Padding day / time values to ensure 3 digit length
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I have a script which renders timestamps in the format YYYY'DDD'TTT where Y = year, D = day out of 365 and T = time in 1000th's of the day:
#!/bin/bash
clear
s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432))
d=$(date +%j)
y=$(date +%Y)
printf "$y`$d`$t" "$y" "$d" "$t"
but I need to modify it so that D and T will always produce a three-character value (ie, '001' instead of '1' for Jan 1, '001' instead of '1' for the first minute of the day) - and I have no idea how.
Any help is hugely appreciated
date time
edited Sep 29 '11 at 15:33
manatwork
22.2k38386
asked Sep 29 '11 at 14:52
Jack AndersJack Anders
262
add a comment |
I have a script which renders timestamps in the format YYYY'DDD'TTT where Y = year, D = day out of 365 and T = time in 1000th's of the day:
#!/bin/bash
clear
s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432))
d=$(date +%j)
y=$(date +%Y)
printf "$y`$d`$t" "$y" "$d" "$t"
but I need to modify it so that D and T will always produce a three-character value (ie, '001' instead of '1' for Jan 1, '001' instead of '1' for the first minute of the day) - and I have no idea how.
Any help is hugely appreciated
date time
edited Sep 29 '11 at 15:33
manatwork
22.2k38386
asked Sep 29 '11 at 14:52
Jack AndersJack Anders
262
add a comment |
I have a script which renders timestamps in the format YYYY'DDD'TTT where Y = year, D = day out of 365 and T = time in 1000th's of the day:
#!/bin/bash
clear
s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432))
d=$(date +%j)
y=$(date +%Y)
printf "$y`$d`$t" "$y" "$d" "$t"
but I need to modify it so that D and T will always produce a three-character value (ie, '001' instead of '1' for Jan 1, '001' instead of '1' for the first minute of the day) - and I have no idea how.
Any help is hugely appreciated
date time
edited Sep 29 '11 at 15:33
manatwork
22.2k38386
asked Sep 29 '11 at 14:52
Jack AndersJack Anders
262
I have a script which renders timestamps in the format YYYY'DDD'TTT where Y = year, D = day out of 365 and T = time in 1000th's of the day:
#!/bin/bash
clear
s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432))
d=$(date +%j)
y=$(date +%Y)
printf "$y`$d`$t" "$y" "$d" "$t"
but I need to modify it so that D and T will always produce a three-character value (ie, '001' instead of '1' for Jan 1, '001' instead of '1' for the first minute of the day) - and I have no idea how.
Any help is hugely appreciated
date time
date time
edited Sep 29 '11 at 15:33
manatwork
22.2k38386
asked Sep 29 '11 at 14:52
Jack AndersJack Anders
262
edited Sep 29 '11 at 15:33
manatwork
22.2k38386
asked Sep 29 '11 at 14:52
Jack AndersJack Anders
262
edited Sep 29 '11 at 15:33
manatwork
22.2k38386
edited Sep 29 '11 at 15:33
manatwork
22.2k38386
edited Sep 29 '11 at 15:33
manatwork
22.2k38386
22.2k38386
asked Sep 29 '11 at 14:52
Jack AndersJack Anders
262
asked Sep 29 '11 at 14:52
Jack AndersJack Anders
262
asked Sep 29 '11 at 14:52
Jack AndersJack Anders
262
262
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Perhaps you could take a look at the printf format string. The first parameter of printf should be a format string which includes placeholders for each of the arguments that follow.
You can include %d to represent an argument in signed decimal format, and you can prefix d by 0n for n characters of zero padding.
printf "%d/%d/%d" 2011 2 3
Will output 2011/2/3
printf "%04d/%03d/%05d" 2011 2 45
Should output 2011/002/00045
edited 1 hour ago
Rui F Ribeiro
41.9k1483142
answered Sep 29 '11 at 15:14
digitalseandigitalsean
16113
Thanks! That does help, but my real problem is that I can't just insert additional zeros, since after the first 99 values (days, ect), the value gains a third character on it's own. My script needs to pad values < 100 with one zero, and values < 10 with two zeros, if that makes sense...
