CODEWITHSUNDEEP
bash
user3178889
user3178889

Reputation: 729

Value too great for base (error token is "09")

When running this part of my bash script am getting an error

Script

value=0
for (( t=0; t <= 4; t++ ))
do
d1=${filedates[$t]}
d2=${filedates[$t+1]}
((diff_sec=d2-d1))
SEC=$diff_sec
compare=$((${SEC}/(60*60*24)))
value=$((value+compare))
done

Output

jad.sh: line 28: ((: 10#2014-01-09: value too great for base (error token is "09")
jad.sh: line 30: /(60*60*24): syntax error: operand expected (error token is "/(60*60*24)")

d1 and d2 are dates in that form 2014-01-09 and 2014-01-10

Any solution please?

Upvotes: 66

Views: 46797

Answers (8)

Jortstek
Jortstek

Reputation: 363

Quick fix using string substitution to get rid of any leading zero in a variable:

var=01
var=${var#0}

Gone.

Upvotes: 0

enharmonic
enharmonic

Reputation: 2090

As others have said, the error results from Bash interpreting digit sequences with leading zeros as octal numbers. If you have control over the process creating the date values and you're using date, you can prefix the output format string with a hyphen to remove leading zero padding.

Without prefixing date format with hyphen:

$ (( $(date --date='9:00' +%H) < 10 )) && echo true || echo oops
-bash: ((: 09: value too great for base (error token is "09")
oops

With prefixing date format with hyphen:

$ (( $(date --date='9:00' +%-H) < 10 )) && echo true || echo oops
true

From the date man page:

By default, date pads numeric fields with zeroes. The following optional flags may follow '%':

  -      (hyphen) do not pad the field

Upvotes: 11

MrPotatoHead
MrPotatoHead

Reputation: 1144

Posting some tips here related to the title of this question, but not directly related to the details of the original question. I realize that's a bit controversial action on Stack Overflow, however these related questions:

convert octal to decimal in bash [duplicate]

Value too great for base (error token is "08") [duplicate]

point to this one, and yet they are closed and hence, I could not post this answer there. Therefore, this seemed like a logical place (at least to me) to post this information that may help others in a similar situation, especially new-to-BaSH programmers.

An alternative approach to ensuring a number is treated as a 10-base integer is to use printf. This command instructs printf to treat $num as an integer and round it to 0 decimal places.

num="$(printf "%.0f" "$num")"

Or, if you want to also ensure there are no non-numeric characters in the string, you can do this:

num="$(printf "%.0f" "${num//[!0-9]/}")"

Both commands will strip out leading zeroes and round decimal values to the nearest whole number. Note the first (simpler) solution works with negative numbers, but the second does not (it will always return absolute value).

Note that printf rounds down, meaning .01 to 0.5 is rounded down to 0, while .51 to .99 is rounded up to 1. Basically, the difference between rounding up versus down in this case is that printf rounds down 0.5 and any below. I mention this because 0.5 rounded up is a more common practice.

Now, addressing the OP's specific scenario.... Combining printf with awk allows arithmetic expressions not possible with printf alone.

This

compare=$((${SEC}/(606024)))

could be alternatively be expressed as

compare=$(awk -v sec=$SEC 'BEGIN { print int(sec/(60*60*24))}')

or

compare="$(printf "%.0f" "$(awk "BEGIN { print ( $SEC / ( 60 * 60 * 24 ) ) }")")"

Meanwhile,

value=$((value+compare))

Could be calculated as

value="$(printf "%.0f" "$(awk "BEGIN { print ( $value + $compare ) }")")"

Upvotes: 2

Dick Guertin
Dick Guertin

Reputation: 769

For 'mm' and 'dd' values in dates, I use this trick:

mm="1${date:5,2}"  # where 5 is the offset to mm in the date
let mm=$mm-100     # turn 108 into 8, and 109 into 9

Upvotes: 0

glenn jackman
glenn jackman

Reputation: 246837

d1 and d2 are dates in that form 2014-01-09 and 2014-01-10

and then 

((diff_sec=d2-d1))

What do you expect to get? ((diffsec=2014-01-09-2014-01-10)) ??

You need to convert the dates to seconds first:

d1=$( date -d "${filedates[$t]}" +%s )
d2=$( date -d "${filedates[$t+1]}" +%s )
(( compare = (d2 - d1) / (60*60*24) ))
(( value += compare ))

Upvotes: 4

rojomoke
rojomoke

Reputation: 4025

Prepend the string "10#" to the front of your variables. That forces bash to treat them as decimal, even though the leading zero would normally make them octal.

Upvotes: 88

dogbane
dogbane

Reputation: 274612

What are d1 and d2? Are they dates or seconds?

Generally, this error occurs if you are trying to do arithmetic with numbers containing a zero-prefix e.g. 09.

Example:

$ echo $((09+1))
-bash: 09: value too great for base (error token is "09")

In order to perform arithmetic with 0-prefixed numbers you need to tell bash to use base-10 by specifying 10#:

$ echo $((10#09+1))
10

Upvotes: 23

hek2mgl
hek2mgl

Reputation: 158020

You don't need the $ and the {} in an arithmetic expansion expression. It should look like this:

compare=$((SEC/(60*60*24)))

Upvotes: 0

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