I have this code:
printf -v s '%(%S)T' -1 # grab the current second
if ((s == 0)); then
# at the top of the minute, run some code
fi
This code throws an error on the eighth and ninth second of every minute:
bash: ((: 08: value too great for base (error token is "08")
bash: ((: 09: value too great for base (error token is "09")
How can I rectify this? Basically, we need to suppress the leading zero in the date output generated by printf.
Use a - prefix in the format string, thus:
printf -v s '%(%-S)T' -1
This suppresses the leading zero.
A more generic way of solving this is to specify the base in Bash arithmetic this way, while keeping the printf command unchanged:
if ((10#$s == 0)); then
Related post on Unix & Linux Stack Exchange:
Removing leading zeros from date
Related
I am trying to calculate the sum of numbers entered via command line to my script file. Here is my code
#!/bin/bash
for ((i=0;i<=$#;i++))
do
sum=$(($i + $sum))
done
echo $sum | bc
My terminal input is
bash file.sh 1 2
So the output should be 3 but I am getting
syntax error: operand expected (error token is "+ ")
The actual error reason is because of uninitialized variable sum going through the first iteration of the loop. Initialize the variable before entering the loop.
Also a major logical flaw is that you are not even iterating over the input arguments, but just over the counter i which will produce incorrect results if you pass arguments other than 1 2 from the command-line.
You need to pass over the actual arguments argc and argv (arg count and arg vector: for understanding purposes only) and you don't need bc at all
argc=$#
argv=("$#")
sum=0
for ((i=0; i<${argc}; i++)); do
sum=$((${argv[i]} + $sum))
done
To loop over all command line arguments, you can use the simplified form of the shell for statement:
sum=0
for i do
((sum += i))
done
((sum+=i)) is accepted by bash and many other shells; for a Posix-compatible shell, you can use arithmetic expansion with the : builtin:
: $((sum += i))
I saw couple of posts (some depends upon date -d $xyz to verify) but I'm trying to create an until loop where the user should be re-prompted to enter the value of a date format until it matches the custom date format.
My date format (what I need for Splunk) is m/d/yyyy:h:m:s or mm/dd/yyyy:hh:mm:ss
which means, if m (month number) is a single digit lets say 1 for January, then both 1 or 01 values are possible for date format but 0 or 00 is NOT a valid value. Value range is 01-to->12 or 1-to->12 but not greater than 12.
Similarly, the same rule applies to d (day number), it can be 01-to->10-to->31 or 1-to->31 but not 00 or more than 31 and all other yyyy (year), h (hour), m (minute), s (second) part.
What could be a minimal code (obfuscated is fine) to do this verification in BASH? It seems like date -d ??? doesn't provides this custom kind of verification for date/times!
OK, I can write one verifyDateFormatfunc() to do this, but I know there are people who have already written a one-liner / minimal snippet to verify this for sure. grep -f .. (where bunch of regex are listed line by line for all possible combinations, again the main code will look very minimal if I follow this? as the patterns sitting in -f file for grep will be transparent to a user) -or creating a map funcation (based on delimiters) for value ranges?
Possible values:
1/03/2017:23:0:15
02/4/2017:0:1:2
09/05/2017:10:10:0
10/6/2017:12:14:16
Here's an unholy extended regular expression (POSIX ERE):
^([1-9]|1[0-2]|0[1-9])/([1-9]|0[1-9]|[12][0-9]|3[01])/[0-9]{4}:([0-9]|0[0-9]|1[0-9]|2[0-3]):([0-9]|[0-5][0-9]):([0-9]|[0-5][0-9])$
that will test for the date/time patterns you specified (m/d/yyyy:h:m:s and mm/dd/yyyy:hh:mm:ss), with:
month: 1-12, 01-12
day: 1-31, 01-31
year: 0000-9999
hour: 0-23, 00-23
minute: 0-59, 00-59
second: 0-59, 00-59
You can use in an awk program that will exit with success (exit code 0) if the (first) line is a valid date/time (wrapped in a shell function that tests the first argument, for convenience):
#!/bin/bash
is_datetime_valid() {
awk '{exit $0!~"^([1-9]|1[0-2]|0[1-9])/([1-9]|0[1-9]|[12][0-9]|3[01])/[0-9]{4}:([0-9]|0[0-9]|1[0-9]|2[0-3]):([0-9]|[0-5][0-9]):([0-9]|[0-5][0-9])$"}' <<<"$1"
}
Or, if you prefer a pure bash solution (with ERE support in bash v3.0+):
#!/bin/bash
is_datetime_valid() {
local pat='^([1-9]|1[0-2]|0[1-9])/([1-9]|0[1-9]|[12][0-9]|3[01])/[0-9]{4}:([0-9]|0[0-9]|1[0-9]|2[0-3]):([0-9]|[0-5][0-9]):([0-9]|[0-5][0-9])$'
[[ $1 =~ $pat ]]
}
You can use it like:
if is_datetime_valid "1/03/2017:23:0:15"; then
# yup, it's valid
else
# ney, it's invalid
fi
Tested on a few examples:
#!/bin/bash
samples=(
"1/03/2017:23:0:15" "02/4/2017:0:1:2" "09/05/2017:10:10:0" "10/6/2017:12:14:16"
"00/03/2017:23:0:15" "1/33/2017:23:0:15"
)
for dt in "${samples[#]}"; do
if is_datetime_valid "$dt"; then
echo "$dt is valid"
else
echo "$dt is invalid"
fi
done
Gives:
1/03/2017:23:0:15 is valid
02/4/2017:0:1:2 is valid
09/05/2017:10:10:0 is valid
10/6/2017:12:14:16 is valid
00/03/2017:23:0:15 is invalid
1/33/2017:23:0:15 is invalid
I do not know whether using BSD date is an option for you, but it has what you are looking for.
