Value too great for base (error token is “0925”)

Question

I have the following logic in my bash script:

#!/bin/bash
local_time=$(date +%H%M)

if (( ( local_time > 1430  && local_time < 2230 ) || ( local_ti         


        

Answer 1

bash is treating your numbers as octal because of the leading zero

From man bash

Constants with a leading 0 are interpreted as octal numbers. A leading 0x or 0X denotes hexadecimal. Otherwise, numbers take the form [base#]n, where base is a decimal number between 2 and 64 represent- ing the arithmetic base, and n is a number in that base. If base# is omitted, then base 10 is used.

To fix it, specify the base-10 prefix

#!/bin/bash
local_time="10#$(date +%H%M)"

if (( ( local_time > 1430  && local_time < 2230 ) || ( local_time > 0300 && local_time < 0430 ) )); then
 # do something
fi

Answer 2

Solving the issue within the conditional test

One may be forced to keep the variable as it is for a variety of reasons (e.g. file naming issues). If this is the case, solve the issue within the conditional test by explicitly specifying base 10#:

#!/bin/bash
local_time=$(date +%H%M)

if (( ( 10#${local_time} > 1430  && 10#${local_time} < 2230 ) || ( 10#${local_time} > 0300 && 10#${local_time} < 0430 ) )); then
 # do something
fi

Answer 3

Following the advice from this blog, this works:

#!/bin/bash
local_time=`date +%H%M`
local_time="$(( 10#$local_time ))"

if (( ( local_time > 1430  && local_time < 2230 ) || ( local_time > 0300 && local_time < 0430 ) )); then
    echo "it is time!"
fi