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开发者_运维知识库. I've written a simple script:
while read line
do
echo $line | exec 'md5sum'
done
./g.sh < directory_listnitg.txt
How do I remove the first dot from each line?
myString="${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 $line | cut -c2- | md5sum
done
./g.sh < directory_listnitg.txt
remove first n characters from a line or string
method 1) using bash
str="price: 9832.3"
echo "${str:7}"
method 2) using cut
str="price: 9832.3"
cut -c8- <<< $str
method 3) using sed
str="price: 9832.3"
sed 's/^.\{7\}//' <<< $str
method 4) 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 -
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
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
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