CODEWITHSUNDEEP
bashloops
Stackoverflow
Stackoverflow

Reputation: 467

Loop BASH with counter in one line

I wait program finish and print nothing; but this intro in loop:

while [ 0 > 0 ]; do echo 1; done

logically 0 no't is > to 0... why get loop?

How I can get nothing by screen? and program finish fine?

After of my progam "nothing to do":

while [ 0 > 0 ]; do echo 1; done

I like in one line:

q = 0; while [ q < 9 ]; q ++; do echo q; done

it´s possible in one line?

Thanks

Upvotes: 6

Views: 11354

Answers (2)

user unknown
user unknown

Reputation: 36229

q = 0; while [ q < 9 ]; q ++; do echo q; done

Just 5 errors in one line.

  • q=0 may not have white space around the assignment operator
  • q ++ has to be in double round parens
  • q < 9 has to be -lt (less than)
  • in single brackets [ $q -lt 9 ] q needs to be $q
  • the do should follow the while

Possible solution:

q=0; while [[ q -lt 3 ]]; do ((q++)); echo $q; done
1
2
3

echoing values can be done with 

echo {1..9}

too, but is not flexible, so you can't use variable expansion inside, like echo {1..$n}. The canonical way of doing initialization, increment and threshold check, is a for loop:

 for (( q=1; q < 4; ++q)); do  echo $q ; done

There is the external program seq, which is not so much recommended, for that reason: 

seq 1 3

First question:

while [ 0 > 0 ]; do echo 1; done

Look for a file 0 where you used file redirection (instead of -gt), like in echo foo > 0.file.

Instead of 

while [ 0 -gt 0 ]; do echo 1; done

because it does nothing. It doesn't wait for anything. Either your program is sequential, then it is finished at that point anyhow. Or there is a program/command in the background running, which doesn't care about this anti loop anyhow.

Upvotes: 13

Nikita Malyavin
Nikita Malyavin

Reputation: 2107

For your second question, use for (( expr1 ; expr2 ; expr3 )) ; do list ; done

Also refer to man bash

Upvotes: 0

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