Following some issues with scp (it did not like the presence of the bash bind command in my .bashrc file, apparently), I followed the advice of a clever guy on the Internet (I just cannot find that post right now) that put at the top of its .bashrc file this:
[[ ${-#*i} != ${-} ]] || return
in order to make sure that the bash initialization is NOT executed unless in interactive session.
Now, that works. However, I am not able to figure how it works. Could you enlighten me?
According to this answer, the $- is the current options set for the shell and I know that the ${} is the so-called "substring" syntax for expanding variables.
However, I do not understand the ${-#*i} part. And why $-#*i is not the same as ${-#*i}.
${parameter#word}
${parameter##word}
The word is expanded to produce a pattern just as in filename
expansion. If the pattern matches the
beginning 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. If parameter is ‘#’ or ‘’, the pattern removal
operation is applied to each positional parameter in turn, and the
expansion is the resultant list. If parameter is an array variable
subscripted with ‘#’ or ‘’, the pattern removal operation is applied
to each member of the array in turn, and the expansion is the
resultant list.
Source: http://www.gnu.org/software/bash/manual/html_node/Shell-Parameter-Expansion.html
So basically what happens in ${-#*i} is that *i is expanded, and if it matches the beginning of the value of $-, then the result of the whole expansion is $- with the shortest matching pattern between *i and $- deleted.
Example
VAR="baioasd";
echo ${VAR#*i};
outputs oasd.
In your case
If shell is interactive, $- will contain the letter 'i', so when you strip the variable $- of the pattern *i you will get a string that is different from the original $- ( [[ ${-#*i} != ${-} ]] yelds true).
If shell is not interactive, $- does not contain the letter 'i' so the pattern *i does not match anything in $- and [[ ${-#*i} != $- ]] yelds false, and the return statement is executed.
See this:
To determine within a startup script whether or not Bash is running interactively, test the value of the ‘-’ special parameter. It contains i when the shell is interactive
Your substitution removes the string up to, and including the i and tests if the substituted version is equal to the original string. They will be different if there is i in the ${-}.
Related
To rephrase - I want to use Bash command substitution and string substitution in the same line.
My actual commands are longer, but the ridiculous use of echo here is just a "substitution" for shortness and acts the same - with same errors ;)
I know we can use a Bash command to produce it's output string as a parameter for another command like this:
echo "$(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 ${var/'</ddd>'}
I simply want a function (or just a 1-liner) to find/replace a string inside a variable, and not worry if the variables contain crazy characters.
Pseudo-code:
findReplace () {
#what goes here?
}
myLongVar="some long \crazy/ text my_placeholder bla"
replace="my_placeholder"
replaceWith="I like hamburgers/fries"
myFinalVar=$(findReplace $myLongVar $replace $replaceWith)
All similar questions seem complicated and use files
You can define the function like this:
findReplace1() {
printf "%s" "${1/"$2"/$3}"
}
And then run it like this:
myFinalVar=$(findReplace "$myLongVar" "$replace" "$replaceWith")
Note the double-quotes -- they're very important, because without them bash will split the variables' values into separate words (e.g. "some long \crazy/ text..." -> "some" "long" "\crazy/" "text...") and also try to expand anything that looks like a wildcard into a list of matching filenames. It's ok to leave them off on the right side of an assignment (myFinalVar=...), but that's one of the few places where it's ok. Also, note that within the function I put double-quotes around $2 -- in that case again it's to keep it from being treated as a wildcard pattern, but here it'd a string-match wildcard rather than filenames. Oh, and I used printf "%s" instead of echo because some versions of echo do weird things with strings that contain backslashes and/or start with "-".
And, of course, you can just skip the function and do the replacement directly:
myFinalVar=${myLongVar/"$replace"/$replaceWith}
Try:
myFinalVar=${myLongVar/$replace/$replaceWith}
If your want to replace all occurrences of $replace, not just the first, use:
myFinalVar=${myLongVar//$replace/$replaceWith}
Documentation
From man bash:
${parameter/pattern/string}
Pattern substitution. The pattern is expanded to produce a pattern
just as in pathname expansion. Parameter is expanded and the longest
match of pattern against its value is replaced with
string. If pattern begins with /, all matches of pattern are
replaced with string. Normally only the first match is replaced. If
pattern begins with #, it must match at the beginning of
the expanded value of parameter. If pattern begins with %, it must
match at the end of the expanded value of parameter. If string is
null, matches of pattern are deleted and the / following pattern may
be omitted. If the nocasematch shell option is enabled, the match is
performed without regard to the case of alphabetic
characters. If parameter is # or *, the substitution operation is
applied to each positional parameter in turn, and the
expansion is the resultant list. If parameter is an array variable
subscripted with # or *, the substitution operation is applied to each
member of the array in turn, and the expansion is the
resultant list.
