I am trying to parse the last 4 characters of Mac serial numbers from terminal. I can grab the serial with this command:
serial=$(ioreg -l |grep "IOPlatformSerialNumber"|cut -d ""="" -f 2|sed -e s/[^[:alnum:]]//g)
but I need to output just the last 4 characters.
Found it in a linux forum echo ${serial:(-4)}
Using a shell parameter expansion to extract the last 4 characters after the fact works, but you could do it all in one step:
ioreg -k IOPlatformSerialNumber | sed -En 's/^.*"IOPlatformSerialNumber".*(.{4})"$/\1/p'
ioreg -k IOPlatformSerialNumber returns much fewer lines than ioreg -l, so it speeds up the operation considerably (about 80% faster on my machine).
The sed command matches the entire line of interest, and replaces it with the last 4 characters before the " that ends the line; i.e., it returns the last 4 chars. of the value.
Note: The ioreg output line of interest looks something like this:
| "IOPlatformSerialNumber" = "A02UV13KDNMJ"
As for your original command: cut -d ""="" is the same as cut -d = - the shell simply removes the empty strings around the = before cut sees the value. Note that cut only accepts a single delimiter char.
You can also do: grep -Eo '.{4}$' <<< "$serial"
I don't know how the output of ioreg -l looks like, but it looks to me that you are using so many pipes to do something that awk alone could handle:
use = as field separator
vvv
awk -F= '/IOPlatformSerialNumber/ { #match lines containing IOPlatform...
gsub(/[^[:alnum:]]/, "", $2) # replace all non alpha chars from 2nd field
print substr($2, length($2)-3, length($2)) # print last 4 characters
}'
Or even sed (a bit ugly one since the repetition of command): catch the first 4 alphanumeric characters occuring after the first =:
sed -rn '/IOPlatformSerialNumber/{
s/^[^=]*=[^a-zA-Z0-9]*([a-zA-Z0-9])[^a-zA-Z0-9]*([a-zA-Z0-9])[^a-zA-Z0-9]*([a-zA-Z0-9])[^a-zA-Z0-9]*([a-zA-Z0-9]).*$/\1\2\3\4/;p
}'
Test
$ cat a
aaa
bbIOPlatformSerialNumber=A_+23B/44C//55=ttt
IOPlatformSerialNumber=A_+23B/44C55=ttt
asdfasd
The last 4 alphanumeric characters between the 1st and 2nd = are 4C55:
$ awk -F= '/IOPlatformSerialNumber/ {gsub(/[^[:alnum:]]/, "", $2); print substr($2, length($2)-3, length($2))}' a
4C55
4C55
Without you posting some sample output of ioreg -l this is untested and a guess but it looks like all you need is something like:
ioreg -l | sed -r -n 's/IOPlatformSerialNumber=[[:alnum:]]+([[:alnum:]]{4})/\1/'
Related
I'm trying to create a simple shell script showing how many times a user has logged in to their linux machine for at least one week. The output of the shell script should be like this:
2021-12-16
****
2021-12-15
**
2021-12-14
*******
I have tried this so far but it shows only numeric but i want showing * symbols.
user="$1"
last -F | grep "${user}" | sed -E "s/${user}.*(Mon|Tue|Wed|Thu|Fri|Sat|Sun) //" | awk '{print $1"-"$2"-"$4}' | uniq -c
Any help?
You might want to refactor all of this into a simple Awk script, where repeating a string n times is also easy.
user="$1"
last -F |
awk -v user="$1" 'BEGIN { split("Jan:Feb:Mar:Apr:May:Jun:Jul:Aug:Sep:Oct:Nov:Dec", m, ":");
for(i=1; i<=12; i++) mon[m[i]] = sprintf("%02i", i) }
$1 == user { ++count[$8 "-" mon[$5] "-" sprintf("%02i", $6)] }
END { for (date in count) {
padded = sprintf("%-" count[date] "s", "*");
gsub(/ /, "*", padded);
print date, padded } }'
The BEGIN block creates an associative array mon which maps English month abbreviations to month numbers.
sprintf("%02i", number) produces the value of number with zero padding to two digits (i.e. adds a leading zero if number is a single digit).
The $1 == user condition matches the lines where the first field is equal to the user name we passed in. (Your original attempt had two related bugs here; it would look for the user name anywhere in the line, so if the user name happened to match on another field, it would erroneously match on that; and the regex you used would match a substring of a longer field).
When that matches, we just update the value in the associative array count whose key is the current date.
Finally, in the END block, we simply loop over the values in count and print them out. Again, we use sprintf to produce a field with a suitable length. We play a little trick here by space-padding to the specified width, because sprintf does that out of the box, and then replace the spaces with more asterisks.
