i need to add 2 string variables into a positioning array , repeatedly - in loop.
First string in $2, second to $3 then up to the desired count incrementing the "position".
Using set -- alone does not increment the count so I end up with 2 variables in the array.
How do I increment the positioning index ?
Code:
echo " # add loop"
arrayTEST=( "$@" )
for i in {0..16..2}
do
set -- "CLI command" "CLI command description " # assign variables to ar$
# next position in array ??
done
echo "# number of elements in array "
echo "$#" # number of elements in array
echo " # content of array"
echo "$@" # contennt of array
The positional parameters are limited in the sense that it can only be filled in one go, either by passing them to a script of function or using the set command.
So you can only manipulate the this to set the positional parameters, for example:
Code:
set -- "" "String in position 2" "Another string in pos 3"
The empty string as the first parameter put the other strings in position 2 and 3.
But then the number of elements is not accurate, because echo "$#" will report 3
But why don't you use a regular array, available in bash or ksh? Then you can put strings in any position you like.
Code:
arr[2]="String in element 2"
arr[3]="Another string in element 3"
In this case the the elements arr[0] and arr[1] remain unused
Code:
$ echo "${#arr[@]}"
2
Last edited by Scrutinizer; 08-26-2018 at 04:00 AM..
Minor problem - now I need to copy the positioning array to a file.
This simple verification code does reads the array in reverse.
It would make things easier if I could add the lines into the file starting with 1 and go up.
I suppose I can change the while[condition] and check for # for != 16 in this case.
Or can I start from $# = 1 and shift "up" ?
Code:
testLoopAddition(){
55 for i in {0..16..2}
56 do
57 set -- "$@" "CLI command" "CLI command description "
58 echo "index $i"
59 done
62 echo "In reverse start with $# positional parameters"
63 # while positional parameter not empty
64 while [ "$1" != "" ]; do
65 echo "Parameter $# equals $1"
copy $1 to file here
68 shift
69 done
How do you intend to copy these to a file? Is it one per line to the same file? If so, you could look at the bash built-in printf:
Code:
printf "%s\n" $* > file
This will print your parameter list as a list of space-separated strings, one per line, until your list is exhausted. Try this noddy program:
Code:
#!/bin/bash
printf "%s\n" $*
This might not be what you want, of course. As we were putting spaces in the parameters in the above examples you may prefer this:
Code:
#!/bin/bash
printf "%s\n" "$@"
Basically, if you give printf more arguments than it needs it will use as many as it can and then repeat with the next set until the list is exhausted. So if you wanted to print two parameters per line, use
I am still trying to understand how bash code is executed.
Perhaps my perception of code line executing "left to right" is basically wrong.
Look at line #277
Every "manual" I read about using "wc" states it "prints".
I do not need "print" - I need to process what I am used to call "function return value".
OK, I understand 'wc" "prints" most likely to some "stdxxx" - as output of "pipe" .
So how does lineCount value , being at the "front" of the pipe get assigned?
And why is the file to be analyzed at the "end" of the code line?
Code:
echo "$LINENO test for actual file contents"
277 lineCount=$(wc -l < "$PWD$DEBUG_DIR$DEBUG_MENU")
278 echo "line count $lineCount"
279 if [ lineCount = 0 ]
280 then
281 echo "$LINENO File $DEBUG_MENU empty "
282 #build defaukt menu in file
283 defaultMenu
284 pause
285 else
286 echo "$LINENO File $DEBUG_MENU has data "
287 pause
288 fi
I may have asked this before, so forgive OF.
Problem:
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Code:
DynamicEntry=$(whiptail \
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