How to convert a DateTime variable in awk to seconds?
I have a variable in which stores the DateTime in awk
test=$(f["DateTime"])
print "Printing Test variable:",test
Output:
Printing Test variable:,"2018-12-18 18:36:55"
I want to convert the value in test to seconds
Could anybody tell me how to do that in awk
Using mktime() is giving -1 as output
print mktime(gensub(/[ :-]/," ","g",test))
My Input data (CSV File)is like below
DateTime,Dealer,Some Value,State,CallEndTime,Some Value,TotalDuration,,..
"2019-01-07 11:35:42","Car","fab","foo",,"bar","100","boo",..
I got the DateTime value from this,
command-line awk
asked Jan 23 at 3:23
mittumittu
83
add a comment |
I have a variable in which stores the DateTime in awk
test=$(f["DateTime"])
print "Printing Test variable:",test
Output:
Printing Test variable:,"2018-12-18 18:36:55"
I want to convert the value in test to seconds
Could anybody tell me how to do that in awk
Using mktime() is giving -1 as output
print mktime(gensub(/[ :-]/," ","g",test))
My Input data (CSV File)is like below
DateTime,Dealer,Some Value,State,CallEndTime,Some Value,TotalDuration,,..
"2019-01-07 11:35:42","Car","fab","foo",,"bar","100","boo",..
I got the DateTime value from this,
command-line awk
asked Jan 23 at 3:23
mittumittu
83
add a comment |
I have a variable in which stores the DateTime in awk
test=$(f["DateTime"])
print "Printing Test variable:",test
Output:
Printing Test variable:,"2018-12-18 18:36:55"
I want to convert the value in test to seconds
Could anybody tell me how to do that in awk
Using mktime() is giving -1 as output
print mktime(gensub(/[ :-]/," ","g",test))
My Input data (CSV File)is like below
DateTime,Dealer,Some Value,State,CallEndTime,Some Value,TotalDuration,,..
"2019-01-07 11:35:42","Car","fab","foo",,"bar","100","boo",..
I got the DateTime value from this,
command-line awk
asked Jan 23 at 3:23
mittumittu
83
I have a variable in which stores the DateTime in awk
test=$(f["DateTime"])
print "Printing Test variable:",test
Output:
Printing Test variable:,"2018-12-18 18:36:55"
I want to convert the value in test to seconds
Could anybody tell me how to do that in awk
Using mktime() is giving -1 as output
print mktime(gensub(/[ :-]/," ","g",test))
My Input data (CSV File)is like below
DateTime,Dealer,Some Value,State,CallEndTime,Some Value,TotalDuration,,..
"2019-01-07 11:35:42","Car","fab","foo",,"bar","100","boo",..
I got the DateTime value from this,
command-line awk
command-line awk
asked Jan 23 at 3:23
mittumittu
83
asked Jan 23 at 3:23
mittumittu
83
edited Jan 23 at 4:04
mittu
asked Jan 23 at 3:23
mittumittu
83
asked Jan 23 at 3:23
mittumittu
83
asked Jan 23 at 3:23
mittumittu
83
83
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
If you have GNU awk (gawk) you can use its mktime function - however the input must be a space-separated datespec of the form "YYYY MM DD HH MM SS [DST]". The date fields in your variable are in the right order, but you will need to replace the delimiters with spaces:
$ gawk -v test="2018-12-18 18:36:55" 'BEGIN{print mktime(gensub(/[ :-]/," ","g",test))}'
1545176215
See for example Time Functions in the GNU Awk User's Guide
If your string is enclosed in double quotes, you will need to remove those as well:
$ gawk -v test='"2018-12-18 18:36:55"' 'BEGIN{print mktime(gensub(/[ ":-]/," ","g",test))}'
1545176215
answered Jan 23 at 3:45
steeldriversteeldriver
68.8k11113184
I tried this,but its giving me -1 as O/P, Please check the edited question
– mittu
Jan 23 at 4:05
@mittu please see update
– steeldriver
Jan 23 at 4:09
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
If you have GNU awk (gawk) you can use its mktime function - however the input must be a space-separated datespec of the form "YYYY MM DD HH MM SS [DST]". The date fields in your variable are in the right order, but you will need to replace the delimiters with spaces:
$ gawk -v test="2018-12-18 18:36:55" 'BEGIN{print mktime(gensub(/[ :-]/," ","g",test))}'
1545176215
See for example Time Functions in the GNU Awk User's Guide
If your string is enclosed in double quotes, you will need to remove those as well:
$ gawk -v test='"2018-12-18 18:36:55"' 'BEGIN{print mktime(gensub(/[ ":-]/," ","g",test))}'
1545176215
answered Jan 23 at 3:45
steeldriversteeldriver
68.8k11113184
I tried this,but its giving me -1 as O/P, Please check the edited question
– mittu
Jan 23 at 4:05
@mittu please see update
– steeldriver
Jan 23 at 4:09
add a comment |
If you have GNU awk (gawk) you can use its mktime function - however the input must be a space-separated datespec of the form "YYYY MM DD HH MM SS [DST]". The date fields in your variable are in the right order, but you will need to replace the delimiters with spaces:
$ gawk -v test="2018-12-18 18:36:55" 'BEGIN{print mktime(gensub(/[ :-]/," ","g",test))}'
1545176215
See for example Time Functions in the GNU Awk User's Guide
If your string is enclosed in double quotes, you will need to remove those as well:
$ gawk -v test='"2018-12-18 18:36:55"' 'BEGIN{print mktime(gensub(/[ ":-]/," ","g",test))}'
1545176215
answered Jan 23 at 3:45
steeldriversteeldriver
68.8k11113184
I tried this,but its giving me -1 as O/P, Please check the edited question
– mittu
Jan 23 at 4:05
@mittu please see update
– steeldriver
Jan 23 at 4:09
add a comment |
If you have GNU awk (gawk) you can use its mktime function - however the input must be a space-separated datespec of the form "YYYY MM DD HH MM SS [DST]". The date fields in your variable are in the right order, but you will need to replace the delimiters with spaces:
$ gawk -v test="2018-12-18 18:36:55" 'BEGIN{print mktime(gensub(/[ :-]/," ","g",test))}'
1545176215
See for example Time Functions in the GNU Awk User's Guide
If your string is enclosed in double quotes, you will need to remove those as well:
$ gawk -v test='"2018-12-18 18:36:55"' 'BEGIN{print mktime(gensub(/[ ":-]/," ","g",test))}'
1545176215
answered Jan 23 at 3:45
steeldriversteeldriver
68.8k11113184
If you have GNU awk (gawk) you can use its mktime function - however the input must be a space-separated datespec of the form "YYYY MM DD HH MM SS [DST]". The date fields in your variable are in the right order, but you will need to replace the delimiters with spaces:
$ gawk -v test="2018-12-18 18:36:55" 'BEGIN{print mktime(gensub(/[ :-]/," ","g",test))}'
1545176215
See for example Time Functions in the GNU Awk User's Guide
If your string is enclosed in double quotes, you will need to remove those as well:
$ gawk -v test='"2018-12-18 18:36:55"' 'BEGIN{print mktime(gensub(/[ ":-]/," ","g",test))}'
1545176215
answered Jan 23 at 3:45
steeldriversteeldriver
68.8k11113184
edited Jan 23 at 4:09
answered Jan 23 at 3:45
steeldriversteeldriver
68.8k11113184
answered Jan 23 at 3:45
steeldriversteeldriver
68.8k11113184
answered Jan 23 at 3:45
steeldriversteeldriver
68.8k11113184
68.8k11113184
I tried this,but its giving me -1 as O/P, Please check the edited question
– mittu
Jan 23 at 4:05
@mittu please see update
– steeldriver
Jan 23 at 4:09
add a comment |
I tried this,but its giving me -1 as O/P, Please check the edited question
– mittu
Jan 23 at 4:05
@mittu please see update
– steeldriver
Jan 23 at 4:09
I tried this,but its giving me -1 as O/P, Please check the edited question
– mittu
Jan 23 at 4:05
I tried this,but its giving me -1 as O/P, Please check the edited question
– mittu
Jan 23 at 4:05
@mittu please see update
– steeldriver
Jan 23 at 4:09
@mittu please see update
– steeldriver
Jan 23 at 4:09
add a comment |
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