I'm trying to understand a bash script whose first four lines are:
#!/bin/sh
SCRIPT="`basename $0 | sed 's/\..*$//'`"
CONFIG=${1:-$HOME/.$SCRIPT}
DIR=${2:-$HOME/Documents}
I understand that the last two lines are doing parameter substitution on paths input as script arguments 1 and 2, but I've been unable to figure out how this works (e.g. here). What does the ":-" part mean? Sorry for the newbie question.
From man bash:
${parameter:-word}
Use Default Values. If parameter is unset or null, the expansion of word is substituted. Other‐
wise, the value of parameter is substituted.
Very easy to find, with man bash, and then /:-. The slash introduces a search, and :- is just the content to search for. Else, searching in bash can get very boring, because it is huge, but here it is the first hit.
Related
To rephrase - I want to use Bash command substitution and string substitution in the same line.
My actual commands are longer, but the ridiculous use of echo here is just a "substitution" for shortness and acts the same - with same errors ;)
I know we can use a Bash command to produce it's output string as a parameter for another command like this:
echo "$(echo "aahahah</ddd>")"
aahahah</ddd>
I also know we can remove last known part of a string like this:
var="aahahah</ddd>"; echo "${var%</ddd>}"
aahahah
I am trying to write a command where one command gives a string output, where I want to remove last part, which is known.
echo "${$(echo "aahahah</ddd>")%</ddd>}"
-bash: ${$(echo "aahahah</ddd>")%</ddd>}: bad substitution
It might be the order of things happening or substitution only works on variables or hardcoded strings. But I suspect just me missing something and it is possible.
How do I make it work?
Why doesn't it work?
When a dollar sign as in $word or equivalently ${word} is used, it asks for word's content. This is called parameter expansion, as per man bash.
You may write var="aahahah</ddd>"; echo "${var%</ddd>}": That expands var and performs a special suffix operation before returning the value.
However, you may not write echo "${$(echo "aahahah</ddd>")%</ddd>}" because there is nothing to expand once $(echo "aahahah</ddd>") is evaluated.
From man bash (my emphasis):
${parameter%word}
Remove matching suffix pattern. The word is expanded to produce a
pattern just as in pathname expansion. If the pattern
matches a trailing portion of the expanded value of parameter, then
the result of the expansion is the expanded value of parameter
with the shortest matching pattern (the ''%'' case) or the longest matching pattern (the ''%%'' case) deleted.
Combine your commands like this
var=$(echo "aahahah</ddd>")
echo ${var/'</ddd>'}
I'm looking for a one liner to replace any character in a variable string at a variable position with a variable substitute. I came up with this working solution:
echo "$string" | sed "s/./${replacement}/${position}"
An example usage:
string=aaaaa
replacement=b
position=3
echo "$string" | sed "s/./${replacement}/${position}"
aabaa
Unfortunately, when I run shellcheck with a script which contains my current solution it tells me:
SC2001: See if you can use ${variable//search/replace} instead.
I'd like to use parameter expansion like it's suggesting instead of piping to sed, but I'm unclear as to the proper formatting when using a position variable. The official documentation doesn't seem to discuss positioning within strings at all.
Is this possible?
Bash doesn't have a general-case replacement for all sed facilities (the shellcheck wiki page for warning SC2001 acknowledges as much), but in some specific scenarios -- including the case posed -- parameter expansions can be combined to achieve the desired effect:
string=aaaaa
replacement=b
position=3
echo "${string:0:$(( position - 1 ))}${replacement}${string:position}"
Here, we're splitting the value up into substrings: ${string:0:$(( position - 1 ))} is the text preceding the content to be replaced, and ${string:position} is the text following that point.
Does anyone know the bash construct where you specify the delimeter and you get in two variables ( $# or $! or something like that I think ) the values?
For example:
--option=false should be specified with '=' delimeter and in one variable there will be the word 'option' whereas in the other variable the word 'false' is stored. Anyone know?
P.S.: No sed, awk or IFS solutions please. I am aware of them but I am requesting the specific bash construct :)
As found in the Bash man page, you can use Parameter Expansion to solve your problems.
# split first argument on equal sign (left=right)
left=${1%%=*}
right=${1#*=}
I believe the GNU getopt program is appropriate here (if available, of course).
It supports long options.
A contrived example... given
FOO="/foo/bar/baz"
this works (in bash)
BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1} # result is BAZ="b"
this doesn't
BAZ=${$(basename $FOO):0:1} # result is bad substitution
My question is which rule causes this [subshell substitution] to evaluate incorrectly? And what is the correct way, if any, to do this in 1 hop?
