Bash从字符串中剪切第一个和/或最后一个字符，但前提是它是某个特定字符

Question

In bash I need to shave a first and/or last character from string, but only if it is a certain character.

If I have | I need

/foo/bar/hah/   =>   foo/bar/hah

foo/bar/hah     =>   foo/bar/hah

You can downvote me for not listing everything I've tried. But the fact is I've tried at least 35 differents sed strings and bash character stuff, many of which was from stack overflow. I simply cannot get this to happen.

Answer 1

In pure bash :

$ var=/foo/bar/hah/
$ var=${var%/}
$ echo ${var#/}
foo/bar/hah
$ 

Check bash parameter expansion

or with sed :

$ sed -r 's@(^/|/$)@@g' file

Answer 2

what's the problem with the simple one?

sed "s/^\///;s/\/$//"

Output is

foo/bar/hah
foo/bar/hah

Answer 3

这个怎么样：

echo "$x" | sed -e 's:^/::' -e 's:/$::'

Answer 4

Further to @sputnick's answer and from this answer , here's a function that would do it:

STR="/foo/bar/etc/";
STRB="foo/bar/etc";

function trimslashes {
    STR="$1"
    STR=${STR#"/"}
    STR=${STR%"/"}
    echo "$STR"
}

trimslashes $STR
trimslashes $STRB

# foo/bar/etc
# foo/bar/etc

Answer 5

 echo '/foo/bar/hah/' | sed 's#^/##' | sed 's#/$##'

Answer 6

assuming the / character is the only one you're trying to remove, then sed -E 's_^[/](.*)_\\1_' should do the job:

$ echo "$var1"; echo "$var2"
/foo/bar/hah
foo/bar/hah


$ echo "$var1" | sed -E 's_^[/](.*)_\1_'
foo/bar/hah


$ echo "$var2" | sed -E 's_^[/](.*)_\1_'
foo/bar/hah

if you also need to replace other characters at the start of the line, add it to the [/] class. for example, if you need to replace / or - , it would be sed -E 's_^[/-](.*)_\\1_'

Answer 7

Here is an awk version:

echo "/foo/bar/hah/" | awk '{gsub(/^\/|\/$/,"")}1'
foo/bar/hah