Variable print problem

Shell Programming and Scripting
1

Hi,
I'm facing some problem with printing the values in the variables

I have passed all the database values into the variable, but when i echo the variable i get the values printed in one line

var=`echo <query> | database`
echo $var
or 
awk 'BEGIN {printf("%s\n","'"${var}"'")}'

this prints

table1 <value> table2 <value> table3 <value> table4 <value>

But the above needs to be printed as

table1 <value>
table2 <value>
table3 <value> and so on

Can anyone please help me

2

It would help to see the output from

echo <query> | database

Your problem is that Shell Environment Variables are not files and the line break character is lost.

But. assuming that the database writes the output in pairs (with a line terminator), we need to read them as pairs:

echo <query> | database | while read table value
do
       echo "${table} ${value}"
done
3

Thanks Methyl

The output i get from the command is

echo <query> | database

table1 <value>
table2 <value>
table3 <value>

but if i pass the values to the variable then i'm getting

table1 <value> table2 <value> table3 <value> table4 <value>
4

Try echo "$var". Without that, shell expansion is going to squeeze all the whitespace together.

5

Try my solution from post #2 . Should work.