Problem with dynamic variables outside the loop


 
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# 1  
Old 08-21-2009
Problem with dynamic variables outside the loop

Hi All,

Shell is ksh

I've given portion of the script here to explain the problem.

It will accept 2 input parameters .

Code:
in_file1=$1 
in_file2=$2 
outbound_dir=/home/outbound 

for i in 1 2 
do 
  eval file$i=$outbound_dir/\$in_file$i 
  eval echo "filename is \$file$i" 
  eval temp_file=$outbound_dir/\$in_file$i 
  eval FILE$i_LINE_COUNT=`wc -l < $temp_file` 
  eval echo "Total lines in file$i are \$FILE$i_LINE_COUNT" 
done 

echo "file1 name outside loop is $file1" 
echo "file1 count outside loop is $FILE1_LINE_COUNT"

When i am displaying file1 and $FILE1_LINE_COUNT variables inside the loop its giving correct values.

But ouside the loop getting correct value for only file1 variable not getting for $FILE1_LINE_COUNT variable.

can you pls help me.

Last edited by Franklin52; 08-21-2009 at 02:38 PM.. Reason: Adding code tags and reformat code
# 2  
Old 08-21-2009
First off, please format your post so it is readable. Not reading your post very closely, but based on the last sentence, have a look at https://www.unix.com/shell-programmin...s-however.html
# 3  
Old 08-21-2009
Hi Peterro,Formatted now. Can you pls have a look at it now.
# 4  
Old 08-21-2009
Better, please use the code tags for code.

In any case, this has nothing to do with the loop specifically. Replace

Code:
eval FILE$i_LINE_COUNT=`wc -l < $temp_file`

with

Code:
eval FILE${i}_LINE_COUNT=`wc -l < $temp_file`

You were assigning the word count output to $FILE since $i_LINE_COUNT was undefined. Surround the 'i' with curly braces to distinguish it from the rest of the line.
# 5  
Old 08-23-2009
thanks for your reply....it works fine with curly braces.

eval file$i=$outbound_dir/\$in_file$i
eval echo "filename is \$file$i"
eval temp_file=$outbound_dir/\$in_file$i

eval FILE${i}_LINE_COUNT=`wc -l < $temp_file`

Can i use $file$i instead of $temp_file in the above line ??
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