This question already has answers here:
Dynamic variable names in Bash
(19 answers)
Closed 6 years ago.
I was able to do the following in batch, but for the life of me, I cannot figure out how to do this in bash and could use some help. Basically, I used a for loop and delayed expansion to set variables as the for loop iterated through an array. It looked something like this:
FOR /L %%A in (1,1,10) DO (
SET someVar = !inputVar[%%A]!
)
The brackets are merely for clarity.
I now have a similar problem in bash, but cannot figure out how "delayed expansion" (if that's even what it is called in bash) works:
for (( a=1; a<=10; a++ )); do
VAR${!a}= some other thing
done
Am I completely off base here?
Update:
So it seems that I was completely off base and #muru's hint of the XY problem made me relook at what I was doing. The easy solution to my real question is this:
readarray -t array < /filepath
I can now easily use the needed lines.
I think, that eval could help in this case. Not sure, if it's the best option, but could work.
INPUT_VAR=(fish cat elephant)
SOME_VAR=
for i in `seq 0 3`;do
SOME_VAR[$i]='${INPUT_VAR['"$i"']}'
done
echo "${SOME_VAR[2]}" # ${INPUT_VAR[2]}
eval echo "${SOME_VAR[2]}" # elephant
Nice eval explanation:
eval command in Bash and its typical uses
Working with arrays in bash, would be helpful too:
http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_10_02.html
Note, that arrays are supported only at new version of bashs.
Related
This question already has answers here:
Backticks vs braces in Bash
(3 answers)
Brackets ${}, $(), $[] difference and usage in bash
(1 answer)
Closed 4 years ago.
I have two questions and could use some help understanding them.
What is the difference between ${} and $()? I understand that ()
means running command in separate shell and placing $ means passing
the value to variable. Can someone help me in understanding
this? Please correct me if I am wrong.
If we can use for ((i=0;i<10;i++)); do echo $i; done and it works fine then why can't I use it as while ((i=0;i<10;i++)); do echo $i; done? What is the difference in execution cycle for both?
The syntax is token-level, so the meaning of the dollar sign depends on the token it's in. The expression $(command) is a modern synonym for `command` which stands for command substitution; it means run command and put its output here. So
echo "Today is $(date). A fine day."
will run the date command and include its output in the argument to echo. The parentheses are unrelated to the syntax for running a command in a subshell, although they have something in common (the command substitution also runs in a separate subshell).
By contrast, ${variable} is just a disambiguation mechanism, so you can say ${var}text when you mean the contents of the variable var, followed by text (as opposed to $vartext which means the contents of the variable vartext).
The while loop expects a single argument which should evaluate to true or false (or actually multiple, where the last one's truth value is examined -- thanks Jonathan Leffler for pointing this out); when it's false, the loop is no longer executed. The for loop iterates over a list of items and binds each to a loop variable in turn; the syntax you refer to is one (rather generalized) way to express a loop over a range of arithmetic values.
A for loop like that can be rephrased as a while loop. The expression
for ((init; check; step)); do
body
done
is equivalent to
init
while check; do
body
step
done
It makes sense to keep all the loop control in one place for legibility; but as you can see when it's expressed like this, the for loop does quite a bit more than the while loop.
Of course, this syntax is Bash-specific; classic Bourne shell only has
for variable in token1 token2 ...; do
(Somewhat more elegantly, you could avoid the echo in the first example as long as you are sure that your argument string doesn't contain any % format codes:
date +'Today is %c. A fine day.'
Avoiding a process where you can is an important consideration, even though it doesn't make a lot of difference in this isolated example.)
$() means: "first evaluate this, and then evaluate the rest of the line".
Ex :
echo $(pwd)/myFile.txt
will be interpreted as
echo /my/path/myFile.txt
On the other hand ${} expands a variable.
Ex:
MY_VAR=toto
echo ${MY_VAR}/myFile.txt
will be interpreted as
echo toto/myFile.txt
Why can't I use it as bash$ while ((i=0;i<10;i++)); do echo $i; done
I'm afraid the answer is just that the bash syntax for while just isn't the same as the syntax for for.
your understanding is right. For detailed info on {} see bash ref - parameter expansion
'for' and 'while' have different syntax and offer different styles of programmer control for an iteration. Most non-asm languages offer a similar syntax.
