Bash script while with boolean and counter

Question

Can somebody help me with this... I can't seem to get a working while statement (it never enters the loop)... Tried many different syntaxes and I'm stuck.

tries=0
success=false

while (!${success} && ${tries} -lt 10); do
  {
    echo "Trying..." &&
    myCommand &&
    success=true &&
    echo "Success"
  } || {
    success=false &&
    echo "Failed"
  }
  let tries=tries+1
done

score 2 · Accepted Answer · edited May 23 '17 at 11:49

2

Just a little change

tries=0
success=false

while (( !${success} && ${tries} < 10 )); do
  {
    echo "Trying..." &&
    myCommand &&
    success=true &&
    echo "Success"
  } || {
    success=false &&
    echo "Failed"
  }
  let tries=tries+1
done

Difference in Bash between IF statements with parenthesis and square brackets

edited May 23 '17 at 11:49

Community

1
1

answered Jan 14 '14 at 11:11

Raul Andres

3,766
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24

Ok, that got me a little bit further. Thank you... However, there seems to be something wrong with my success=true variable assignment. The loop doesn't exit on that condition, it executes 10 times whether the command succeeds or not. Any idea? – ddx001 Jan 14 '14 at 12:52
Trivially, `success=0` and `while (( ${success} < 1 && ${tries} < 10 ));` do ... shoud work – Raul Andres Jan 14 '14 at 13:03

score 2 · Answer 2 · answered Jan 14 '14 at 15:12

You appear to be trying to write C code in bash. Specifically, your use of success as both a Boolean flag and an executable program is a bit awkward. Try

while (( tries < 10 )); do
    { myCommand && echo Success && break; } || { echo Failed && let tries=tries+1; }
done

Using an explicit if statement would also be more readable:

while (( tries < 10 )); do
    if myCommand; then
        echo Success
        break
    else
        echo Failed
        let tries=tries+1
    fi
done

Bash script while with boolean and counter

2 Answers2