– Jack Anders
Sep 29 '11 at 15:19
1
That is what the printf "%03d" is doing. The '3' is 'minimum' width and the '0' means to pad with a zero character ('0') to fill the minimum width. See the examples that digitalsean gave. Notice that the "2011" did not have any zeros added, because it met the minimum width.
– Arcege
Sep 29 '11 at 16:15
add a comment |
Let date do the work!
date +%3j
Then, for the milliday part, a simple trick is to compute 1000 plus the number of millidays and strip away the leading 1.
So for your script:
s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432 + 1000))
date "+%Y'%3j'${t#1}"
answered Sep 29 '11 at 22:40
GillesGilles
546k12911101624
For using date alone, +1, but as shown,2011-01-01 00:02produces2011'001'1without the zero-padding .... Here is a tweaked variant:t=00$(($(($(date +"%H*3600+%M*60+%S"))) * 5 / 432)); date "+%Y'%3j'${t:(-3)}"
– Peter.O
Sep 30 '11 at 3:35
@fred Oops, good spot. I've edited to a more portable way of padding the number.
– Gilles
Sep 30 '11 at 7:20
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Perhaps you could take a look at the printf format string. The first parameter of printf should be a format string which includes placeholders for each of the arguments that follow.
You can include %d to represent an argument in signed decimal format, and you can prefix d by 0n for n characters of zero padding.
printf "%d/%d/%d" 2011 2 3
Will output 2011/2/3
printf "%04d/%03d/%05d" 2011 2 45
Should output 2011/002/00045
edited 1 hour ago
Rui F Ribeiro
41.9k1483142
answered Sep 29 '11 at 15:14
digitalseandigitalsean
16113
Thanks! That does help, but my real problem is that I can't just insert additional zeros, since after the first 99 values (days, ect), the value gains a third character on it's own. My script needs to pad values < 100 with one zero, and values < 10 with two zeros, if that makes sense...
– Jack Anders
Sep 29 '11 at 15:19
1
That is what the printf "%03d" is doing. The '3' is 'minimum' width and the '0' means to pad with a zero character ('0') to fill the minimum width. See the examples that digitalsean gave. Notice that the "2011" did not have any zeros added, because it met the minimum width.
– Arcege
Sep 29 '11 at 16:15
add a comment |
Perhaps you could take a look at the printf format string. The first parameter of printf should be a format string which includes placeholders for each of the arguments that follow.
You can include %d to represent an argument in signed decimal format, and you can prefix d by 0n for n characters of zero padding.
printf "%d/%d/%d" 2011 2 3
Will output 2011/2/3
printf "%04d/%03d/%05d" 2011 2 45
Should output 2011/002/00045
edited 1 hour ago
Rui F Ribeiro
41.9k1483142
answered Sep 29 '11 at 15:14
digitalseandigitalsean
16113
Thanks! That does help, but my real problem is that I can't just insert additional zeros, since after the first 99 values (days, ect), the value gains a third character on it's own. My script needs to pad values < 100 with one zero, and values < 10 with two zeros, if that makes sense...
– Jack Anders
Sep 29 '11 at 15:19
1
That is what the printf "%03d" is doing. The '3' is 'minimum' width and the '0' means to pad with a zero character ('0') to fill the minimum width. See the examples that digitalsean gave. Notice that the "2011" did not have any zeros added, because it met the minimum width.
– Arcege
Sep 29 '11 at 16:15
add a comment |
Perhaps you could take a look at the printf format string. The first parameter of printf should be a format string which includes placeholders for each of the arguments that follow.
You can include %d to represent an argument in signed decimal format, and you can prefix d by 0n for n characters of zero padding.
printf "%d/%d/%d" 2011 2 3
Will output 2011/2/3
printf "%04d/%03d/%05d" 2011 2 45
Should output 2011/002/00045
edited 1 hour ago
Rui F Ribeiro
41.9k1483142
answered Sep 29 '11 at 15:14
digitalseandigitalsean
16113
Perhaps you could take a look at the printf format string. The first parameter of printf should be a format string which includes placeholders for each of the arguments that follow.