There the date checker function can look like this
is_datetime_valid() {
date -j -f "%m/%d/%Y:%T" $1 1> /dev/null 2>&1
return $?
}
I want to have the day of week in the variable DOW.
So I use the following bash-script:
DOM=$(date +%d)
DOW=($($DOM % 7) ) | sed 's/^0*//'
Unfortunately there I get this error: bash: 09: command not found. The expected result is 2 ( 9 % 7 = 2) in the variable $DOW.
How can I get this working?
It works for the days 1-8 but because of the C-hex, there is no number over 8 available and the following message appears: bash: 09: value too great for base (error token is "09").
Use %u. Like this:
DOW=$(date +%u)
From the man page:
%u day of week (1..7); 1 is Monday
You can also use to return the day name
date +'%A'
Using a different %-specifier is the real answer to your question. The way to prevent bash from choking on invalid octal numbers is to tell it that you actually have a base-10 number:
$ DOM=09
$ echo $(( DOM % 7 ))
bash: 09: value too great for base (error token is "09")
$ echo $(( 10#$DOM % 7 ))
2
This works fine here
#!/bin/sh
DOW=$(date +"%a")
echo $DOW
The format option %u returns weekday number [1..7]. Monday is 1.
With date in string:
$ date -d "2022-07-17" +%u
7
Also check return value in $?.
For invalid date, $? is nonzero.
You can use the - flag:
DOM=$(date +%-d)
^
which would prevent the day from being padded with 0.
From man date:
- (hyphen) do not pad the field
Observe the difference:
$ DOM=$(date +%d)
$ echo $((DOM % 7))
bash: 09: value too great for base (error token is "09")
$ DOM=$(date +%-d)
$ echo $((DOM % 7))
2
When running this part of my bash script am getting an error
Script
value=0
for (( t=0; t <= 4; t++ ))
do
d1=${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?
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.
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
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:
$ (( $(date --date='9:00' +%-H) > 10 )) && echo true || echo oops
true
From the date man page:
By default, date pads numeric fields with zeroes. The following
optional flags may follow '%':
- (hyphen) do not pad the field
d1 and d2 are dates in that form 2014-01-09 and 2014-01-10
and then
((diff_sec=d2-d1))
What do you expect to get? ((diffsec=2014-01-09-2014-01-10)) ??
You need to convert the dates to seconds first:
d1=$( date -d "${filedates[$t]}" +%s )
d2=$( date -d "${filedates[$t+1]}" +%s )
(( compare = (d2 - d1) / (60*60*24) ))
(( value += compare ))
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 ) }")")"
You don't need the $ and the {} in an arithmetic expansion expression. It should look like this:
compare=$((SEC/(60*60*24)))
For 'mm' and 'dd' values in dates, I use this trick:
mm="1${date:5,2}" # where 5 is the offset to mm in the date
let mm=$mm-100 # turn 108 into 8, and 109 into 9
In a text file, test.txt, I have the next information:
sl-gs5 desconnected Wed Oct 10 08:00:01 EDT 2012 1001
I want to extract the hour of the event by the next command line:
hour=$(grep -n sl-gs5 test.txt | tail -1 | cut -d' ' -f6 | awk -F ":" '{print $1}')
and I got "08". When I try to add 1,
14 echo $((hour+1))
I receive the next error message:
./test2.sh: line 14: 08: value too great for base (error token is "08")
If variables in Bash are untyped, why?
See ARITHMETIC EVALUATION in man bash:
Constants with a leading 0 are interpreted as octal numbers.
You can remove the leading zero by parameter expansion:
hour=${hour#0}
or force base-10 interpretation:
$((10#$hour + 1))
what I'd call a hack, but given that you're only processing hour values, you can do
hour=08
echo $(( ${hour#0} +1 ))
9
hour=10
echo $(( ${hour#0} +1))
11
with little risk.
IHTH.
You could also use bc
hour=8
result=$(echo "$hour + 1" | bc)
echo $result
9
Here's an easy way, albeit not the prettiest way to get an int value for a string.
hour=`expr $hour + 0`
Example
bash-3.2$ hour="08"
bash-3.2$ hour=`expr $hour + 0`
bash-3.2$ echo $hour
8
In Short: In order to deal with "Leading Zero" numbers (any 0 digit that comes before the first non-zero) in bash
- Use bc An arbitrary precision calculator language
Example:
a="000001"
b=$(echo $a | bc)
echo $b
Output: 1
From Bash manual:
"bc is a language that supports arbitrary precision numbers with interactive execution
of statements. There are some similarities in the syntax to the C programming lan-
guage. A standard math library is available by command line option. If requested, the
math library is defined before processing any files. bc starts by processing code from
all the files listed on the command line in the order listed. After all files have
been processed, bc reads from the standard input. All code is executed as it is read.
(If a file contains a command to halt the processor, bc will never read from the standard input.)"
Since hours are always positive, and always 2 digits, you can set a 1 in front of it and subtract 100:
echo $((1$hour+1-100))
which is equivalent to
echo $((1$hour-99))
Be sure to comment such gymnastics. :)
The leading 0 is leading to bash trying to interpret your number as an octal number, but octal numbers are 0-7, and 8 is thus an invalid token.
If I were you, I would add some logic to remove a leading 0, add one, and re-add the leading 0 if the result is < 10.
How about sed?
hour=`echo $hour|sed -e "s/^0*//g"`