BASH count with two digit even its a one digit task - bash

i'am currently trying to count +1 to a variable which could contain either
02
15
i use the following line to count one up, but the output seems not to be two digit. So if 02 is the $version and count one up, it will give the output "3" and not "03". Sure if its already two digit, it works.
$((echo $version) + 01))
how can I get bash to count with two digits and output it correctly?

Use printf to format a number. Also, note that 08 fails in numeric expressions, as numbers starting with 0 are interpreted as octal.
#!/bin/bash
shopt -s extglob
for n in 02 08 15 001; do
i=${n##+(0)} # Remove leading zeros. Needs the extglob.
length=${#n}
printf "%0${length}d\\n" $((i + 1))
done

Related

Bash: checking substring increments with modular arithmetic

I have a list of files with file names that contain a substring of 6 numbers that represents HHMMSS, HH: 2 digits hour, MM: 2 digits minutes, SS: 2 digits seconds.
If the list of files is ordered, the increments should be in steps of 30 minutes, that is, the first substring should be 000000, followed by 003000, 010000, 013000, ..., 233000.
I want to check that no file is missing iterating the list of files and checking that neither of these substrings is missing. My approach:
string_check=000000
for file in ${file_list[#]}; do
if [[ ${file:22:6} == $string_check ]]; then
echo "Ok"
else
echo "Problem: an hour (file) is missing"
exit 99
fi
string_check=$((string_check+3000)) #this is the key line
done
And the previous to the last line is the key. It should be formatted to 6 digits, I know how to do that, but I want to add time like a clock, or, in more specific words, modular arithmetic modulo 60. How can that be done?
Assumptions:
all 6-digit strings are of the format xx[03]0000 (ie, has to be an even 00 or 30 minutes and no seconds)
if there are strings like xx1529 ... these will be ignored (see 2nd half of answer - use of comm - to address OP's comment about these types of strings being an error)
Instead of trying to do a bunch of mod 60 math for the MM (minutes) portion of the string, we can use a sequence generator to generate all the desired strings:
$ for string_check in {00..23}{00,30}00; do echo $string_check; done
000000
003000
010000
013000
... snip ...
230000
233000
While OP should be able to add this to the current code, I'm thinking we might go one step further and look at pre-parsing all of the filenames, pulling the 6-digit strings into an associative array (ie, the 6-digit strings act as the indexes), eg:
unset myarray
declare -A myarray
for file in ${file_list}
do
myarray[${file:22:6}]+=" ${file}" # in case multiple files have same 6-digit string
done
Using the sequence generator as the driver of our logic, we can pull this together like such:
for string_check in {00..23}{00,30}00
do
[[ -z "${myarray[${string_check}]}" ]] &&
echo "Problem: (file) '${string_check}' is missing"
done
NOTE: OP can decide if the process should finish checking all strings or if it should exit on the first missing string (per OP's current code).
One idea for using comm to compare the 2 lists of strings:
# display sequence generated strings that do not exist in the array:
comm -23 <(printf "%s\n" {00..23}{00,30}00) <(printf "%s\n" "${!myarray[#]}" | sort)
# OP has commented that strings not like 'xx[03]000]` should generate an error;
# display strings (extracted from file names) that do not exist in the sequence
comm -13 <(printf "%s\n" {00..23}{00,30}00) <(printf "%s\n" "${!myarray[#]}" | sort)
Where:
comm -23 - display only the lines from the first 'file' that do not exist in the second 'file' (ie, missing sequences of the format xx[03]000)
comm -13 - display only the lines from the second 'file' that do not exist in the first 'file' (ie, filenames with strings not of the format xx[03]000)
These lists could then be used as input to a loop, or passed to xargs, for additional processing as needed; keeping in mind the comm -13 output will display the indices of the array, while the associated contents of the array will contain the name of the original file(s) from which the 6-digit string was derived.
Doing this easy with POSIX shell and only using built-ins:
#!/usr/bin/env sh
# Print an x for each glob matched file, and store result in string_check
string_check=$(printf '%.0sx' ./*[0-2][0-9][03]000*)
# Now string_check length reflects the number of matches
if [ ${#string_check} -eq 48 ]; then
echo "Ok"
else
echo "Problem: an hour (file) is missing"
exit 99
fi
Alternatively:
#!/usr/bin/env sh
if [ "$(printf '%.0sx' ./*[0-2][0-9][03]000*)" \
= 'xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx' ]; then
echo "Ok"
else
echo "Problem: an hour (file) is missing"
exit 99
fi

Iterate Through List with Seq and Variable

I am attempting to loop through a list of integers starting out like so:
start=000
for i in $(seq -w $start 48 006);
However, when I try this code above, the loop seems to loop once and then quit.
What do I need to modify? (The leading zeroes need to stay)
Could you please try following.
start=0
diff=6
for i in $(seq $start $diff 48);
do
printf '%03d\n' $i
done
Output will be as follows.
000
006
012
018
024
030
036
042
048
Problem in OP's tried code:
I believe you have given wrong syntax in seq it should be startpoint then increment_number then endpoint eg-->(seq(start_point increment end_point)). Since you have given them wrongly thus it is printing them only once in loop.
In your attempt it is taking starting point as 0 and should run till 6 with difference of 48 which is NOT possible so it is printing only very first integer value which is fair enough.
EDIT: As per #Cyrus sir's comment adding BASH builtin solution here without using seq.
for ((i=0; i<=48; i=i+6)); do printf '%03d\n' $i; done
seq's input takes a start, increment-by, and finish.
You've reversed the increment-by with finish: seq -w $start 48 006 means start at zero, increment by 48 to finish at 6. The simple fix is seq -w $start 6 48. Note: 006 is not needed, just 6 since seq will equalize the widths of the numbers to two places.

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?
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

Convert string into integer in bash script - "Leading Zero" number error

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"`

12345678 = 123 45 67 8 in bash

I have a script that takes in one big 17 digit number as input from the command line. I want to separate it into 5 different numbers, each with different no. of digits. Like so:
Input: 23063080434560228
Output of the program:
Number1: 23063
Number2: 08
Number3: 04
Number4: 3456
Number5: 0228
Now the output of the program (in terms of digits per number) is fixed, i.e, Number2 will always have 2 digits and so on. Given this sort of a scheme of the output, I am not sure if division is even an option. Is there some bash command/utility that I can use? I have looked it over the net and not come across much.
Thanks,
Sriram.
You can use Substring Extraction:
${string:position:length}
Extracts length characters of substring from string starting at position.
For example:
INPUT=23063080434560228
num1=${INPUT:0:5}
num2=${INPUT:5:2}
num3=${INPUT:7:2}
num4=${INPUT:9:4}
num5=${INPUT:13:4}
You can put this directly into an array in one shot in Bash 3.2 or greater.
string=23063080434560228
pattern='(.{5})(.{2})(.{2})(.{4})(.{4})'
[[ $string =~ $pattern ]] && substrings=(${BASH_REMATCH[#]:1})
for substring in "${substrings[#]}"; do echo "$substring"; done

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