You can include %d to represent an argument in signed decimal format, and you can prefix d by 0n for n characters of zero padding.
printf "%d/%d/%d" 2011 2 3
Will output 2011/2/3
printf "%04d/%03d/%05d" 2011 2 45
Should output 2011/002/00045
edited 1 hour ago
Rui F Ribeiro
41.9k1483142
answered Sep 29 '11 at 15:14
digitalseandigitalsean
16113
edited 1 hour ago
Rui F Ribeiro
41.9k1483142
edited 1 hour ago
Rui F Ribeiro
41.9k1483142
edited 1 hour ago
Rui F Ribeiro
41.9k1483142
41.9k1483142
answered Sep 29 '11 at 15:14
digitalseandigitalsean
16113
answered Sep 29 '11 at 15:14
digitalseandigitalsean
16113
answered Sep 29 '11 at 15:14
digitalseandigitalsean
16113
16113
Thanks! That does help, but my real problem is that I can't just insert additional zeros, since after the first 99 values (days, ect), the value gains a third character on it's own. My script needs to pad values < 100 with one zero, and values < 10 with two zeros, if that makes sense...
– Jack Anders
Sep 29 '11 at 15:19
1
That is what the printf "%03d" is doing. The '3' is 'minimum' width and the '0' means to pad with a zero character ('0') to fill the minimum width. See the examples that digitalsean gave. Notice that the "2011" did not have any zeros added, because it met the minimum width.
– Arcege
Sep 29 '11 at 16:15
add a comment |
Thanks! That does help, but my real problem is that I can't just insert additional zeros, since after the first 99 values (days, ect), the value gains a third character on it's own. My script needs to pad values < 100 with one zero, and values < 10 with two zeros, if that makes sense...
– Jack Anders
Sep 29 '11 at 15:19
1
That is what the printf "%03d" is doing. The '3' is 'minimum' width and the '0' means to pad with a zero character ('0') to fill the minimum width. See the examples that digitalsean gave. Notice that the "2011" did not have any zeros added, because it met the minimum width.
– Arcege
Sep 29 '11 at 16:15
Thanks! That does help, but my real problem is that I can't just insert additional zeros, since after the first 99 values (days, ect), the value gains a third character on it's own. My script needs to pad values < 100 with one zero, and values < 10 with two zeros, if that makes sense...
– Jack Anders
Sep 29 '11 at 15:19
Thanks! That does help, but my real problem is that I can't just insert additional zeros, since after the first 99 values (days, ect), the value gains a third character on it's own. My script needs to pad values < 100 with one zero, and values < 10 with two zeros, if that makes sense...
– Jack Anders
Sep 29 '11 at 15:19
1
1
That is what the printf "%03d" is doing. The '3' is 'minimum' width and the '0' means to pad with a zero character ('0') to fill the minimum width. See the examples that digitalsean gave. Notice that the "2011" did not have any zeros added, because it met the minimum width.
– Arcege
Sep 29 '11 at 16:15
That is what the printf "%03d" is doing. The '3' is 'minimum' width and the '0' means to pad with a zero character ('0') to fill the minimum width. See the examples that digitalsean gave. Notice that the "2011" did not have any zeros added, because it met the minimum width.
– Arcege
Sep 29 '11 at 16:15
add a comment |
Let date do the work!
date +%3j
Then, for the milliday part, a simple trick is to compute 1000 plus the number of millidays and strip away the leading 1.
So for your script:
s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432 + 1000))
date "+%Y'%3j'${t#1}"
answered Sep 29 '11 at 22:40
GillesGilles
546k12911101624
For using date alone, +1, but as shown,2011-01-01 00:02produces2011'001'1without the zero-padding .... Here is a tweaked variant:t=00$(($(($(date +"%H*3600+%M*60+%S"))) * 5 / 432)); date "+%Y'%3j'${t:(-3)}"
– Peter.O
Sep 30 '11 at 3:35
@fred Oops, good spot. I've edited to a more portable way of padding the number.
