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Replace string with asterisk(*) in variable

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Operating Systems AIX Replace string with asterisk(*) in variable
# 1  
Old 04-19-2010
Replace string with asterisk(*) in variable
I was trying to replace a string ( for eg - @@asterisk@@ to * ) in variable using

Code:
cat $INFILE  | while read LINE
do
    stmt1=`echo $LINE | sed 's/@@asterisk@@/\*/g'`
    stmt=$stmt' '$stmt1
    stmt2=`echo $LINE`
    STATEMENT=$STATEMENT' '$stmt2
done
echo 'Statement with sed -- > ' $stmt 
echo 'Statement without sed -- > ' $STATEMENT

Output I am getting is
=================
Code:
Statement with sed -- >
Statement without sed -- >  INSERT INTO table1 SELECT @@ASTERISK@@ FROM table2

Seems sed is not working.. Appreciate if someone say what is wrong in this.

Thanks in advance
Last edited by Scott; 04-19-2010 at 01:29 PM.. Reason: Code tags, please...
# 2  
Old 04-19-2010
Quote:
Originally Posted by Vaddadi
Seems sed is not working.. Appreciate if someone say what is wrong in this.
Actually nothing is wrong in your replacement, but you fail to protect your variables content until the output:

Code:
variable_with_glob='*'
print - "compare this: $variable_with_glob"
print - "to this:" $variable_with_glob

I hope this helps.

bakunin
# 3  
Old 04-19-2010
Or do something with turning off glob
ksh turns off globbing with set -f or set noglob
Code:
#!/bin/ksh
cat $INFILE | while read LINE 
do 
   set -f
   stmt1=`echo $LINE | sed 's/@@asterisk@@/\*/g'` 
   stmt=$stmt' '$stmt1 stmt2=`echo $LINE` 
   STATEMENT=$STATEMENT ' ' $stmt2 
   set +f
done
print "$STATEMENT"

Make sure you turn globbing back on - it will affect subsequent code. Ver badly -- if it is off.
# 4  
Old 04-19-2010
Thanks ... tried set -f & set +f , but it didn't work ..
# 5  
Old 04-19-2010
It looks like you have this in your input file:
Code:
INSERT INTO table1 SELECT @@ASTERISK@@ FROM table2

Your sed command is looking for "@@asterisk@@" instead of "@@ASTERISK@@". Is that the problem?

Also, make sure you use double-quotes on the last two echos like this or the "*" will be replaced with a list of files in the current directory.
Code:
echo "Statement with sed -- >  $stmt "
echo "Statement without sed -- >  $STATEMENT"



---------- Post updated at 02:28 PM ---------- Previous update was at 02:23 PM ----------

This is what I get now with your script. I named it "test" with the INFILE=test.in

Code:
host:/home:$ ls -l test*
-rwxr-xr-x    1 root     system          267 Apr 19 14:25 test
-rw-r--r--    1 root     system           51 Apr 19 14:24 test.in

host:/home:$ cat ./test.in
INSERT INTO table1 SELECT @@ASTERISK@@ FROM table2

host:/home:$ cat ./test
INFILE=./test.in
cat $INFILE | while read LINE
do
        stmt1=`echo $LINE | sed 's/@@ASTERISK@@/\*/g'`
        stmt=$stmt' '$stmt1
        stmt2=`echo $LINE`
        STATEMENT=$STATEMENT' '$stmt2
done
echo "Statement with sed -- >  $stmt "
echo "Statement without sed -- >  $STATEMENT"

host:/home:$ ./test
Statement with sed -- >   INSERT INTO table1 SELECT * FROM table2
Statement without sed -- >   INSERT INTO table1 SELECT @@ASTERISK@@ FROM table2

host:/home:$

# 6  
Old 04-20-2010
Thanks to all ... it worked like a champ.

Shyam
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