Bash: Getting first digit of a string


 
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# 1  
Old 07-17-2014
Bash: Getting first digit of a string

If i'm given a string like "abc-def-1.2.3", how would I return "1"? I'm new to scripting and got stumped on this problem. Thanks in advance!
# 2  
Old 07-17-2014
Is this a homework assignment?
# 3  
Old 07-17-2014
Nope, it's something that I was interested in doing for a small side project. The 123's represent version numbers but I can't get it to just print out the first digit. I'm off school for now haha
# 4  
Old 07-17-2014
Is that string in a shell variable? Then use bash's parameter expansion/Remove matching prefix/suffix pattern:
Code:
A="abc-def-1.2.3"
echo ${A##*-}
1.2.3
B=${A##*-}
echo ${B%%.*}
1

See "man bash"
This User Gave Thanks to RudiC For This Post:
# 5  
Old 07-17-2014
I am sure sed is the tool to use, but just an idea - tr using character classes to delete al non-numerical characters, such as :

Code:
$ echo abc123 | tr -d [:alpha:]
123
$ echo abc123 | tr -d [:alpha:] | cut -c1
1

This User Gave Thanks to migurus For This Post:
# 6  
Old 07-17-2014
hi,

like RudiC using parameter expansion, but not format dependant:
Code:
$ string="abc-def-1.2.3"
$ digits=${string//[^[:digit:]]/}
$ echo ${digits::1} 
1

This User Gave Thanks to daPeach For This Post:
# 7  
Old 07-17-2014
Yes it is. Thanks a lot! Exactly what I was looking for.

Edit: Thanks for all the answers!
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