CODEWITHSUNDEEP
bashshell
d_luffy_de
d_luffy_de

Reputation: 967

Get Year, Month, and Date from Variable

I have a variable of the format $var_YYYY_MM_DD_HH_MM_SS.txt

Eg: variable=sample_data_2017_01_01_10_22_10.txt

I need to extract the following from this variable:

Year=YYYY
Month=MM
Date=DD

Can you please help?

Upvotes: 1

Views: 1859

Answers (5)

Raman Sailopal
Raman Sailopal

Reputation: 12887

Building on some of the other awk solutions a more complete solution would be: 

echo $variable | awk -F_ '{ printf "Year="$3"\nMonth="$4"\nDate="$5"\n" }'

Upvotes: 0

sat
sat

Reputation: 14949

You can simply use read command with setting IFS='_'.

$ variable=sample_data_2017_01_01_10_22_10.txt
$ IFS='_' read -r tmp tmp Year Month Date tmp <<< "$variable"
$ echo "$Year : $Month : $Date"
2017 : 01 : 01

Upvotes: 1

F.Igor
F.Igor

Reputation: 4390

Using awk

Year=$(echo $variable | awk '{split($0,a,"_"); print a[3]}')
Month=$(echo $variable | awk '{split($0,a,"_"); print a[4]}')
Day=$(echo $variable | awk '{split($0,a,"_"); print a[5]}')

Upvotes: 0

mshildt
mshildt

Reputation: 9362

You could try:

Year=$(echo $variable | cut -d '_' -f3)
Month=$(echo $variable | cut -d '_' -f4)
Date=$(echo $variable | cut -d '_' -f5)

This only works if you are sure your variable is laid out in the exact way you describe in your question though. It splits up the string delimited by the '_' character and then returns the field denoted -f argument to cut.

Upvotes: 1

Inian
Inian

Reputation: 85800

Use the native bash, regex operator ~ and use the captured groups to store them in variables for using it later.

variable="sample_data_2017_01_01_10_22_10.txt"

if [[ $variable =~ ^sample_data_([[:digit:]]{4})_([[:digit:]]{2})_([[:digit:]]{2}).*$ ]]; then
    year="${BASH_REMATCH[1]}"
    month="${BASH_REMATCH[2]}"
    date="${BASH_REMATCH[3]}"
fi

Upvotes: 4

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