Your desired output shows the asterisks on a separate line from the date; obviously, it's easy to change that if you like, but I would advise against it in favor of a format which is easy to sort, grep, etc (perhaps to then reformat into your desired final human-readable form).
If you have GNU sed you're almost there. Just pipe the output of uniq -c to this GNU sed command:
sed -En 's/^\s*(\S+)\s+(\S+).*/printf "\2\n%\1s" ""/e;s/ /*/g;p'
Explanation: in the output of uniq -c we substitute a line like:
6 Dec-15-2021
by:
printf "Dec-15-2021\n%6s" ""
and we use the e GNU sed flag (this is a GNU sed extension so you need GNU sed) to pass this to the shell. The output is:
Dec-15-2021
where the second line contains 6 spaces. This output is copied back into the sed pattern space. We finish by a global substitution of spaces by stars and print:
Dec-15-2021
******
A simple soluction, using tempfile
#!/bin/bash
user="$1"
tempfile="/tmp/last.txt"
IFS='
'
last -F | grep "${user}" | sed -E "s/"${user}".*(Mon|Tue|Wed|Thu|Fri|Sat|Sun) //" | awk '{print $1"-"$2"-"$4}' | uniq -c > $tempfile
for LINE in $(cat $tempfile)
do
qtde=$(echo $LINE | awk '{print $1'})
data=$(echo $LINE | awk '{print $2'})
echo -e "$data "
for ((i=1; i<=qtde; i++))
do
echo -e "*\c"
done
echo -e "\n"
done
I have a delimited file that is separated by octal \036 or Hexadecimal value 1e.
I need to count the number of delimiters on each line using a bash shell script.
I was trying to use awk, not sure if this is the best way.
Sample Input (| is a representation of \036)
Example|Running|123|
Expected output:
3
awk -F'|' '{print NF-1}' file
Change | to whatever separator you like. If your file can have empty lines then you need to tweak it to:
awk -F'|' '{print (NF ? NF-1 : 0)}' file
You can try
awk '{print gsub(/\|/,"")}'
Simply try
awk -F"|" '{print substr($3,length($3))}' OFS="|" Input_file
Explanation: Making field separator -F as | and then printing the 3rd column by doing $3 only as per your need. Then setting OFS(output field separator) to |. Finally mentioning Input_file name here.
This will work as far as I know
echo "Example|Running|123|" | tr -cd '|' | wc -c
Output
3
This should work for you:
awk -F '\036' '{print NF-1}' file
3
-F '\036' sets input field delimiter as octal value 036
Awk may not be the best tool for this. Gnu grep has a cool -o option that prints each matching pattern on a separate line. You can then count how many matching lines are generated for each input line, and that's the count of your delimiters. E.g. (where ^^ in the file is actually hex 1e)
$ cat -v i
a^^b^^c
d^^e^^f^^g
$ grep -n -o $'\x1e' i | uniq -c
2 1:
3 2:
if you remove the uniq -c you can see how it's working. You'll get "1" printed twice because there are two matching patterns on the first line. Or try it with some regular ascii characters and it becomes clearer what the -o and -n options are doing.
If you want to print the line number followed by the field count for that line, I'd do something like:
$grep -n -o $'\x1e' i | tr -d ':' | uniq -c | awk '{print $2 " " $1}'
1 2
2 3
This assumes that every line in the file contains at least one delimiter. If that's not the case, here's another approach that's probably faster too:
$ tr -d -c $'\x1e\n' < i | awk '{print length}'
2
3
0
0
0
This uses tr to delete (-d) all characters that are not (-c) 1e or \n. It then pipes that stream of data to awk which just counts how many characters are left on each line. If you want the line number, add " | cat -n" to the end.
I'm trying to cut a HDD ID's in sed to just contain the serial number of the drive. The ID's looks like:
t10.ATA_____WDC_WD30EFRX2D68EUZN0_________________________WD2DWMC4N2575116
So, I only want to keep the "WD2DWMC4N2575116". Serial numbers are not fixed length so I tried to keep the last character until the first "_" appears. Unfortunately I suck at RegExp :(
To capture all characters after last _, using backreference:
$ sed 's/.*_\(.*\)/\1/' <<< "t10.ATA_____WDC_WD30EFRX2D68EUZN0_________________________WD2DWMC4N2575116"
WD2DWMC4N2575116
Or as pointed out in comment, you can just remove all characters from beginning of the line up to last _:
sed 's/.*_//' file
echo "t10.ATA_____WDC_WD30EFRX2D68EUZN0_________________________WD2DWMC4N2575116" | rev | awk -F '_' '{print $1}' | rev
It works only if the ID is at the end.