First off, note that when you say this:
BAR=$(basename $FOO) # result is BAR="baz"
BAZ=${BAR:0:1} # result is BAZ="b"
the first bit in the construct for BAZ is BAR and not the value that you want to take the first character of. So even if bash allowed variable names to contain arbitrary characters your result in the second expression wouldn't be what you want.
However, as to the rule that's preventing this, allow me to quote from the bash man page:
DEFINITIONS
The following definitions are used throughout the rest of this docu‐
ment.
blank A space or tab.
word A sequence of characters considered as a single unit by the
shell. Also known as a token.
name A word consisting only of alphanumeric characters and under‐
scores, and beginning with an alphabetic character or an under‐
score. Also referred to as an identifier.
Then a bit later:
PARAMETERS
A parameter is an entity that stores values. It can be a name, a num‐
ber, or one of the special characters listed below under Special Param‐
eters. A variable is a parameter denoted by a name. A variable has a
value and zero or more attributes. Attributes are assigned using the
declare builtin command (see declare below in SHELL BUILTIN COMMANDS).
And later when it defines the syntax you're asking about:
${parameter:offset:length}
Substring Expansion. Expands to up to length characters of
parameter starting at the character specified by offset.
So the rules as articulated in the manpage say that the ${foo:x:y} construct must have a parameter as the first part, and that a parameter can only be a name, a number, or one of the few special parameter characters. $(basename $FOO) is not one of the allowed possibilities for a parameter.
As for a way to do this in one assignment, use a pipe to other commands as mentioned in other responses.
Modified forms of parameter substitution such as ${parameter#word} can only modify a parameter, not an arbitrary word.
In this case, you might pipe the output of basename to a dd command, like
BAR=$(basename -- "$FOO" | dd bs=1 count=1 2>/dev/null)
(If you want a higher count, increase count and not bs, otherwise you may get fewer bytes than requested.)
In the general case, there is no way to do things like this in one assignment.
It fails because ${BAR:0:1} is a variable expansion. Bash expects to see a variable name after ${, not a value.
I'm not aware of a way to do it in a single expression.
As others have said, the first parameter of ${} needs to be a variable name. But you can use another subshell to approximate what you're trying to do.
Instead of:
BAZ=${$(basename $FOO):0:1} # result is bad substitution
Use:
BAZ=$(_TMP=$(basename $FOO); echo ${_TMP:0:1}) # this works
A contrived solution for your contrived example:
BAZ=$(expr $(basename $FOO) : '\(.\)')
as in
$ FOO=/abc/def/ghi/jkl
$ BAZ=$(expr $(basename $FOO) : '\(.\)')
$ echo $BAZ
j
${string:0:1},string must be a variable name
for example:
FOO="/foo/bar/baz"
baz="foo"
BAZ=eval echo '${'"$(basename $FOO)"':0:1}'
echo $BAZ
the result is 'f'
I am writing a shell (bash) script and I'm trying to figure out an easy way to accomplish a simple task.
I have some string in a variable.
I don't know if this is relevant, but it can contain spaces, newlines, because actually this string is the content of a whole text file.
I want to replace the last occurence of a certain substring with something else.
Perhaps I could use a regexp for that, but there are two moments that confuse me:
I need to match from the end, not from the start
the substring that I want to scan for is fixed, not variable.
for truncating at the start: ${var#pattern}
truncating at the end ${var%pattern}
${var/pattern/repl} for general replacement
the patterns are 'filename' style expansion, and the last one can be prefixed with # or % to match only at the start or end (respectively)
it's all in the (long) bash manpage. check the "Parameter Expansion" chapter.
amn expression like this
s/match string here$/new string/
should do the trick - s is for sustitute, / break up the command, and the $ is the end of line marker. You can try this in vi to see if it does what you need.
I would look up the man pages for awk or sed.
Javier's answer is shell specific and won't work in all shells.
The sed answers that MrTelly and epochwolf alluded to are incomplete and should look something like this:
MyString="stuff ttto be edittted"
NewString=`echo $MyString | sed -e 's/\(.*\)ttt\(.*\)/\1xxx\2/'`
The reason this works without having to use the $ to mark the end is that the first '.*' is greedy and will attempt to gather up as much as possible while allowing the rest of the regular expression to be true.
This sed command should work fine in any shell context used.
Usually when I get stuck with Sed I use this page,
http://sed.sourceforge.net/sed1line.txt