Bashscript throws command error when populating variable

Question

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.

1. 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
   

2. 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.