With while, you would probably write i=0; while [ $i -lt 10 ]; do echo $i; i=$(( i + 1 )); done in essence manage everything about the iteration yourself
This question already has answers here:
How to use a variable's value as another variable's name in bash [duplicate]
(6 answers)
Closed 5 years ago.
Let's say I have a variable's name stored in another variable:
myvar=123
varname=myvar
Now, I'd like to get 123 by just using $varname variable.
Is there a direct way for that? I found no such bash builtin for lookup by name, so came up with this:
function var { v="\$$1"; eval "echo "$v; }
so
var $varname # gives 123
Which doesn't look too bad in the end, but I'm wondering if I missed something more obvious.
From the man page of bash:
${!varname}
If the first character of parameter is an exclamation point, a level of
variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable;
this variable is then expanded and that value is used in the rest of
the substitution, rather than the value of parameter itself. This is
known as indirect expansion.
There isn't a direct Posix-conforming syntax, only a bashism. I usually do this:
eval t="\$$varname"
This will work on any Posix shell, including those systems where bash is the login shell and /bin/sh is something smaller and faster like ash. I like bash and use it for my login shell but I avoid bashisms in command files.
Note: One problem with writing bash-specific scripts is that even if you can count on bash being installed, it could be anywhere on the path. It might be a good idea in that case to use the fully general /usr/bin/env shebang style, but note that this is still not 100% portable and has security issues.
${!varname} should do the trick
$ var="content"
$ myvar=var
$ echo ${!myvar}
content
I usually look at Advance Bash-Scripting Guide when I need to freshen up my Bash skills.
Regarding your question look at Indirect References
Notation is:
Version < 2
\$$var
Version >= 2
${!varname}
# bmuSetIndirectVar()
# TO DOUBLE CHECK THIS COMMENT AND DEMO
# This function is an helper to read indirect variables.
# i.e. get the content of a variable whose name is saved
# within an other variable. Like:
# MYDIR="/tmp"
# WHICHDIR="MYDIR"
# bmuSetIndirectVar "WHICHDIR" "$MYDIR"
#
bmuSetIndirectVar(){
tmpVarName=$1
locVarName=$1
extVarName=$2
#echo "debug Ind Input >$1< >$2<"
eval tmpVarName=\$$extVarName
#echo "debug Ind Output >$tmpVarName< >$extVarName<"
export $locVarName="${tmpVarName}"
}
I am currently using this little function. I am not fully happy with it, and I have seen different solutions on the web (if I could recall I would write them here), but it seems to work. Within these few lines there is already some redundancy and extra data but it was helpful for debugging.
If you want to see it in place, i.e. where I am using it, check:
https://github.com/mariotti/bmu/blob/master/bin/backmeup.shellfunctions.sh
Of course it is not the best solution, but made me going on with the work, in
the hope I can replace it with something a bit more general soon.
This question already has an answer here:
Expanding a bash array only gives the first element
(1 answer)
Closed 3 years ago.
I'm very new to shell script. I'm learning the basic of it. My very simple for loop is not working. It always stops at the first iteration. I already follow the document to create an array variable and using for loop with super simple code.
#!/bin/bash
LIST=()
LIST+=('aaa')
LIST+=('bbb')
LIST+=('ccc')
for i in $LIST
do
echo '----------'$i'----------'
done
It only show 'aaa' then stop the loop. I really have no idea. Please help.
$LIST expands to the first element in array LIST, it's basically the same thing as ${LIST[0]}. You need to use ${LIST[#]} in double-quotes to get each element as a separate word, like:
#!/bin/bash
LIST=()
LIST+=('aaa')
LIST+=('bbb')
LIST+=('ccc')
for i in "${LIST[#]}"
do
echo '----------'"$i"'----------'
done
c.f. Bash Reference Manual ยง Arrays
This question already has answers here:
A variable modified inside a while loop is not remembered
(8 answers)
Closed 6 years ago.
I want to fill an associative array in bash in somewhat non-trivial setup. I have a pipeline of commands to produce the required input for the array.
