while loop in Linux Shell Script
 


Syntax:

           while [ condition ]
           do
                 command1
                 command2
                 command3
                 ..
                 ....
            done

Loop is executed as long as given condition is true. For e.g..the for loop program can be written using while loop as:

$cat > nt1
#!/bin/sh
#
#Script to test while statement
#
#
if [ $# -eq 0 ]
then
   echo "Error - Number missing form command line argument"
   echo "Syntax : $0 number"
   echo " Use to print multiplication table for given number"
exit 1
fi
n=$1
i=1
while [ $i -le 10 ]
do
  echo "$n * $i = `expr $i \* $n`"
  i=`expr $i + 1`
done

Save it and try as
$ chmod 755 nt1
$./nt1 7
Above loop can be explained as follows:

 
n=$1 Set the value of command line argument to variable n. (Here it's set to 7 )
i=1 Set variable i to 1
while [ $i -le 10 ]                   This is our loop condition, here if value of i is less than 10 then, shell execute all statements between do and done
do  Start loop
echo "$n  *  $i = `expr $i  \*  $n`" Print multiplication table as
7 * 1 = 7
7 * 2 = 14
....
7 * 10 = 70, Here each time value of variable n is multiply be i.
i=`expr $i + 1` Increment i by 1 and store result to i.  ( i.e. i=i+1)
Caution: If you ignore (remove) this statement  than our loop become infinite loop because value of variable i always remain less than 10 and program will only output
7 * 1 = 7
...
...
E (infinite times)
done                                   

Loop stops here if i is not less than 10 i.e. condition of loop is not true. Hence
loop is terminated.

 
 
 
 
 
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