Problem with avoiding new line

jim_geek · 08-30-2011, 03:08 PM

hello i am new to linux(mandriva-konsole) and just starting the shell scripting.i am getting the following output for
1
2
1
3
2
1
4
3
2
1

but i want the following output
1
12
123
1234

the program is

for ((i=0;i < 5;i++))
do
j=$j;
while [ $j -ne 0 ]
do
echo $j \n;
j=$(($j-1));
done
done

the logic of program is correct but only the output has different pattern for display.i suspect problem with echo $j \n; its not giving output for whole loop in one line.because i should get only four line op with pattern given above.please reply quickly.thanks in advance.

MTK358 · 08-30-2011, 04:32 PM

First of all, use code tags.

Quote:

Originally Posted by jim_geek

for ((i=0;i < 5;i++))
do
j=$j;
while [ $j -ne 0 ]
do
echo $j \n;
j=$(($j-1));
done
done

"echo $j \n" clearly won't work. Bash doesn't interpret backslash escape sequences like C does. The are only interpreted that way in a single-quotes string preceeded by a "$" character, like this:

Code:

$'\n'

Also, echo automatically adds a newline, so it isn't necessary in the first place.

Finally, What's th point of the "j=$j" line? I think you mean "j=$i". This is the corrented code:

Code:

for (( i=0; i<5; i++))
do
    for (( j=1; j<=i; j++ ))
    do
        echo -n $j
    done
    echo
done

harry edwards · 08-30-2011, 04:38 PM

Another way closer to your original code:

Code:

j=1
for ((i=1;i < 5;i++))
do
 a=1
 while [ "$a" -le "$j" ]
 do
  echo -n $a;
  a=$(($a+1));
 done
 j=$(($j+1));
done

Tinkster · 08-30-2011, 06:44 PM

Please post your thread in only one forum. Posting a single thread in the most relevant forum will make it easier for members to help you and will keep the discussion in one place. This thread is being closed because it is a duplicate.

http://www.linuxquestions.org/questi...-linux-900297/