UNIX for Dummies Questions & Answers

Hi All

I've very nearly finished this script I'm working on but have hit another idiots problem, after googling I can't see a solution for this one.

I have a while count loop that checks checks two consecutive values then increments the count by two.

What the script has to do is then check all the 'pairs' of values match, if they do the check is true and the script can continue.

so what I thought was.....

f=1
while [ "$f" -le "$totalofchecks" ]; do

blah.....
blah.....
OK test is true so lets save this pass result.

1STCHECK$f="$CHECKVALUE"

done

the script is borking at the 1STCHECK$f line with a "command not found" error.

I've tried to enclose the 1STCHECK$f part in all the usual ' " ( ` options hoping it was a syntax problem, however it appears that I cant set a variable that has a variable as part of the name...is that right?

Could someone perhaps suggest an alternate way of doing this?

Thanks in advance.

---------- Post updated at 07:21 AM ---------- Previous update was at 07:19 AM ----------

Sorry I should add there is a f=`expr $f + 2` before the done

---------- Post updated at 07:25 AM ---------- Previous update was at 07:21 AM ----------

I think I've just realised that I can't start a variable with a number.......Dohhhhhhh

I think you want to use arrays:

Arrays

I'd already read that section but you can't put a variable in the array field either....

i.e.

$1STCHECK[$f] doesn't work

Yes you can.

Code:

$ X=blah
$ Y[$X]=more_blah
$ echo ${Y[$X]}
more_blah

Although you are correct about variables starting with numbers.

OK I'm now back o this question.

I originally sorted it as suggested using arrays and that works fine.

However I now want to post a similar script to an embedded device that uses busybox and doesn't support arrays.

So back to my original question could someone suggest a way to structure something like

Code:

while [ "$COUNT" -le "10" ]; do
  COUNT=1
  $VARIABLE$COUNT="Some stuff read here"
  $OUTPUT$COUNT=`sed -e '/keep this line/!d' -e 's/now delete these bits//g'`
  COUNT=`expr $COUNT + 1`
done

Thanks in advance

---------- Post updated at 12:25 PM ---------- Previous update was at 12:24 PM ----------

Ignore the deliberate error, the COUNT=1 should be before the while....do

I'm not sure but probably this is something what you're looking for:

Code:

#!/bin/ksh

num1="one"
num2="two"

count=1
while [ $count -le 2 ]
do
  eval "val${count}=\$num${count}"
  count=$(( $count + 1 ))
done

echo $val1	# output : one
echo $val2	# output : two