bash while循环比较不评估

Question

I am trying to understand the difference between coding for if and while statements in bash. In the script below, if I use IF I get the expected result using == OR -eq etc. However with the while loop each while seems to evaluate as true for all tests. While loops 1 & 3 should fail, correct?

A != B, 
A == B,
A == C

I have tried different combinations of [ "" "" ] or [[ "" "" ]] etc.

What is the correct syntax for while

thx Art

#!/bin/sh
set -x
myVarA="abc"
myVarB="abc"
myVarC="def"

while [[ ${myVarA} != ${myVarB} ]]
    echo "A does not equal B"
    i=$(($i+1))
    do sleep 1s
    break
done

while [[ ${myVarA} == ${myVarB} ]]
    echo "A does equal B"
    i=$(($i+1))
    do sleep 1s
    break
done

while [[ ${myVarA} == ${myVarC} ]]
    echo "A does equal C"
    i=$(($i+1))
    do sleep 1s
    break
done

renders the following,

+ myVarA=abc
+ myVarB=abc
+ myVarC=def
+ [[ abc != abc ]]
+ echo 'A does not equal B'
A does not equal B
+ i=1
+ sleep 1s
+ break
+ [[ abc == abc ]]
+ echo 'A does equal B'
A does equal B
+ i=2
+ sleep 1s
+ break
+ [[ abc == def ]]
+ echo 'A does equal C'
A does equal C
+ i=3
+ sleep 1s
+ break

Answer 1

Your do keywords are wildly misplaced.

Everything between while and do is the condition list.

So your condition list in the first block is

[[ ${myVarA} != ${myVarB} ]]; echo "A does not equal B"; i=$(($i+1))

which, as you might imagine, evaluates as true since i=$(($i+1)) evaluates as true .

Similarly for the other blocks.

Move the do to the end of the while lines

while [[ ${myVarA} != ${myVarB} ]]; do
    : do something
done