Subtract time using bash?
Clash Royale CLAN TAG#URR8PPP
Is it possible to use bash to subtract variables containing 24-hour time?
#!/bin/bash
var1="23:30" # 11:30pm
var2="20:00" # 08:00pm
echo "$(expr $var1 - $var2)"
Running it produces the following error.
./test
expr: non-integer argument
I need the output to appear in decimal form, for example:
./test
3.5
bash date variable arithmetic
add a comment |
Is it possible to use bash to subtract variables containing 24-hour time?
#!/bin/bash
var1="23:30" # 11:30pm
var2="20:00" # 08:00pm
echo "$(expr $var1 - $var2)"
Running it produces the following error.
./test
expr: non-integer argument
I need the output to appear in decimal form, for example:
./test
3.5
bash date variable arithmetic
Thanks for pointing that you. You're probably right. I've updated the question.
– user328302
Dec 24 '18 at 12:37
1
What are your constraints? Shell builtins only, or external programs likebc
orawk
?
– Mark Plotnick
Dec 24 '18 at 12:39
Related: unix.stackexchange.com/q/24626/315749
– fra-san
Dec 24 '18 at 12:41
@fra-san, the dates are static, in variables, so it's kinda confusing to usedate
to solve this problem. @MarkPlotnick, no constraints, which ever method requires the least code. I'm ready to accept your Answer
– user328302
Dec 24 '18 at 12:47
add a comment |
Is it possible to use bash to subtract variables containing 24-hour time?
#!/bin/bash
var1="23:30" # 11:30pm
var2="20:00" # 08:00pm
echo "$(expr $var1 - $var2)"
Running it produces the following error.
./test
expr: non-integer argument
I need the output to appear in decimal form, for example:
./test
3.5
bash date variable arithmetic
Is it possible to use bash to subtract variables containing 24-hour time?
#!/bin/bash
var1="23:30" # 11:30pm
var2="20:00" # 08:00pm
echo "$(expr $var1 - $var2)"
Running it produces the following error.
./test
expr: non-integer argument
I need the output to appear in decimal form, for example:
./test
3.5
bash date variable arithmetic
bash date variable arithmetic
edited Dec 24 '18 at 12:57
Jeff Schaller
39k1053125
39k1053125
asked Dec 24 '18 at 12:25
user328302
262
262
Thanks for pointing that you. You're probably right. I've updated the question.
– user328302
Dec 24 '18 at 12:37
1
What are your constraints? Shell builtins only, or external programs likebc
orawk
?
– Mark Plotnick
Dec 24 '18 at 12:39
Related: unix.stackexchange.com/q/24626/315749
– fra-san
Dec 24 '18 at 12:41
@fra-san, the dates are static, in variables, so it's kinda confusing to usedate
to solve this problem. @MarkPlotnick, no constraints, which ever method requires the least code. I'm ready to accept your Answer
– user328302
Dec 24 '18 at 12:47
add a comment |
Thanks for pointing that you. You're probably right. I've updated the question.
– user328302
Dec 24 '18 at 12:37
1
What are your constraints? Shell builtins only, or external programs likebc
orawk
?
– Mark Plotnick
Dec 24 '18 at 12:39
Related: unix.stackexchange.com/q/24626/315749
– fra-san
Dec 24 '18 at 12:41
@fra-san, the dates are static, in variables, so it's kinda confusing to usedate
to solve this problem. @MarkPlotnick, no constraints, which ever method requires the least code. I'm ready to accept your Answer
– user328302
Dec 24 '18 at 12:47
Thanks for pointing that you. You're probably right. I've updated the question.
– user328302
Dec 24 '18 at 12:37
Thanks for pointing that you. You're probably right. I've updated the question.
– user328302
Dec 24 '18 at 12:37
1
1
What are your constraints? Shell builtins only, or external programs like
bc
or awk
?– Mark Plotnick
Dec 24 '18 at 12:39
What are your constraints? Shell builtins only, or external programs like
bc
or awk
?– Mark Plotnick
Dec 24 '18 at 12:39
Related: unix.stackexchange.com/q/24626/315749
– fra-san
Dec 24 '18 at 12:41
Related: unix.stackexchange.com/q/24626/315749
– fra-san
Dec 24 '18 at 12:41
@fra-san, the dates are static, in variables, so it's kinda confusing to use
date
to solve this problem. @MarkPlotnick, no constraints, which ever method requires the least code. I'm ready to accept your Answer– user328302
Dec 24 '18 at 12:47
@fra-san, the dates are static, in variables, so it's kinda confusing to use
date
to solve this problem. @MarkPlotnick, no constraints, which ever method requires the least code. I'm ready to accept your Answer– user328302
Dec 24 '18 at 12:47
add a comment |
2 Answers
2
active
oldest
votes
The date
command is pretty flexible about its input. You can use that to your advantage:
#!/bin/bash
var1="23:30"
var2="20:00"
