I'm trying to subtract the first day of class (17Feb13 day#47) from the current day (24Feb13 day#55) The command date +%j yields 055, instead of 55 so the math errors...
todayIn=$(date +%j)................= 055
firstDay=(47)..........................= 47
myVal=$(todayIn-firstDay)....= error (expecting 8)
It seems like it should be enough to just strip the leading zero(es) with:
todayIn=$(date '+%j' | sed -e 's/^0*//')
Or use expr:
$ a=$(expr 055 - 5)
$ echo $a
$ 50
Related
Sorry in advance for the beginner question, but I'm quite stuck and keen to learn.
I am trying to echo a string (in hex) and then cut a piece of that with cut command. It looks like this:
for y in "${Offset}"; do
echo "${entry}" | cut -b 60-$y
done
Where echo ${Offset} results in
75 67 69 129 67 567 69
I would like each entry to be printed, and then cut from the 60th byte until the respective number in $Offset.
So the first entry would be cut 60-75.
However, I get an error:
cut: 67: No such file or directory
cut: 69: No such file or directory
cut: 129: No such file or directory
cut: 67: No such file or directory
cut: 567: No such file or directory
cut: 69: No such file or directory
I tried adding/removing parentheses around each variable but never got the right result.
Any help will be appreciated!
UPDATE: updated the code with changed from markp-fuso. However, this codes still does not work as intended. I would like to print every entry based on the respective offset, but it goes wrong. This prints every entry seven times, where each time is based on seven different offsets. Any ideas on how to fix this?
#!/bin/bash
MESSAGES=$( sqlite3 -csv file.db 'SELECT quote(data) FROM messages' | tr -d "X'" )
for entry in ${MESSAGES}; do
Offset='75 67 69 129 67 567 69'
for y in $Offset; do
echo "${entry:59:(y-59)}"
done
done
echo ${MESSAGES}
Results in seven strings with minimal length 80 bytes and max 600.
My output should be:
String one: cut by first offset
String two: cut by second offset
and so on...
In order for for to iterate over each space-separated "word" in $Offset, you need to get rid of the quotes, which are making it read as a single variable.
for y in ${Offset}; do
echo "${entry}" | cut -b 60-$y
done
To eliminate the sub-process that's going to be invoked due to the | cut ..., we could look at a comparable parameter expansion solution ...
Quick reminder on how to extract a substring from a variable:
${variable:start_position:length}
Keeping in mind that the first character in ${variable} is in position zero/0.
Next, we need to convert each individual offset (y) into a 'length':
length=$((y-60+1))
Rolling these changes into your code (and removing the quotes from around ${Offset}) gives us:
for y in ${Offset}
do
start=$((60-1))
length=$((y-60+1))
echo "${entry:${start}:${length}}"
#echo "${entry:59:(y-59)}"
done
NOTE: You can also replace the start/length/echo with the single commented-out echo.
Using a smaller data set for demo purposes, and using 3 (instead of 60) as the start of our extraction:
# base-10 character position
# 1 2
# 123456789012345678901234567
$ entry='123456789ABCDEFGHIabcdefghi'
$ echo ${#entry} # length of entry?
27
$ Offset='5 8 10 13 20'
$ for y in ${Offset}
do
start=$((3-1))
length=$((y-3+1))
echo "${entry:${start}:${length}}"
done
345 # 3-5
345678 # 3-8
3456789A # 3-10
3456789ABCD # 3-13
3456789ABCDEFGHIab # 3-20
And consolidating the start/length/echo into a single echo:
$ for y in ${Offset}
do
echo "${entry:2:(y-2)}"
done
345 # 3-5
345678 # 3-8
3456789A # 3-10
3456789ABCD # 3-13
3456789ABCDEFGHIab # 3-20
I am new to bash. I have experience in java and python but no experience in bash so I'm struggling with the simplest of tasks.
What I want to achieve is I want to look through the string and find certain sub strings, numbers to be exact. But not all numbers just number that are followed by " xyz". For example:
string="Blah blah boom boom 14 xyz foo bar 12 foo boom 55 XyZ hue hue 15 xyzlkj 45hh."
And I want to find numbers:
14 55 and 15
How would I go about that?
You can use grep with lookahead
echo "$string" | grep -i -P -o '[0-9]+(?= xyz)'
Explanation:
-i – ignore case
-P – interpret pattern as a Perl regular expression
-o – print only matching
[0-9]+(?= xyz) – match one or more numbers followed by xyz
For more information see:
https://linux.die.net/man/1/grep
http://www.regular-expressions.info/lookaround.html
https://github.com/tldr-pages/tldr/blob/master/pages/common/grep.md
grep + cut approach (without PCRE):
echo $string | grep -io '[0-9]* xyz' | cut -d ' ' -f1
The output:
14
55
15
I am new to shell script
I Have tried to multiply two hex numbers in shell script in the following manner.
initial= expr 0x10000 \* 0x22
echo $initial
While running the script,The following error is seen.
expr: non-numeric argument
Can someone point out what might the mistake?
No need to expr, use $(( )) just like this:
$ echo $((0x10000 * 0x22))
2228224
Or you can use bc like this, indicating input is hex (ibase) and desired output also in hex (obase) (as Adobe's deleted answer states):
$ echo "ibase=16; obase=16; 10000*22" | bc
09 11 05 16 20
$ echo "ibase=16; 10000*22" | bc
2228224
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"`
I am trying to convert all negative numbers to positive numbers and have so far come up with this
echo "-32 45 -45 -72" | sed -re 's/\-([0-9])([0-9])\ /\1\2/p'
but it is not working as it outputs:
3245 -45 -72
I thought by using \1\2 I would have got the positive number back ?
Where am I going wrong ?
Why not just remove the -'s?
[root#vm ~]# echo "-32 45 -45 -72" | sed 's/-//g'
32 45 45 72
My first thought is not using sed, if you don't have to. awk can understand that they're numbers and convert them thusly:
echo "-32 45 -45 -72" | awk -vRS=" " -vORS=" " '{ print ($1 < 0) ? ($1 * -1) : $1 }'
-vRS sets the "record separator" to a space, and -vORS sets the "output record separator" to a space. Then it simply checks each value, sees if it's less than 0, and multiplies it by -1 if it is, and if it's not, just prints the number.
In my opinion, if you don't have to use sed, this is more "correct," since it treats numbers like numbers.
This might work for you:
echo "-32 45 -45 -72" | sed 's/-\([0-9]\+\)/\1/g'
Reason why your regex is failing is
Your only doing a single substitution (no g)
Your replacement has no space at the end.
The last number has no space following so it will always fail.
This would work too but less elegantly (and only for 2 digit numbers):
echo "-32 45 -45 -72" | sed -rn 's/-([0-9])([0-9])(\s?)/\1\2\3/gp'
Of course for this example only:
echo "-32 45 -45 -72" | tr -d '-'
You are dealing with numbers as with a string of characters. More appropriate would be to store numbers in an array and use built in Shell Parameter Expansion to remove the minus sign:
[~] $ # Creating and array with an arbitrary name:
[~] $ array17=(-32 45 -45 -72)
[~] $ # Calling all elements of the array and removing the first minus sign:
[~] $ echo ${array17[*]/-}
32 45 45 72
[~] $