Here is a minimal/toy example:
declare -A mapping
seq 10 | while read i; do
key="key_$i"
val="val_$i"
echo "mapping[$key]=$val"
mapping["${key}"]="${val}"
done
echo "${mapping["key_1"]}"
echo "${mapping["key_2"]}"
In this example mapping is changed inside while, but these changes do not propagate into the global namespace. I think this is because while works inside a separate subshell, thus namespaces have diverged.
In order to avoid (what I suggest) the problem with subshells, I came up with the following:
declare -A mapping
while read i; do
key="key_$i"
val="val_$i"
echo "mapping[$key]=$val"
mapping["${key}"]="${val}"
done < <(seq 10)
echo "${mapping["key_1"]}"
echo "${mapping["key_2"]}"
Thus, the generation part explicitly goes into a subshell, while the while loop left at the top-level alone. The construction works.
My questions are: is there any better way to accomplish my goal? And, is my suggestion about subshells correct? If so, why bash uses a subshell in the first case, but not in the second?
EDIT: after little more digging, the question mostly is a duplicate of this one. A good list of options to handle the issue could be found at http://mywiki.wooledge.org/BashFAQ/024
Not sure if this is a better way than your second code snippet, but a way to solve the first one is to use sub shell { ... } right after the pipe:
declare -A mapping
seq 10 | {
while read i; do
key="key_$i"
val="val_$i"
echo "mapping[$key]=$val"
mapping["${key}"]="${val}"
done
echo "${mapping["key_1"]}"
echo "${mapping["key_2"]}"
}
This question already has answers here:
How to use a variable's value as another variable's name in bash [duplicate]
(6 answers)
Closed 5 years ago.
Let's say I have a variable's name stored in another variable:
myvar=123
varname=myvar
Now, I'd like to get 123 by just using $varname variable.
Is there a direct way for that? I found no such bash builtin for lookup by name, so came up with this:
function var { v="\$$1"; eval "echo "$v; }
so
var $varname # gives 123
Which doesn't look too bad in the end, but I'm wondering if I missed something more obvious.
From the man page of bash:
${!varname}
If the first character of parameter is an exclamation point, a level of
variable indirection is introduced. Bash uses the value of the variable formed from the rest of parameter as the name of the variable;
this variable is then expanded and that value is used in the rest of
the substitution, rather than the value of parameter itself. This is
known as indirect expansion.
There isn't a direct Posix-conforming syntax, only a bashism. I usually do this:
eval t="\$$varname"
This will work on any Posix shell, including those systems where bash is the login shell and /bin/sh is something smaller and faster like ash. I like bash and use it for my login shell but I avoid bashisms in command files.
Note: One problem with writing bash-specific scripts is that even if you can count on bash being installed, it could be anywhere on the path. It might be a good idea in that case to use the fully general /usr/bin/env shebang style, but note that this is still not 100% portable and has security issues.
${!varname} should do the trick
$ var="content"
$ myvar=var
$ echo ${!myvar}
content
I usually look at Advance Bash-Scripting Guide when I need to freshen up my Bash skills.
Regarding your question look at Indirect References
Notation is:
Version < 2
\$$var
Version >= 2
${!varname}
# bmuSetIndirectVar()
# TO DOUBLE CHECK THIS COMMENT AND DEMO
# This function is an helper to read indirect variables.
# i.e. get the content of a variable whose name is saved
# within an other variable. Like:
# MYDIR="/tmp"
# WHICHDIR="MYDIR"
# bmuSetIndirectVar "WHICHDIR" "$MYDIR"
#
bmuSetIndirectVar(){
tmpVarName=$1
locVarName=$1
extVarName=$2
#echo "debug Ind Input >$1< >$2<"
eval tmpVarName=\$$extVarName
#echo "debug Ind Output >$tmpVarName< >$extVarName<"
export $locVarName="${tmpVarName}"
}
I am currently using this little function. I am not fully happy with it, and I have seen different solutions on the web (if I could recall I would write them here), but it seems to work. Within these few lines there is already some redundancy and extra data but it was helpful for debugging.
If you want to see it in place, i.e. where I am using it, check:
https://github.com/mariotti/bmu/blob/master/bin/backmeup.shellfunctions.sh
Of course it is not the best solution, but made me going on with the work, in
the hope I can replace it with something a bit more general soon.