
function cnv_date(date, a)
split(date, a, "/")
return a[3] "-" a[2] "-" a[1]
' $file
edited Nov 10 '18 at 17:01
answered Nov 10 '18 at 16:51
oguzismailoguzismail
3,33031025
3,33031025
add a comment |
add a comment |
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2
You could provide a minimal example of your dataset
– miimote
Nov 10 '18 at 16:44
1
awk version 20121220 on a macbook pro late 2017 up-to-date
– Zarkopafilis
Nov 10 '18 at 16:44