This question already has answers here:
What is the meaning of the ${0##...} syntax with variable, braces and hash character in bash?
(4 answers)
Closed 11 months ago.
I found the following code in a shell script, but I am unsure of what the test condition is evaluating for
if test "${SOME_VAR#*$from asdf/qwer}" != "$SOME_VAR"; then
echo "##zxcv[message text='some text.' status='NORMAL']";
fi
The combination #*$ does not mean anything special. There are three special symbols and each of them has its own meaning in this context.
${SOME_VAR#abc} is a parameter expansion. Its value is the value of $SOME_VAR with the shortest prefix that matches abc removed.
In your example, abc is *${from} asdf/qwer. That means anything * followed by the value of variable $from (which is dynamic and replaced when the expression is evaluated), followed by a space and followed by asdf/qwer.
All in all, if the value of $SOME_VAR starts with a string that ends in ${from} asdf/qwer then everything before and including asdf/qwer is removed and the resulting value is passed as the first argument to test.
Type man bash in your terminal to read the documentation of bash or read it online.
Related
This question already has answers here:
Dynamic variable names in Bash
(19 answers)
Closed last year.
I have these variables:
var1=ab
var2=cd
result=${var1}-text-${var2}
ab-text-cd=bingo
I have:
$ echo $result
ab-text-cd
I would like to have:
$ echo $result
bingo
Is it possible and how?
More info:
Var1 and var2 are arguments given to script.
Thanks to #Léa Gris.
I didn't know about "indirect parameter expansion".
"If the first character of PARAMETER is an exclamation point, Bash uses the value of the variable formed from the rest of PARAMETER as the name of the variable."
Solution :
result2=$(echo ${!result1})
You can use eval to achieve this. Be warned though, that using eval is almost always a bad idea, as it has glaring security issues (rooted in its design -- it is meant to execute everything passed to it) and even apart from that, all kinds of things might go wrong when a variable has an unexpected value.
result=${var1}-text-${var2}
eval ${var1}'_text_'${var2}=bingo
echo $ab_text_cd
Also, environment variables cannot have dashes (-) as part of the variable name, so I replaced them by underscores (_) for the example.
This question already has answers here:
Bash expand variable in a variable
(5 answers)
Difference between single and double quotes in Bash
(7 answers)
Closed 1 year ago.
I have a problem that I am not able to solve, i need replace values of text variable in sh by the value of local variables.
Example:
~/Documents$ VERSION=1.0.0
~/Documents$ TEST='This is a test, VERSION: $VERSION'
~/Documents$ echo $TEST
This is a test, VERSION: $VERSION
I would like to transform the TEST string so that it uses the values of the local variables.
I know I could change the ' ' to " " and that would work, but in my problem I get a literal string so I can't just do that.
Edit:
I am trying to convert a literal string into a string with replacement of values by local variables in words beginning with "$", I could do this with a simple regex but I am looking for a better way, I believe that you should achieve this using only simple SH commands.
If TEST must be a literal string, you can use parameter substitution. These are bash docs, but it also works in sh:
${var/Pattern/Replacement}:
First match of Pattern, within var replaced with Replacement.
If Replacement is omitted, then the first match of Pattern is replaced by nothing, that is, deleted.
${var//Pattern/Replacement}: Global replacement. All matches of Pattern, within var replaced with Replacement.
For your given example, it would be ${TEST/\$VERSION/$VERSION}:
$ VERSION=1.0.0
$ TEST='This is a test, VERSION: $VERSION'
$ echo "${TEST/\$VERSION/$VERSION}"
This is a test, VERSION: 1.0.0
The first dollar sign is escaped so the Pattern is \$VERSION which becomes the literal string "$VERSION". Then its Replacement is $VERSION which gets interpolated as "1.0.0".
This question already has answers here:
Quoting vs not quoting the variable on the RHS of a variable assignment
(5 answers)
Closed 4 years ago.
Are the quotes in the below example necessary or superfluous. And why?
#!/bin/bash
arg1="$1"
arg2="$2"
How do you explain the fact when $1 is 123 echo abc, the first assignment is not interpreted as:
arg1=123 echo abc
which is a normal command (echo) call with argument abc and an environment variable (arg) passed to the execution.
From section 2.9.1 of the POSIX shell syntax specification:
Each variable assignment shall be expanded for tilde expansion, parameter expansion, command substitution, arithmetic expansion, and quote removal prior to assigning the value.
String-splitting and globbing (the steps which double quotes suppress) are not in this list.
Thus, the quotes are superfluous -- not just for assignments where the right-and side refers to a positional parameter, but for all assignments barring those where (1) the behavior of single-quoted, not double-quoted, strings are desired; or (2) whitespace or other content in the value would be otherwise parsed as syntactic rather than literal.
(Note that the decision on how to parse a command -- thus, whether it is an assignment, a simple command, a compound command, or something else -- takes place before parameter expansions; thus, var=$1 is determined to be an assignment before the value of $1 is ever considered! Were this untrue, such that data could silently become syntax, it would be far more difficult -- if not impossible -- to write secure code handling untrusted data in bash).
This question already has answers here:
I just assigned a variable, but echo $variable shows something else
(7 answers)
Closed 7 years ago.
i want to read whole file in a variable.
for example my file is.
file name : Q.txt
My name is Naeem Rehmat.
I am a student of FAST university.
Now i am in 4th semester.
I am learning bash script.
I have this problem.
code:
text=$(cat Q.txt)
echo $text
out put should be like this:
My name is Naeem Rehmat.
I am a student of FAST university.
Now i am in 4th semester.
I am learning bash script.
I have this problem.
Presumably the problem is that your whitespace is incorrect. Use double quotes:
echo "$text"
When you write echo $text without quotes, bash evaluates the string and performs what is known as "field splitting" or "word splitting" before generating the command. To simplify the case, suppose text is the string foo bar. Bash splits that into two "words" and passes "foo" as the first argument to echo, and "bar" as the second. If you use quotes, bash passes only one argument to echo, and that argument contains multiple spaces which echo will print.
Note that it is probably good style to also use quotes in the assignment of text (ie, text="$(cat Q.txt)"), although it is not necessary since field splitting does not occur in a variable assignment.
This question already has answers here:
explain the linux regex for getting filename
(2 answers)
Closed 8 years ago.
I am tring to understand the bash script.
I am seeing ##* / expression with bash variable.
i.e ${foo##*/}
Can someone please tell me why we use that expression?
It's called "Parameter expansion". The variable $foo is searched for a substring matching the pattern */ (i.e. anything up to a slash) from the beginning (#), and what remains in the variable is returned. Doubling the #-sign makes the matching greedy, i.e. it tries to find the longest possible match.