What is $opt variable for command line parameter in bash

Question

I have below bash script, I am a bit confuse about what is $opt meaning. and I do not find detail information about it after search.

test.sh:

echo

Answer 1

From the bash(1) man page:

   for name [ [ in [ word ... ] ] ; ] do list ; done
          The list of words following in is expanded, generating a list of
          items.  The variable name is set to each element of this list in
          turn, and list is executed each time.  If the in word  is  omit‐
          ted,  the  for  command  executes  list once for each positional
          parameter that is set (see PARAMETERS below).  The return status
          is  the  exit  status of the last command that executes.  If the
          expansion of the items following in results in an empty list, no
          commands are executed, and the return status is 0.

So, if the in word is missing the for loop iterates over the positional arguments to the script, i.e. $1, $2, $3, ....

There is nothing special about the name opt, any legal variable name can be used. Consider this script and its output:

#test.sh
echo $*    # output all positional parameters $1, $2,...
for x
do
    echo $x
done

$ ./test.sh -a -b hello
-a -b hello
-a
-b
hello

Answer 2

the for statement goes generally like this:

for x in a b c; do
  ...
done

$x has the value a on the first iteration, then b, then c. the a b c part need not be written out literally, the list to iterate through can be given by a variable, like the $@, which has the list of all arguments to the current scope:

for x in "$@"; do
  ...
done

further, in "$@" is assumed if you provide no list explicitly:

for x; do
  ...
done

Answer 3

$opt is not a built-in or special variable. If it's mentioned in the script (outside of for opt without the in something part, which @mhawke already explained) it should either be defined earlier in the script or it's expected that it is exported before running the script.