Integer expression expected bash 2

Question

I'm just learning bash scripting. Tried to sort the array, it says "integer expression expected for lines 10 and 15. What am I doing wrong? Here is my script:

#!/bin/bash array=('5' '9' '0' '20' '2' '15' '6' '25' '1') b=0 n=${#array[@]} i=0 while [ "$i" -lt "$n" ] do c=${array[$i]} d=${array[$i+1]} if [ "$c" -lt "$d" ]; then j=0 while [ "$j" -le "$i" ] do f=${b[$j]} if [ "$f" -gt "$c" ]; then b[$j]=$c echo "${b[$j]}" fi j=$(( j+1 )) done fi i=$(( i+1 )) done 

Answer 1

You're calling the [ command with the -lt/-gt decimal integer comparison operators on operands that are not always decimal integers.

You can see what happens if you run the script with bash -x. You'll see things like:

+ f= + '[' '' -gt 0 ']' ./myscript: line 15: [: : integer expression expected 

With:

while [ "$i" -lt "$n" ] do [...] d=${array[$i+1]} 

On the last pass in that loop, you'll try to access beyond the last element of the array, so $d will be empty.

You're also initialising $b as a 0 string, and later on accessing it as an array. See also how f=${b[$j]} will get you an empty $f except for when $j is 0.

I don't know what you're trying to do with that code, but it seems like you need to go back to the drawing board.