I want to print the numbers in the following pattern using WHILE loop.
0
01
012
0123
.....
......
0123456789
#!/bin/sh
a=0
b=0
while [ $a -le 10 ]
do
while [ $b -le $a ]
do
echo -n "$b"
b=`expr $b + 1`
done
echo
a=`expr $a + 1`
done
Getting output:
0
1
2
3
4
5
6
7
8
9
10
As you simply append the latest count to the line output, simply do so as text.
#!/bin/bash
a=0
out=''
while [ $a -lt 10 ]
do
out=$out$a
echo $out
a=`expr $a + 1`
done
Also, le is less or equal, so you end up with 10. Use lt 10 or le 9.
a=$(( a + 1 )) instead of expr? It should work in any POSIX-compliant that isn't too old.If you use bashshell, you can take advantage of sequence expressions of the form {x..y} and use the special parameter $_ which usually expands to the last argument to the previous command.
#/bin/bash
i=
for i in {0..9}
do
echo "$_$i"
done
a=0
b=1
while [ $a -lt 10 ]
do
a=0
while [ $a -lt $b ]
do
echo -n $a
a=`expr $a + 1 `
done
echo
b=`expr $b + 1`
done
bvalue which is incremented in each iteration? In other words, have you thought about the algorithm?012). Then you should append it the value ofband print it and increase the counter.whileloop? What shell are you using?expris a non-standard, old-fashioned way of doing arithmetic in the shell.bash, not any otherUNIX shellas you have tagged this question?