When responding to this comment:
Now I got the the two ":"s are independent, and that's why I couldn't find any document about them. Is the first one needed in this case?
I noticed this paragraph in the spec for the first time:
In the parameter expansions shown previously, use of the <colon> in the format shall result in a test for a parameter that is unset or null; omission of the <colon> shall result in a test for a parameter that is only unset. If parameter is '#' and the colon is omitted, the application shall ensure that word is specified (this is necessary to avoid ambiguity with the string length expansion).
I've seen the matching explanation in the bash reference manual:
When not performing substring expansion, using the form described below (e.g., ‘:-’), Bash tests for a parameter that is unset or null. Omitting the colon results in a test only for a parameter that is unset. Put another way, if the colon is included, the operator tests for both parameter’s existence and that its value is not null; if the colon is omitted, the operator tests only for existence.
before and I understand what the difference is with the colon versions of these expansions.
What confused me just now is this sentence from the spec:
If parameter is '#' and the colon is omitted, the application shall ensure that word is specified (this is necessary to avoid ambiguity with the string length expansion).
I don't understand what ambiguity is possible here if word is unspecified.
None of the expansion sigils are valid in shell variable names so they cannot possibly start a single-character variable name. If they could then using a parameter of # would always be ambiguous without a colon since you could never tell if ${#+foo} meant the length of the variable foo or an alternate expansion on #, etc.
What am I missing here? What ambiguity requires ensuring that word exist? (I mean not having word in this expansion is clearly not useful but that's not the same thing.)
- is also a shell special parameter, whose value is a string indicating which shell options are currently set. For example,
$ echo $-
himBH
${#parameter} is the syntax for the length of a parameter.
$ foo=bar
$ echo ${#foo}
3
The expression ${#-}, therefore is ambiguous: is it the length of the value of $-, or is does it expand to the empty string if $# is empty? (Unlikely, since $# is always an integer and cannot be unset, but syntactically legal.) I interpret the spec to meant that ${#-} should resolve the ambiguity by expanding to the length of $- (which is what most shells seem to do).
This is code
a=''
b=john
c=${a-$b}
echo $c
And the output is empty line
And for similar code where first variable is not initialized
b1=doe
c1=${a1-$b1}
echo $c1
And the output is
doe
I do not understand how bash deals with expanding of variables leading to different results.
There are two variants of the ${var-value} notation, one without a colon, as shown, and one with a colon: ${var:-value}.
The first version, without colon, means 'if $var is set to any value (including an empty string), use it; otherwise, use value instead'.
The second version, with colon, means 'if $var is set to any value except the empty string, use it; otherwise, use value instead'.
This pattern holds for other variable substitutions too, notably:
${var:=value}
if $var is set to any non-empty string, leave it unchanged; otherwise, set $var to value.
${var=value}
if $var is set to any value (including an empty string), leave it unchanged; otherwise, set $var to value.
${var:?message}
if $var is set to any non-empty string, do nothing; otherwise, complain using the given message' (where a default message is supplied if message is itself empty).
${var?message}
if $var is set to any value (including an empty string), do nothing; otherwise, complain using the given message'.
These notations all apply to any POSIX-compatible shell (Bourne, Korn, Bash, and others). You can find the manual for the bash version online — in the section Shell Parameter Expansion. Bash also has a number of non-standard notations, many of which are extremely useful but not necessarily shared with other shells.
This question already has answers here:
What does $# mean in a shell script?
(8 answers)
Closed 6 years ago.
I reckon that the handle $# in a shell script is an array of all arguments given to the script. Is this true?
I ask because I normally use search engines to gather information, but I can't google for $# and I have grown too accustomed to easily getting served everything.
Yes. Please see the man page of Bash (the first thing you go to) under Special Parameters:
Special Parameters
The shell treats several parameters specially. These parameters may only be referenced; assignment to them is not allowed.
* Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, it expands to a single word with the value of each parameter separated by the first character of the IFS special variable. That is, "$*" is equivalent to "$1c$2c...", where c is the first character of the value of the IFS variable. If IFS is unset, the parameters are separated by spaces. If IFS is null, the parameters are joined without intervening separators.
# Expands to the positional parameters, starting from one. When the expansion occurs within double quotes, each parameter expands to a separate word. That is, "$#" is equivalent to "$1" "$2" ... If the double-quoted expansion occurs within a word, the expansion of the first parameter is joined with the beginning part of the original word, and the expansion of the last parameter is joined with the last part of the original word. When there are no positional parameters, "$#" and $# expand to nothing (i.e., they are removed).
Just from reading that I would have never understood that "$#" expands into a list of separate parameters. Whereas, "$*" is one parameter consisting of all the parameters added together.
If it still makes no sense, do this.
Bash special parameters explained with four example shell scripts