Another in awk, this time using sub:
Data:
$ cat file
t10.ATA_____WDC_WD30EFRX2D68EUZN0_________________________WD2DWMC4N2575116
Code + result:
$ awk 'sub(/^.*_/,"")' file
WD2DWMC4N2575116
ie. replace everything from the first character to the last _. As sub returns the number of substitutions made, that value is used to trigger the implicit output. If you have several records to process and not all of them have _s, add ||1 after the sub:
$ cat foo >> file
$ awk 'sub(/^.*_/,"") || 1' file
WD2DWMC4N2575116
foo
Ok guys I'm really at a dead end here, don't know what else to try...
I am writing a script for some e-mail statistics, one of the things it needs to do is calculate the complete size of all messages in the maillog, this is what I wrote so far:
egrep ' HOSTNAME sendmail\[.*.from=.*., size=' maillog | awk '{print $8}' |
tr "," "+" | tr -cd '[:digit:][=+=]' | sed 's/^/(/;s/+$/)\/1048576/' |
bc -ql | awk -F "." '{print $1}'
And here is a sample line from my maillog:
Nov 15 09:08:48 HOSTNAME sendmail[3226]: oAF88gWb003226:
from=<name.lastname#domain.com>, size=40992, class=0, nrcpts=24,
msgid=<E08A679A54DA4913B25ADC48CC31DD7F#domain.com>, proto=ESMTP,
daemon=MTA1, relay=[1.1.1.1]
So I'll try to explain it step by step:
First I grep through the file to find all the lines containing the actual "size", next i print the 8th field, in this case "size=40992,".
Next I replace all the comma characters with a plus sign.
Then I delete everything except the digits and the plus sign.
Then I replace the beginning of the line with a "(", and I replace the last extra plus sign with a ")" followed by "/1048576". So i get a huge expression looking like this:
"(1+2+3+4+5...+n)/1048576"
Because I want to add up all the individual message sizes and divide it so I get the result in MB.
The last awk command is when I get a decimal number I really don't care for precision so i just print the part before the decimal point.
The problem is, this doesn't work... And I could swear it was working at one point, could it be my expression is too long for bc to handle?
Thanks if you took the time to read through :)
I think a one-line awk script will work too. It matches any line that your egrep pattern matches, then for those lines it splits the eighth record by the = sign and adds the second part (the number) to the SUM variable. When it sees the END of the file it prints out the value of SUM/1048576 (or the byte count in Mibibytes).
awk '/ HOSTNAME sendmail\[.*.from=.*., size=/{ split($8,a,"=") ; SUM += a[2] } END { print SUM/1048576 }' maillog
bc chokes if there is no newline in its input, as happens with your expression. You have to change the sed part to:
sed 's/^/(/;s/+$/)\/1048576\n/'
The final awk will happily eat all your output if the total size is less than 1MB and bc outputs something like .03333334234. If you are not interested in the decimal part remove that last awk command and the -l parameter from bc.
I'd do it with this one-liner:
grep ' HOSTNAME sendmail[[0-9][0-9]*]:..*:.*from=..*, size=' maillog | sed 's|.*, size=\([0-9][0-9]*\), .*|\1+|' | tr -d '\n' | sed 's|^|(|; s|$|0)/1048576\n|' | bc
For example, I might want to:
tail -f logfile | grep org.springframework | <command to remove first N characters>
I was thinking that tr might have the ability to do this but I'm not sure.
Use cut. Eg. to strip the first 4 characters of each line (i.e. start on the 5th char):
tail -f logfile | grep org.springframework | cut -c 5-
sed 's/^.\{5\}//' logfile
and you replace 5 by the number you want...it should do the trick...
EDIT
if for each line
sed 's/^.\{5\}//g' logfile
You can use cut:
cut -c N- file.txt > new_file.txt
-c: characters
file.txt: input file
new_file.txt: output file
N-: Characters from N to end to be cut and output to the new file.
Can also have other args like: 'N' , 'N-M', '-M' meaning nth character, nth to mth character, first to mth character respectively.
This will perform the operation to each line of the input file.
Here is simple function, tested in bash. 1st param of function is string, 2nd param is number of characters to be stripped
function stringStripNCharsFromStart {
echo ${1:$2:${#1}}
}
Usage:
$ stringStripNCharsFromStart "12abcdefgh-" 2
# 2abcdefgh-
Screenshot:
tail -f logfile | grep org.springframework | cut -c 900-
would remove the first 900 characters
cut uses 900- to show the 900th character to the end of the line
however when I pipe all of this through grep I don't get anything
I think awk would be the best tool for this as it can both filter and perform the necessary string manipulation functions on filtered lines:
tail -f logfile | awk '/org.springframework/ {print substr($0, 6)}'
or
tail -f logfile | awk '/org.springframework/ && sub(/^.{5}/,"",$0)'
x=hello
echo ${x:1}
returns ello
replace 1 with N as required