This question already has answers here:
Why does shell ignore quoting characters in arguments passed to it through variables? [duplicate]
(3 answers)
Closed 5 years ago.
I wrote some script:
#!/bin/sh
dialog --menu \
"Please select a partition from the following list to use for your \
root (/) Linux partition." 13 70 3 \
"/dev/hda2" "Linux native 30724312K" "/dev/hda4" "Linux native 506047K"
DISKS='"disk1" "50 Gb" "disk2" "100 Gb"'
dialog --menu \
"Please select a partition from the following list to use for your \
root (/) Linux partition." 13 70 3 \
$DISKS
The first call looks good.
The second call looks bad.
I need to transmit information about the drives through the variable. Please tell me why the second version of the call does not work as expected?
Let's ask shellcheck:
$ shellcheck myscript
In myscript line 9:
DISKS='"disk1" "50 Gb" "disk2" "100 Gb"'
^-- SC2089: Quotes/backslashes will be treated literally. Use an array.
In myscript line 14:
$DISKS
^-- SC2090: Quotes/backslashes in this variable will not be respected.
The detailed error page offers an explanation:
Bash does not interpret data as code. Consider almost any other languages, such as Python:
print 1+1 # prints 2
a="1+1"
print a # prints 1+1, not 2
Here, 1+1 is Python syntax for adding numbers. However, passing a literal string containing this expression does not cause Python to interpret it, see the + and produce the calculated result.
Similarly, "My File.txt" is Bash syntax for a single word with a space in it. However, passing a literal string containing this expression does not cause Bash to interpret it, see the quotes and produce the tokenized result.
The solution is to use an array instead, whenever possible.
Ok, let's try that:
#!/bin/bash
DISKS=("disk1" "50 Gb" "disk2" "100 Gb")
dialog --menu \
"Please select a partition from the following list to use for your \
root (/) Linux partition." 13 70 3 \
"${DISKS[#]}"
This works.
However, this is a bash specific solution. If you want it to work for sh, in this case, you can pick a delimiter other than whitespace by setting the Internal Field Separator:
#!/bin/sh
IFS=":" # Split on colons
set -f # Prevent globbing
DISKS='disk1:50 Gb:disk2:100 Gb'
dialog --menu \
"Please select a partition from the following list to use for your \
root (/) Linux partition." 13 70 3 \
$DISKS
This also works as expected. If your script is longer than this, you'd want to set IFS back to its original value.
Related
When inserting a shell script inside a Makefile we have (?) to use a double dollar sign ($$) to make reference to variables. Why is that so?
for number in 1 2 3 4 ; do \
echo $$number ; \
done
As per gnu make official doc:
Variable and function references in recipes have identical syntax and
semantics to references elsewhere in the makefile. They also have the
same quoting rules: if you want a dollar sign to appear in your
recipe, you must double it (‘$$’). For shells like the default shell,
that use dollar signs to introduce variables, it’s important to keep
clear in your mind whether the variable you want to reference is a
make variable (use a single dollar sign) or a shell variable (use two
dollar signs).
So in short:
makefile variable => use a single dollar sign
shell variable => use two dollar signs
$$ is the PID(process ID) of your current process.
$ ps -ef|grep $$
user 208465 200620 0 10:30 pts/4 00:00:00 bash
$ for number in 1 2 3 4 ; do \
echo $$number ; \
done
208465number
208465number
208465number
208465number
Not directly applicable to this example -- except if the code shown is executed via $(shell ...) instead of being a rule:
With secondary expansion enabled, make might also interpret the double dollar itself in the second processing phase, when it occurrs in the prequisites list. (First phase: read file, set variables; second phase: find and invoke dependency targets, execute rules)
This is used to allow dynamically specifying dependency targets, when the variable with the targets name is only available later in the file.
See https://www.gnu.org/software/make/manual/html_node/Secondary-Expansion.html.
This question already has answers here:
How do I set a variable to the output of a command in Bash?
(15 answers)
Bash variable from command with pipes, quotes, etc
(2 answers)
Variable variable assignment error -"command not found"
(1 answer)
Closed 1 year ago.
i have the following two lines in a batch script
iperf_options=" -O 10 -V -i 10 --get-server-output -P " $streams
$iperf_options=$iperf_options $proto
and
$streams = 2
$proto = -u
but when i run this i get the following error.
./bandwidth: line 116: -O: command not found
I am simply trying to wrote a string and then append it to a variable so why does it throw the error on the -O?
I have looked about the web but i jsut seem to find stuff about spaces around the "="
any help greatfully recived.
Thankyou
code block to show error
proto=-u
streams=2
iperf_options=" -O 10 -V -i 10 --get-server-output -P " $streams
$iperf_options=$iperf_options $proto
running this will give this out put
./test
./test: line 3: 2: command not found
./test: line 4: =: command not found
There are two main mistakes here, in a variety of combinations.