# Convert to epoch time and calculate difference.
difference=$(( $(date -d "$var1" "+%s") - $(date -d "$var2" "+%s") ))
# Divide the difference by 3600 to calculate hours.
echo "scale=2 ; $difference/3600" | bc
Output:
$ ./test.bash
3.50
add a comment |
Using only bash
, with no external programs, you could do so something like this:
#!/bin/bash
# first time is the first argument, or 23:30
var1=$1:-23:30
# second time is the second argument, or 20:00
var2=$2:-20:00
# Split variables on `:` and insert pieces into arrays
IFS=':' read -r -a t1 <<< "$var1"
IFS=':' read -r -a t2 <<< "$var2"
# strip leading zeros (so it's not interpreted as octal
t1=("$t1[@]##0")
t2=("$t2[@]##0")
# check if the first time is before the second one
if (( t1[0] > t2[0] || ( t1[0] == t2[0] && t1[1] > t2[1]) ))
then
# if the minutes on the first time are less than the ones on the second time
if (( t1[1] < t2[1] ))
then
# add 60 minutes to time 1
(( t1[1] += 60 ))
# and subtract an hour
(( t1[0] -- ))
fi
# now subtract the hours and the minutes
echo $((t1[0] -t2[0] )):$((t1[1] - t2[1]))
# to get a decimal result, multiply the minutes by 100 and divide by 60
echo $((t1[0] -t2[0] )).$(((t1[1] - t2[1])*100/60))
else
echo "Time 1 should be after time 2" 2>&1
fi
Test:
$ ./script.sh
3:30
3.50
$ ./script.sh 12:10 11:30
0:40
0.66
$ ./script.sh 12:00 11:30
0:30
0.50
If you want more complex time differences, that could span different days etc, then it's probably best to use GNU date.
Wow, thanks. Would you be willing to add a few quick comments above some of the lines of code that explain what's happening?
– user328302
Dec 24 '18 at 12:55
@user328302: I added some comments, let me know if anything is still unclear. I also fixed a bug that wouldn't allow times like09:08
(it was treated as octal)
– user000001
Dec 24 '18 at 13:17
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The date
command is pretty flexible about its input. You can use that to your advantage:
#!/bin/bash
var1="23:30"
var2="20:00"
# Convert to epoch time and calculate difference.
difference=$(( $(date -d "$var1" "+%s") - $(date -d "$var2" "+%s") ))
# Divide the difference by 3600 to calculate hours.
echo "scale=2 ; $difference/3600" | bc
Output:
$ ./test.bash
3.50
add a comment |
The date
command is pretty flexible about its input. You can use that to your advantage:
#!/bin/bash
var1="23:30"
var2="20:00"
# Convert to epoch time and calculate difference.
difference=$(( $(date -d "$var1" "+%s") - $(date -d "$var2" "+%s") ))
# Divide the difference by 3600 to calculate hours.
echo "scale=2 ; $difference/3600" | bc
Output:
$ ./test.bash
3.50
add a comment |
The date
command is pretty flexible about its input. You can use that to your advantage:
#!/bin/bash
var1="23:30"
var2="20:00"
# Convert to epoch time and calculate difference.
difference=$(( $(date -d "$var1" "+%s") - $(date -d "$var2" "+%s") ))
# Divide the difference by 3600 to calculate hours.
echo "scale=2 ; $difference/3600" | bc
Output:
$ ./test.bash
3.50
The date
command is pretty flexible about its input. You can use that to your advantage:
#!/bin/bash
var1="23:30"
var2="20:00"
# Convert to epoch time and calculate difference.
difference=$(( $(date -d "$var1" "+%s") - $(date -d "$var2" "+%s") ))
