Subtract time using bash?

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

edited Dec 24 '18 at 12:57

Jeff Schaller

43.1k1159138

asked Dec 24 '18 at 12:25

user328302

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 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

@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 |

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

edited Dec 24 '18 at 12:57

Jeff Schaller

43.1k1159138

asked Dec 24 '18 at 12:25

user328302

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 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

@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 |

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

edited Dec 24 '18 at 12:57

Jeff Schaller

43.1k1159138

asked Dec 24 '18 at 12:25

user328302

262

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

edited Dec 24 '18 at 12:57

Jeff Schaller

43.1k1159138

asked Dec 24 '18 at 12:25

user328302

262

edited Dec 24 '18 at 12:57

Jeff Schaller

43.1k1159138

asked Dec 24 '18 at 12:25

user328302

262

edited Dec 24 '18 at 12:57

Jeff Schaller

43.1k1159138

edited Dec 24 '18 at 12:57

Jeff Schaller

43.1k1159138

edited Dec 24 '18 at 12:57

Jeff Schaller

43.1k1159138

asked Dec 24 '18 at 12:25

user328302

262

asked Dec 24 '18 at 12:25

user328302

262

asked Dec 24 '18 at 12:25

user328302

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 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

@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 |

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 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

@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

Thanks for pointing that you. You're probably right. I've updated the question.

– user328302
Dec 24 '18 at 12:37

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

@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

answered Dec 24 '18 at 12:59

Haxiel

3,0901919

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.

edited Dec 24 '18 at 14:33

answered Dec 24 '18 at 12:50

user000001

922713

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

add a comment |

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2 Answers
2

active

oldest

votes

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

answered Dec 24 '18 at 12:59

Haxiel

3,0901919

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

answered Dec 24 '18 at 12:59

Haxiel

3,0901919

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

answered Dec 24 '18 at 12:59

Haxiel

3,0901919

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

3,0901919

answered Dec 24 '18 at 12:59

Haxiel

3,0901919

answered Dec 24 '18 at 12:59

Haxiel

3,0901919

answered Dec 24 '18 at 12:59

Haxiel

3,0901919

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.

edited Dec 24 '18 at 14:33

answered Dec 24 '18 at 12:50

user000001

922713

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

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.

edited Dec 24 '18 at 14:33

answered Dec 24 '18 at 12:50

user000001

922713

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

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.

edited Dec 24 '18 at 14:33

answered Dec 24 '18 at 12:50

user000001

922713

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

922713

edited Dec 24 '18 at 14:33

answered Dec 24 '18 at 12:50

user000001

922713

answered Dec 24 '18 at 12:50

user000001

922713

answered Dec 24 '18 at 12:50

user000001

922713

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

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 like 09:08 (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

@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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