Use $ to get the value of a variable, never when setting the variable (or changing its properties):
$var=value # Bad
var=value # Good
var=$othervar # Also good
Spaces are critical delimiters in shell syntax; adding (or removing) them can change the meaning of a command in unexpected ways:
var = value # Runs `var` as a command, passing "=" and "value" as arguments
var=val1 val2 # Runs `val2` as a command, with var=val1 set in its environment
var="val1 val2" # Sets `var1` to `val1 val2`
So, in this command:
iperf_options=" -O 10 -V -i 10 --get-server-output -P " $streams
The space between iperf_options="..." and $streams means that it'll expand $streams and try to run it as a command (with iperf_options set in its environment). You want something like:
iperf_options=" -O 10 -V -i 10 --get-server-output -P $streams"
Here, since $streams is part of the double-quoted string, it'll be expanded (variable expand inside double-quotes, but not in single-quoted), and its value included in the value assigned to iperf_options.
There's actually a third mistake (or at least dubious scripting practice): building lists of options as simple string variables. This works in simple cases, but fails when things get complex. If you're using a shell that supports arrays (e.g. bash, ksh, zsh, etc, but not dash), it's better to use those instead, and store each option/argument as a separate array element, and then expand the array with "${arrayname[#]}" to get all of the elements out intact (yes, all those quotes, braces, brackets, etc are actually needed).
proto="-u" # If this'll always have exactly one value, plain string is ok
streams=2 # Same here
iperf_options=(-O 10 -V -i 10 --get-server-output -P "$streams")
iperf_options=("${iperf_options[#]}" "$proto")
# ...
iperf "${iperf_options[#]}"
Finally, I recommend shellcheck.net to sanity-check your scripts for common mistakes. A warning, though: it won't catch all errors, since it doesn't know your intent. For instance, if it sees var=val1 val2 it'll assume you meant to run val2 as a command and won't flag it as a mistake.
I'm a beginner and not a native english speaker please excuse my clumsiness.
I'm trying to make a linux install script for personal use (and to learn more about linux and bash scripting) but I'm struggling on finding a way to create a disk selection menu :
I wish to make a list witch would look like that :
NAME SIZE DEVICES
sda 256gib intel-ssdx
sdb 1000gib TLxxxxxxxx
nvme0n1 128gib WDxxxxxxxx
So far i've tried to echo fdisk -l and lsblk in text file and use cat to prompt it
Code :
lsblk
Set DiskLayout=("Automatic Install" "Manual Install" "Check pending change" "Quit")
select DiskLayoutopt in "${DiskLayout[#]}"
do
case $DiskLayoutopt in
"Automatic Install")
read Sdsk -p "Select drive"
;;
"Manual Install")
parted -a optimal
;;
"Check pending change")
echo ""
"Quit")
exit 1
;;
*) echo "invalid option $REPLY";;
esac
done
The following code will get your menu:
#!/usr/bin/env bash
disk=()
size=()
name=()
while IFS= read -r -d $'\0' device; do
device=${device/\/dev\//}
disk+=($device)
name+=("`cat "/sys/class/block/$device/device/model"`")
size+=("`cat "/sys/class/block/$device/size"`")
done < <(find "/dev/" -regex '/dev/sd[a-z]\|/dev/vd[a-z]\|/dev/hd[a-z]' -print0)
for i in `seq 0 $((${#disk[#]}-1))`; do
echo -e "${disk[$i]}\t${name[$i]}\t${size[$i]}"
done
This is some tough bash script... Hope you'll learn quick.
Here's some help:
First line is a shebang to tell your system which interpreter is needed for that script. Indeed, this script only works with bash.
Try running with bash myscript.sh on systems that don't work (ie BSD).
variable=() is an array.
Adding something to that array is done by variable+=("my value")
The while loop reads variable device from what it gets from find command
while read device; do
something
done < <(find)
The find command uses a regular expression that says anything like /dev/sdX where X goes from a to z, or anything like /dev/vdX or anything like /dev/hdX (where X still goes from a to z).
The or operator is a pipe | which has to be escaped with an antislash, hence giving \|.
The devices read by the while look look like '/dev/sda' so we need so strip '/dev/' out of it using the following:
device=${device/\/dev\//}
This is a bash substitution which works the following way:
variable="my foo function"
echo ${variable/foo/bar}
This outputs my bar function.
Indeed, we still need to escape / since this is the separator character for the substition, so it becomes \/.
Getting the disk name via
"`cat "/sys/class/block/$device/device/model"`"
cat "/sys/class/block/sda/device/model" gives the disk model.
In order to get the result into a variable, we'll need to quote it with ` sign, eg:
myvar=`cat /var/file`
Last but not least, the for loop part:
for i in seq 0 $((${#disk[#]}-1)); do
echo -e "${disk[$i]}\t${name[$i]}\t${size[$i]}"
done
${#disk[#]} is the number of elements in array disk.
Actually ${#var} is the number of elements in var, which when being a string, is the number of characters. ${var[#]} means all elements of an array.
seq 0 X returns a sequence of 0 to X numbers, in order to construct the for loop.
