如何在Bash Shell脚本的while语句中编写if语句？

Question

I'm currently having simple syntax issues with the following bash shell script. I'm not sure what the syntax is for nesting an if statement into a while statement, or if it's even possible with bash shell scripting (new to all things linux):

#!/bin/bash

myCombo=$((RANDOM%99999+10000));
echo ${myCombo};

myCracker=00000;

while [ (($myCracker<=99999)) ]; do
    if [ $myCracker -eq myCombo ]
    then
        echo "The combination is " ${myCracker} " !"
    else [ $myCracker = $myCracker + 1 ]
    fi
done;

Answer 1

There were quite a few things wrong with your loop. I've made a number of improvements below:

while (( myCracker <= 99999 )); do
    if (( myCracker == myCombo )); then
        echo "The combination is $myCracker !"
        break
    else
        (( ++myCracker ))
    fi
done

As you're using bash, you can make use of (( arithmetic contexts )) , which don't need enclosing in [ . Within them, variable names are expanded, so do not require prefixing with $ .

Note that your original logic will either loop indefinitely or never echo , which is probably a bug. If myCracker == myCombo is ever true, myCracker won't be incremented so the loop will never terminate. The break statement deals with this.

I left the else branch in deliberately to show the syntax but you could also remove it entirely:

while (( myCracker++ <= 99999 )); do
    if (( myCracker == myCombo )); then
        echo "The combination is $myCracker !"
        break
    fi
done

The break is still useful as it prevents the loop from continuing unnecessarily.

Answer 2

You can also use extended tests instead of arithmetic contexts if you do simple comparisions. They should be a little faster, and are more common.

while [[ $myCracker -le 99999 ]]; do
    if [[ $myCracker -eq $myCombo ]]; then
        echo "The combination is ${myCracker}!"
        break
    else
        ((myCracker++))
    fi
done