– Gilles
Sep 30 '11 at 7:20
add a comment |
Let date do the work!
date +%3j
Then, for the milliday part, a simple trick is to compute 1000 plus the number of millidays and strip away the leading 1.
So for your script:
s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432 + 1000))
date "+%Y'%3j'${t#1}"
answered Sep 29 '11 at 22:40
GillesGilles
546k12911101624
For using date alone, +1, but as shown,2011-01-01 00:02produces2011'001'1without the zero-padding .... Here is a tweaked variant:t=00$(($(($(date +"%H*3600+%M*60+%S"))) * 5 / 432)); date "+%Y'%3j'${t:(-3)}"
– Peter.O
Sep 30 '11 at 3:35
@fred Oops, good spot. I've edited to a more portable way of padding the number.
– Gilles
Sep 30 '11 at 7:20
add a comment |
Let date do the work!
date +%3j
Then, for the milliday part, a simple trick is to compute 1000 plus the number of millidays and strip away the leading 1.
So for your script:
s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432 + 1000))
date "+%Y'%3j'${t#1}"
answered Sep 29 '11 at 22:40
GillesGilles
546k12911101624
Let date do the work!
date +%3j
Then, for the milliday part, a simple trick is to compute 1000 plus the number of millidays and strip away the leading 1.
So for your script:
s=$(($(date +"%H*3600+%M*60+%S")))
t=$(($s * 5 / 432 + 1000))
date "+%Y'%3j'${t#1}"
answered Sep 29 '11 at 22:40
GillesGilles
546k12911101624
edited Sep 30 '11 at 7:20
answered Sep 29 '11 at 22:40
GillesGilles
546k12911101624
answered Sep 29 '11 at 22:40
GillesGilles
546k12911101624
answered Sep 29 '11 at 22:40
GillesGilles
546k12911101624
546k12911101624
For using date alone, +1, but as shown,2011-01-01 00:02produces2011'001'1without the zero-padding .... Here is a tweaked variant:t=00$(($(($(date +"%H*3600+%M*60+%S"))) * 5 / 432)); date "+%Y'%3j'${t:(-3)}"
– Peter.O
Sep 30 '11 at 3:35
@fred Oops, good spot. I've edited to a more portable way of padding the number.
– Gilles
Sep 30 '11 at 7:20
add a comment |
For using date alone, +1, but as shown,2011-01-01 00:02produces2011'001'1without the zero-padding .... Here is a tweaked variant:t=00$(($(($(date +"%H*3600+%M*60+%S"))) * 5 / 432)); date "+%Y'%3j'${t:(-3)}"
– Peter.O
Sep 30 '11 at 3:35
@fred Oops, good spot. I've edited to a more portable way of padding the number.
– Gilles
Sep 30 '11 at 7:20
For using date alone, +1, but as shown,
2011-01-01 00:02 produces 2011'001'1 without the zero-padding .... Here is a tweaked variant: t=00$(($(($(date +"%H*3600+%M*60+%S"))) * 5 / 432)); date "+%Y'%3j'${t:(-3)}"– Peter.O
Sep 30 '11 at 3:35
For using date alone, +1, but as shown,
2011-01-01 00:02 produces 2011'001'1 without the zero-padding .... Here is a tweaked variant: t=00$(($(($(date +"%H*3600+%M*60+%S"))) * 5 / 432)); date "+%Y'%3j'${t:(-3)}"– Peter.O
Sep 30 '11 at 3:35
@fred Oops, good spot. I've edited to a more portable way of padding the number.
– Gilles
Sep 30 '11 at 7:20
@fred Oops, good spot. I've edited to a more portable way of padding the number.
– Gilles
Sep 30 '11 at 7:20
add a comment |
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