# Divide the difference by 3600 to calculate hours.
echo "scale=2 ; $difference/3600" | bc
Output:
$ ./test.bash
3.50
answered Dec 24 '18 at 12:59
Haxiel
1,337310
1,337310
add a comment |
add a comment |
Using only bash
, with no external programs, you could do so something like this:
#!/bin/bash
# first time is the first argument, or 23:30
var1=$1:-23:30
# second time is the second argument, or 20:00
var2=$2:-20:00
# Split variables on `:` and insert pieces into arrays
IFS=':' read -r -a t1 <<< "$var1"
IFS=':' read -r -a t2 <<< "$var2"
# strip leading zeros (so it's not interpreted as octal
t1=("$t1[@]##0")
t2=("$t2[@]##0")
# check if the first time is before the second one
if (( t1[0] > t2[0] || ( t1[0] == t2[0] && t1[1] > t2[1]) ))
then
# if the minutes on the first time are less than the ones on the second time
if (( t1[1] < t2[1] ))
then
# add 60 minutes to time 1
(( t1[1] += 60 ))
# and subtract an hour
(( t1[0] -- ))
fi
# now subtract the hours and the minutes
echo $((t1[0] -t2[0] )):$((t1[1] - t2[1]))
# to get a decimal result, multiply the minutes by 100 and divide by 60
echo $((t1[0] -t2[0] )).$(((t1[1] - t2[1])*100/60))
else
echo "Time 1 should be after time 2" 2>&1
fi
Test:
$ ./script.sh
3:30
3.50
$ ./script.sh 12:10 11:30
0:40
0.66
$ ./script.sh 12:00 11:30
0:30
0.50
If you want more complex time differences, that could span different days etc, then it's probably best to use GNU date.
Wow, thanks. Would you be willing to add a few quick comments above some of the lines of code that explain what's happening?
– user328302
Dec 24 '18 at 12:55
@user328302: I added some comments, let me know if anything is still unclear. I also fixed a bug that wouldn't allow times like09:08
(it was treated as octal)
– user000001
Dec 24 '18 at 13:17
add a comment |
Using only bash
, with no external programs, you could do so something like this:
#!/bin/bash
# first time is the first argument, or 23:30
var1=$1:-23:30
# second time is the second argument, or 20:00
var2=$2:-20:00
# Split variables on `:` and insert pieces into arrays
IFS=':' read -r -a t1 <<< "$var1"
IFS=':' read -r -a t2 <<< "$var2"
# strip leading zeros (so it's not interpreted as octal
t1=("$t1[@]##0")
t2=("$t2[@]##0")
# check if the first time is before the second one
if (( t1[0] > t2[0] || ( t1[0] == t2[0] && t1[1] > t2[1]) ))
then
# if the minutes on the first time are less than the ones on the second time
if (( t1[1] < t2[1] ))
then
# add 60 minutes to time 1
(( t1[1] += 60 ))
# and subtract an hour
(( t1[0] -- ))
fi
# now subtract the hours and the minutes
echo $((t1[0] -t2[0] )):$((t1[1] - t2[1]))
# to get a decimal result, multiply the minutes by 100 and divide by 60
echo $((t1[0] -t2[0] )).$(((t1[1] - t2[1])*100/60))
else
echo "Time 1 should be after time 2" 2>&1
fi
Test:
$ ./script.sh
3:30
3.50
$ ./script.sh 12:10 11:30
0:40
0.66
$ ./script.sh 12:00 11:30
0:30
0.50
If you want more complex time differences, that could span different days etc, then it's probably best to use GNU date.
Wow, thanks. Would you be willing to add a few quick comments above some of the lines of code that explain what's happening?
– user328302
Dec 24 '18 at 12:55
@user328302: I added some comments, let me know if anything is still unclear. I also fixed a bug that wouldn't allow times like09:08
(it was treated as octal)
– user000001
Dec 24 '18 at 13:17
add a comment |
Using only bash
, with no external programs, you could do so something like this:
#!/bin/bash
# first time is the first argument, or 23:30
var1=$1:-23:30
# second time is the second argument, or 20:00
var2=$2:-20:00
# Split variables on `:` and insert pieces into arrays
IFS=':' read -r -a t1 <<< "$var1"
IFS=':' read -r -a t2 <<< "$var2"
# strip leading zeros (so it's not interpreted as octal
t1=("$t1[@]##0")
t2=("$t2[@]##0")
# check if the first time is before the second one
if (( t1[0] > t2[0] || ( t1[0] == t2[0] && t1[1] > t2[1]) ))
then
# if the minutes on the first time are less than the ones on the second time
if (( t1[1] < t2[1] ))
then
# add 60 minutes to time 1
(( t1[1] += 60 ))
# and subtract an hour
(( t1[0] -- ))
fi
# now subtract the hours and the minutes
echo $((t1[0] -t2[0] )):$((t1[1] - t2[1]))
# to get a decimal result, multiply the minutes by 100 and divide by 60
echo $((t1[0] -t2[0] )).$(((t1[1] - t2[1])*100/60))
else
echo "Time 1 should be after time 2" 2>&1
fi
Test:
$ ./script.sh
3:30
3.50
$ ./script.sh 12:10 11:30
0:40
0.66
$ ./script.sh 12:00 11:30
0:30
0.50
If you want more complex time differences, that could span different days etc, then it's probably best to use GNU date.
