Remove first character of a string in Bash

I need to calculate md5sum of one string (pathfile) per line in my ls dump, directory_listing_file:

./r/g4/f1.JPG ./r/g4/f2.JPG ./r/g4/f3.JPG ./r/g4/f4.JPG 

But that md5sum should be calculated without the initial dot. I've written a simple script:

while read line do echo $line | exec 'md5sum' done ./g.sh < directory_listnitg.txt 

How do I remove the first dot from each line?

8 Answers

myString="${myString:1}" 

Starting at character number 1 of myString (character 0 being the left-most character) return the remainder of the string. The "s allow for spaces in the string. For more information on that aspect look at $IFS.

You can pipe it to

cut -c2- 

Which gives you

while read line do echo $line | cut -c2- | md5sum done ./g.sh < directory_listnitg.txt 

remove first n characters from a line or string

#method1) using bash

 str="price: 9832.3" echo "${str:7}" 

#method2) using cut

 str="price: 9832.3" cut -c8- <<< $str 

#method3) using sed

 str="price: 9832.3" cut -c8- <<< $str 

#method4) using awk

 str="price: 9832.3" awk '{gsub(/^.{7}/,"");}1' <<< $str 

There ia a very easy way to achieve this:

Suppose that we don't want the prefix "i-" from the variable

$ ROLE_TAG=role $ INSTANCE_ID=i-123456789 

You just need to add '#'+[your_exclusion_pattern], e.g:

$ MYHOSTNAME="${ROLE_TAG}-${INSTANCE_ID#i-}" 

$ echo $MYHOSTNAME role-123456789 

Set the field separator to the path separator and read everything except the stuff before the first slash into $name:

while IFS=/ read junk name do echo $name done < directory_listing.txt 

Different approach, using sed, which has the benefit that it can handle input that doesn't start with a dot. Also, you won't run into problems with echo appending a newline to the output, which will cause md5sum to report bogus result.

#!/bin/bash while read line do echo -n $line | sed 's/^.//' | md5sum done < input 

compare these:

$ echo "a" | md5sum 60b725f10c9c85c70d97880dfe8191b3 - $ echo -n "a" | md5sum 0cc175b9c0f1b6a831c399e269772661 - 

You can do the entire thing like this:

% sh -c `sed 's@^.\(.*\)@md5sum \1@' <./dirlist.txt` 

Really, I'm thinking you can make this a lot more efficient, but I don't know what is generating your list. If you can pipe it from that, or run that command through a heredoc to keep its output sane, you can do this whole job streamed, probably.

EDIT:

OK, you say it's from an "ls dump." Well, here's something a little flexible:

% ls_dump() { > sed 's@^.\(.*\)$@md5sum \1@' <<_EOF_ | sh -s >> `ls ${@}` >> _EOF_ > } % ls_dump -all -args -you /would/normally/give/ls <desired output> 

I think this calls only a single subshell in total. It should be pretty good, but in my opinion, find ... -exec md5sum {} ... + is probably safer, faster, and all you need.

EDIT2:

OK, so now I will actually answer the question. To remove the first character of a string in any POSIX compatible shell you need only look to parameter expansion like:

${string#?} 

-Mike

Or like this

myString="${myString/.}" 

Testing on Ubuntu 18.04.4 LTS, bash 4.4.20:

$ lsb_release -a No LSB modules are available. Distributor ID: Ubuntu Description: Ubuntu 18.04.4 LTS Release: 18.04 Codename: bionic $ echo $BASH_VERSION 4.4.20(1)-release $ myString="./r/g4/f1.JPG" $ myString="${myString/.}" $ echo $myString /r/g4/f1.JPG