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Shell Programming and Scripting

Replace var if conditon is met

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# 1  
Old 07-25-2007
Replace var if conditon is met
HPUX 11i v2 #!/bin/sh

Hi all.

I am using a read to split a file into variables:

cat filename | while read VAR1 VAR2 ETC ETC
then a do to work each line

$VAR2 is always 9 characters. If the first character is NOT an A or a 9, then I need to mask the first 5 characters with *****.

Any ideas?

TIA!!!!!
# 2  
Old 07-25-2007
Code:
awk -F" " '{ x=substr($2, 0, 1); if ( x !~ "A" && x !~ "9" ) { $2 = "*****" substr($2, 6, length - 5); print } }' filename

# 3  
Old 07-25-2007
Code:
mVar2='123456789'
mFirst=`echo $mVar2 | cut -c1`
if [ "$mFirst" != "A" -a "$mFirst" != "9" ]; then
  mOutLine='*****'`echo $mVar2 | cut -c6-`
else
  mOutLine=$mVar2
fi
echo "OutLine = "$mOutLine

# 4  
Old 07-26-2007
why don't you build your variable file like so:

Code:
var1="this"
var2="that"
var3="the other"

then all you need to do is:

. var_file

job done
# 5  
Old 07-26-2007
Quote:
Originally Posted by matrixmadhan
Code:
awk -F" " '{ x=substr($2, 0, 1); if ( x !~ "A" && x !~ "9" ) { $2 = "*****" substr($2, 6, length - 5); print } }' filename

This works great, except....if VAR2 begins with A or 9 it doesn't print the line at all. How can I keep the the line intact if VAR2 begins with A or 9?

TIA!!!!!!!!
# 6  
Old 07-26-2007
I would say that was not mentioned before.

Else part needs to be added

this would do Smilie

Code:
awk -F" " '{ x=substr($2, 0, 1); if ( x !~ "A" && x !~ "9" ) { $2 = "*****" substr($2, 6, length - 5); print } else { print } }' filename

# 7  
Old 07-26-2007
Matrix
You're Awksome!

Thanks!!!!
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