How do I count the number of dots in a string in BASH? For example
VAR="s454-da4_sd.fs_84-df.f-sds.a_as_d.a-565sd.dasd"
# Variable VAR contains 5 dots
str. count('\. ')
Cut string and display three dots : String Util « Data Types « C# / C Sharp.
You can do it combining grep and wc commands:
echo "string.with.dots." | grep -o "\." | wc -l
Explanation:
grep -o # will return only matching symbols line/by/line
wc -l # will count number of lines produced by grep
Or you can use only grep for that purpose:
echo "string.with.dots." | grep -o "\." | grep -c "\."
VAR="s454-da4_sd.fs_84-df.f-sds.a_as_d.a-565sd.dasd"
echo $VAR | tr -d -c '.' | wc -c
tr -d deletes given characters from the input. -c takes the inverse of given characters. together, this expression deletes non '.' characters and counts the resulting length using wc.
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