Using echo -e translates escaped characters into litterals. In our case '\t' become tabs.
Last but not least, showing ${disk[$i]} is disk array value of index $i where $i is an integer.
Btw, bash is quite limited to do these tasks, but really fun to learn in the first place.
Harder tasks might be better accomplished in a higher level scripting language like Python. Anyway, have fun learning bash, it's a life saver in sysadmin's career.
I have a Bash script which generates, stores and modifies values in an array. These values are later used as arguments for a command.
For a MCVE I thought of an arbitrary command bash -c 'echo 0="$0" ; echo 1="$1"' which explains my problem. I will call my command with two arguments -option1=withoutspace and -option2="with space". So it would look like this
> bash -c 'echo 0="$0" ; echo 1="$1"' -option1=withoutspace -option2="with space"
if the call to the command would be typed directly into the shell. It prints
0=-option1=withoutspace
1=-option2=with space
In my Bash script, the arguments are part of an array. However
#!/bin/bash
ARGUMENTS=()
ARGUMENTS+=('-option1=withoutspace')
ARGUMENTS+=('-option2="with space"')
bash -c 'echo 0="$0" ; echo 1="$1"' "${ARGUMENTS[#]}"
prints
0=-option1=withoutspace
1=-option2="with space"
which still shows the double quotes (because they are interpreted literally?). What works is
#!/bin/bash
ARGUMENTS=()
ARGUMENTS+=('-option1=withoutspace')
ARGUMENTS+=('-option2=with space')
bash -c 'echo 0="$0" ; echo 1="$1"' "${ARGUMENTS[#]}"
which prints again
0=-option1=withoutspace
1=-option2=with space
What do I have to change to make ARGUMENTS+=('-option2="with space"') work as well as ARGUMENTS+=('-option2=with space')?
(Maybe it's even entirely wrong to store arguments for a command in an array? I'm open for suggestions.)
Get rid of the single quotes. Write the options exactly as you would on the command line.
ARGUMENTS+=(-option1=withoutspace)
ARGUMENTS+=(-option2="with space")
Note that this is exactly equivalent to your second option:
ARGUMENTS+=('-option1=withoutspace')
ARGUMENTS+=('-option2=with space')
-option2="with space" and '-option2=with space' both evaluate to the same string. They're two ways of writing the same thing.
(Maybe it's even entirely wrong to store arguments for a command in an array? I'm open for suggestions.)
It's the exact right thing to do. Arrays are perfect for this. Using a flat string would be a mistake.
See my previous question on assembling a specific string here.
I was given an answer to that question, but unfortunately the information didn't actually help me accomplish what I was trying to achieve.
Using the info from that post, I have been able to assemble the following set of strings: gnuplot -e "filename='output_N.csv'" 'plot.p' where N is replaced by the string representation of an integer.
The following loop will explain: (Actually, there is probably a better way of doing this loop, which you may want to point out - hopefully the following code won't upset too many people...)
1 #!/bin/bash
2 n=0
3 for f in output_*.csv
4 do
5 FILE="\"filename='output_"$n".csv'\""
6 SCRIPT="'plot.p'"
7 COMMAND="gnuplot -e $FILE $SCRIPT"
8 $COMMAND
9 n=$(($n+1))
10 done
Unfortunately this didn't work... gnuplot does run, but gives the following error message:
"filename='output_0.csv'"
^
line 0: invalid command
"filename='output_1.csv'"
^
line 0: invalid command
"filename='output_2.csv'"
^
line 0: invalid command
"filename='output_3.csv'"
^
line 0: invalid command
...
So, as I said before, I'm no expert in bash. My guess is that something isn't being interpreted correctly - either something is being interpreted as a string where it shouldn't or it is not being interpreted as a string where it should? (Just a guess?)
How can I fix this problem?
The first few (relevant) line of my gnuplot script are the following:
(Note the use of the variable filename which was entered as a command line argument. See this link.)
30 fit f(x) filename using 1:4:9 via b,c,e
31
32 plot filename every N_STEPS using 1:4:9 with yerrorbars title "RK45 Data", f(x) title "Landau Model"
Easy fix - I made a mistake with the quotation marks. ("")
Essentially, the only reason why the quotation marks " and " are required around the text filename='output_"$n".csv' is so that this string is interpreted correctly by bash, before executing the command! So indeed it is correct that the program runs when the command gnuplot -e "filename='output_0.csv'" 'plot.p' is entered into the terminal directly, but the quotation marks are NOT required when assembling the string beforehand. (This is a bit difficult to explain, but hopefully it is clear in your mind the difference between the 2.)
So the corrected version of the above code is:
1 #!/bin/bash
2 n=0
3 for f in output_*.csv
4 do
5 FILE="filename='output_"$n".csv'"
6 SCRIPT='plot.p'
7 COMMAND="gnuplot -e $FILE $SCRIPT"
8 $COMMAND
9 n=$(($n+1))
10 done
That is now corrected and working. Note the removal of the escaped double quotes.