Using only bash
, with no external programs, you could do so something like this:
#!/bin/bash
# first time is the first argument, or 23:30
var1=$1:-23:30
# second time is the second argument, or 20:00
var2=$2:-20:00
# Split variables on `:` and insert pieces into arrays
IFS=':' read -r -a t1 <<< "$var1"
IFS=':' read -r -a t2 <<< "$var2"
# strip leading zeros (so it's not interpreted as octal
t1=("$t1[@]##0")
t2=("$t2[@]##0")
# check if the first time is before the second one
if (( t1[0] > t2[0] || ( t1[0] == t2[0] && t1[1] > t2[1]) ))
then
# if the minutes on the first time are less than the ones on the second time
if (( t1[1] < t2[1] ))
then
# add 60 minutes to time 1
(( t1[1] += 60 ))
# and subtract an hour
(( t1[0] -- ))
fi
# now subtract the hours and the minutes
echo $((t1[0] -t2[0] )):$((t1[1] - t2[1]))
# to get a decimal result, multiply the minutes by 100 and divide by 60
echo $((t1[0] -t2[0] )).$(((t1[1] - t2[1])*100/60))
else
echo "Time 1 should be after time 2" 2>&1
fi
Test:
$ ./script.sh
3:30
3.50
$ ./script.sh 12:10 11:30
0:40
0.66
$ ./script.sh 12:00 11:30
0:30
0.50
If you want more complex time differences, that could span different days etc, then it's probably best to use GNU date.
edited Dec 24 '18 at 14:33
answered Dec 24 '18 at 12:50
user000001
902713
902713
Wow, thanks. Would you be willing to add a few quick comments above some of the lines of code that explain what's happening?
– user328302
Dec 24 '18 at 12:55
@user328302: I added some comments, let me know if anything is still unclear. I also fixed a bug that wouldn't allow times like09:08
(it was treated as octal)
– user000001
Dec 24 '18 at 13:17
add a comment |
Wow, thanks. Would you be willing to add a few quick comments above some of the lines of code that explain what's happening?
– user328302
Dec 24 '18 at 12:55
@user328302: I added some comments, let me know if anything is still unclear. I also fixed a bug that wouldn't allow times like09:08
(it was treated as octal)
– user000001
Dec 24 '18 at 13:17
Wow, thanks. Would you be willing to add a few quick comments above some of the lines of code that explain what's happening?
– user328302
Dec 24 '18 at 12:55
Wow, thanks. Would you be willing to add a few quick comments above some of the lines of code that explain what's happening?
– user328302
Dec 24 '18 at 12:55
@user328302: I added some comments, let me know if anything is still unclear. I also fixed a bug that wouldn't allow times like
09:08
(it was treated as octal)– user000001
Dec 24 '18 at 13:17
@user328302: I added some comments, let me know if anything is still unclear. I also fixed a bug that wouldn't allow times like
09:08
(it was treated as octal)– user000001
Dec 24 '18 at 13:17
add a comment |
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Thanks for pointing that you. You're probably right. I've updated the question.
– user328302
Dec 24 '18 at 12:37
1
What are your constraints? Shell builtins only, or external programs like
bc
orawk
?– Mark Plotnick
Dec 24 '18 at 12:39
Related: unix.stackexchange.com/q/24626/315749
– fra-san
Dec 24 '18 at 12:41
@fra-san, the dates are static, in variables, so it's kinda confusing to use
date
to solve this problem. @MarkPlotnick, no constraints, which ever method requires the least code. I'm ready to accept your Answer– user328302
Dec 24 '18 at 12:47