korn shell variable help

4play · 10-20-2008, 04:07 PM

Im trying to declare a variable that will contain the contents of another variable.

Code:

#!/bin/ksh

query1="select * from table1"
query2="select * from table2"
query3="select * from table3"


for i in 1 2 3
do

VAR=`echo query${i}`

echo ${VAR}

done

This is the basic idea of what im trying to do but instead of VAR being populated with the sql it contain query1,2 or 3.

How exactly do i get to stick the sql statement into another variable.

colucix · 10-20-2008, 04:20 PM

You have to use indirect reference through the eval var1=\$$var2 mechanism

Code:

#!/bin/ksh
query1="select * from table1"
query2="select * from table2"
query3="select * from table3"
for i in 1 2 3
do
   eval VAR=\$query$i
   echo "$VAR"
done

I'm not sure this is what you're looking for, anyway.

4play · 10-20-2008, 04:30 PM

Quote:

Originally Posted by colucix

You have to use indirect reference through the eval var1=\$$var2 mechanism

Code:

#!/bin/ksh
query1="select * from table1"
query2="select * from table2"
query3="select * from table3"
for i in 1 2 3
do
   eval VAR=\$query$i
   echo "$VAR"
done

I'm not sure this is what you're looking for, anyway.

That is exactly what i am after, thank you very much.

can i ask what the eval VAR=\$query bit is actually doing.

why should i have to escape the first $ ?

colucix · 10-20-2008, 04:36 PM

If you escape the first $ it will be interpreted literally from eval, that is you have to evaluate the following:

Code:

VAR=$query1

in the eval statement the first escaped $ will remain untouched, while the $i variable will be substituted by its value:

Code:

eval VAR=\$query$i

4play · 10-20-2008, 04:40 PM

Thanks alot for that.

I have been bashing my head against that problem for about 4 hours today.

allanf · 11-23-2011, 01:44 AM

Quote:

Originally Posted by 4play

Im trying to declare a variable that will contain the contents of another variable.

Code:

#!/bin/ksh

query1="select * from table1"
query2="select * from table2"
query3="select * from table3"


for i in 1 2 3
do

VAR=`echo query${i}`

echo ${VAR}

done

This is the basic idea of what im trying to do but instead of VAR being populated with the sql it contain query1,2 or 3.

How exactly do i get to stick the sql statement into another variable.

You want to do:

Code:

#!/bin/ksh
#

one=1
two=2
three=3
four=4
five=5

for DIGIT in one two three four five; do
  nameref NUMBER=${DIGIT}
  echo ${DIGIT} ${NUMBER}
done

Which results in:

Code:

one 1
two 2
three 3
four 4
five 5

makyo · 11-23-2011, 06:38 AM

Hi.

Old thread, but possibly a useful note from man page for ksh:

Quote:

If a nameref is used as the index of a for loop, a name
reference is established for each item in the list.

See an example at post #4 at http://www.linuxforums.org/forum/pro...once-only.html ... cheers, makyo

David the H. · 11-23-2011, 09:23 AM

Sigh...

No, no, no! This is not the correct way to do it at all. This is a classic case of a situation where an array* should be used.

Also, you need to make sure your parameters are properly quoted to keep the "*" globbing character from expanding to a list of filenames in the current directory.

Code:

#!/bin/ksh


query[1]='select * from table1'
query[2]='select * from table2'
query[3]='select * from table3'

for i in 1 2 3; do

	VAR=${query[i]}
	echo "$VAR"

done

*Why do people ignore arrays so often? Please forget all about trying to generate dynamic variable names; in the vast majority of cases you don't need them. If you have a list of things to access, store it in an array, and call it by number.

makyo · 11-23-2011, 09:39 AM

Hi.

I agree with David the H. that for simple cases of lists like this, arrays are usually the best solution. If numbers are not desired, associative arrays (hashes in most newer perl documentation) might be used. However, it is useful to know alternate solutions, as well as other uses for namerefs. Had it not been for this thread, I think I might not have found out about this from the ksh man page:

Quote:

A nameref provides
a convenient way to refer to the variable inside a function whose name
is passed as an argument to a function ...

Best wishes ... cheers, makyo

David the H. · 11-23-2011, 12:03 PM

The ironic thing is, the same shells that have neat features like that for easily handling indirect references are also the ones that offer other features that make using them generally unnecessary. And vice-versa.

A good synopsis on indirect referencing in both bash and ksh can be found here:
http://mywiki.wooledge.org/BashFAQ/006