在 Bash 中删除字符串的第一个字符
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Remove first character of a string in Bash
提问by JosiP
I need to calculate md5sum of one string (pathfile) per line in my lsdump, directory_listing_file:
我需要计算在我的每行一个字符串（pathfile）的的md5sumls转储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:
但是 md5sum 应该在没有初始点的情况下计算。我写了一个简单的脚本：
while read line
do
echo $line | exec 'md5sum'
done
./g.sh < directory_listnitg.txt
How do I remove the first dot from each line?
如何从每行中删除第一个点？
采纳答案by l0b0
Set the field separator to the path separator and read everything except the stuff before the first slash into $name:
将字段分隔符设置为路径分隔符，并读取除第一个斜杠之前的内容之外的所有内容$name：
while IFS=/ read junk name
do
echo $name
done < directory_listing.txt
回答by LiXCE
myString="${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.
从 myString 的第 1 个字符（字符 0 是最左边的字符）开始返回字符串的其余部分。"s 允许字符串中有空格。有关这方面的更多信息，请查看 $IFS。
回答by fulmicoton
You can pipe it to
你可以用管道把它
cut -c2-
Which gives you
这给你
while read line
do
echo $line | cut -c2- | md5sum
done
./g.sh < directory_listnitg.txt
回答by mikeserv
You can do the entire thing like this:
你可以像这样完成整个事情：
% sh -c `sed 's@^.\(.*\)@md5sum @' <./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 pipeit from that, or run that command through a heredocto keep its output sane, you can do this whole job streamed, probably.
真的，我认为您可以提高效率，但我不知道是什么生成了您的列表。如果您可以pipe这样做，或者通过 a 运行该命令heredoc以保持其输出正常，那么您可能可以通过流式传输完成整个工作。
EDIT:
编辑：
OK, you say it's from an "ls dump." Well, here's something a little flexible:
好吧，你说它来自“ls dump”。好吧，这里有一些灵活的东西：
% ls_dump() {
> sed 's@^.\(.*\)$@md5sum @' <<_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.
我认为这总共只调用了一个子shell。它应该相当不错，但在我看来，find ... -exec md5sum {} ... +它可能更安全、更快，并且满足您的所有需求。
EDIT2:
编辑2：
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:
好的，现在我来实际回答这个问题。要在任何 POSIX 兼容 shell 中删除字符串的第一个字符，您只需要查看如下参数扩展：
${string#?}
-Mike
-麦克风
回答by Fredrik Pihl
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 echoappending a newline to the output, which will cause md5sum to report bogus result.
不同的方法，使用 sed，它的好处是它可以处理不以点开头的输入。此外，您不会遇到echo在输出中附加换行符的问题，这将导致 md5sum 报告虚假结果。
#!/bin/bash
while read line
do
echo -n $line | sed 's/^.//' | md5sum
done < input
compare these:
比较这些：
$ echo "a" | md5sum
60b725f10c9c85c70d97880dfe8191b3 -
$ echo -n "a" | md5sum
0cc175b9c0f1b6a831c399e269772661 -
回答by Rafael Oliveira
There ia a very easy way to achieve this:
有一种非常简单的方法可以实现这一点：
Suppose that we don't want the prefix "i-" from the variable
假设我们不想要变量中的前缀“i-”
$ ROLE_TAG=role
$ INSTANCE_ID=i-123456789
You just need to add '#'+[your_exclusion_pattern], e.g:
您只需要添加 '#'+[your_exclusion_pattern]，例如：
$ MYHOSTNAME="${ROLE_TAG}-${INSTANCE_ID#i-}"
$ echo $MYHOSTNAME
role-123456789
