Bashscript throws command error when populating variable [duplicate] - linux

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How do I set a variable to the output of a command in Bash?
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Variable variable assignment error -"command not found"
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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

Answer 1

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.