No. of underscores in a file name

Shell Programming and Scripting
1

Hi

I want to filter my files which has more than 1 underscores in it.how can i achieve this.
Ex: if my file name is a_b_c, a_b.
my command should result only a_b_c

Thanks
Pracheth

2

Try:

find . -name "*_*_*"
1 Like
3

Hi
Thank you for your quick response.
Adding to that..how can i find whether a particular variable has two underscores in it.
suppose var=a_b_c
i want to know if var has two underscores or not

4 
echo "$var" | grep ".*_.*_.*" > /dev/null && echo "$var has two or more underscores"
5

A longhand version using an OSX 10.7.5 terminal...

Last login: Fri Sep 13 12:52:40 on ttys000
AMIGA:barrywalker~> text="a_b_c"; count=0; for n in $( seq 0 1 ${#text} ); do if [ "${text:$n:1}" == "_" ]; then count=$[ ( $count + 1 ) ]; fi; done; echo""; echo "Underscores = $count"

Underscores = 2
AMIGA:barrywalker~> _
6

If your shell supports substring expansion ( ${text:$n:1} ), then you can probably replace all of that with a much simpler, pattern substitution alternative:

u=${text//[!_]/}
echo Underscores = ${#u}

Regards,
Alister

1 Like
7

Hi alister...

Thanks for that one...

I read it that the OP wanted a number that can be used for detecting further in a script.
Yours shortening works except I changed your "u" variable for count, but...

Last login: Fri Sep 13 19:30:46 on ttys000
AMIGA:barrywalker~> text="a_b_c"
AMIGA:barrywalker~> count=${text//[!_]/}
AMIGA:barrywalker~> echo "$count"
__
AMIGA:barrywalker~> echo ${#count}
2
AMIGA:barrywalker~> count=$(echo -n ${#count})
AMIGA:barrywalker~> echo "$count"
2
AMIGA:barrywalker~> _

Re-using the "count" variable...

count=$(echo -n ${#count})

...keeps the variables to two including the "text" test variable otherwise there would have to be a third variable. The "count" variable is now a number...

8

There are two threads started by this user with for the same problem and both threads have several responses. You may be interested to see the comments there: How to find no of underscores in a variable?

9

Just out of curiosity, why:

count=$(echo -n ${#count})

instead of:

count=${#count}

The echo is not needed, and the -n option to echo (which isn't needed for this example even if you use echo) won't do what you want on most UNIX systems.

1 Like
10

Hi Don Cragun...

I live and learn. Thanks for correcting me.
I chose the echo method because I knew it worked and why. I did not know about your shortened version. That is now logged in the _old_ grey matter. ;o)
As for the "-n" I was thinking I was creating a file, my error. I will leave my post as is as you have already pointed it out and shown a better alternative anyhow...

However my variable(s) point still holds true...

Thanks a lot...

11

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