Applescript need to trim printline - applescript

I have script line
set usedSpace to (do shell script "df -hl | grep 'disk1' | awk print $5}'") as integer
it gives me number with % but i want only the number without %

You cannot convert a text value to a number in AppleScript if there are non-numeral characters in the string. You first must remove the unwanted characters, then convert:
set usedSpace to (do shell script "df -hl | grep 'disk1' | awk '{ print $5}'")
set numberOnly to characters 1 thru -2 of usedSpace as text as number

Related

Show with star symbols how many times a user have logged in

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

printing comma-separated integer size of directory and contents

The following will print out the size in bytes of a directory and its contents:
ls -lR | grep -v '^d' | awk '{bytes += $5} END {print "Total bytes: " bytes}'
The output looks like this:
Total bytes: 1088328265
How can I most-simply modify my command so that the output has comma-separated numbers, like this:
Total bytes: 1,088,328,265
$ awk 'BEGIN{printf "Total bytes: %\047d\n", 1088328265}'
Total bytes: 1,088,328,265
So puting aside the usual advice to not parse the output of ls and getting rid of the grep since you never need grep when you're using awk, we can make your whole command:
ls -lR | awk '!/^d/{bytes += $5} END{printf "Total bytes: %\047d\n", bytes}'
\047 is how to represent a single-quote in a single-quote-delimited awk script and then from the GNU awk manual:
A single quote or apostrophe character is a POSIX extension to ISO C. It indicates that the integer part of a floating-point value, or the entire part of an integer decimal value, should have a thousands-separator character in it. This only works in locales that support such characters. For example:
$ cat thousands.awk Show source program
-| BEGIN { printf "%'d\n", 1234567 }
$ LC_ALL=C gawk -f thousands.awk
-| 1234567 Results in "C" locale
$ LC_ALL=en_US.UTF-8 gawk -f thousands.awk
-| 1,234,567 Results in US English UTF locale
For more information about locales and internationalization issues, see Locales.
Using Perl instead of awk:
perl -lane '$bytes += $F[4];
END { substr $bytes, -3 * $_, 0, ","
for reverse 1 .. (length($bytes)-1)/3;
print "Total bytes: $bytes"}'
-l removes newlines from input and adds them to prints
-n reads the input line by line
-a splits the input on whitespace into the #F array
substr inserts spaces to each position; we use negative positions which count from the right, but we start from the leftmost position so the numbers don't change as we add the commas

Get last four characters from a string

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/'

how can I get the index of a character in a given concurrence which is repeated several times in a TEXT line using SHELL (BASH) script

I have a Text string like below
"/path/to/log/file/LOG_FILE.log.2013-10-02-15:2013-10-02 15:46:57.809 INFO - TTT005|Receive|0000293|N~0000284~YOS~TTT005~ ~000~YC~|YOS TYOS-YCUPDT1-H 20131002154657669284YCARR TTT005 Y0TD04 |1|0150520106050|001|051052020603|003|015030010101502702060510520101|000||000|| "
Here "|" is repeated several times within the string and I need to get the index of 4th occurrence of "|" character using shell-script (BASH) command. I tried to find a way using grep command's options.
Thanks.
Using awk you can do:
awk -F '|' '{print index($0, $5)-1}' file
This will print character position of fourth pipe in the file.
grep can print the byte-offset; when used with -o it prints the byte-offset of the matching part.
$ string="/path/to/log/file/LOG_FILE.log.2013-10-02-15:2013-10-02 15:46:57.809 INFO - TTT005|Receive|0000293|N~0000284~YOS~TTT005~ ~000~YC~|YOS TYOS-YCUPDT1-H 20131002154657669284YCARR TTT005 Y0TD04 |1|0150520106050|001|051052020603|003|015030010101502702060510520101|000||000||"
$ grep -ob "[^|]*" <<< "${string}" | sed '5!d' | cut -d: -f1
132
Alternatively, without using grep:
$ newstring=$(echo "${string}" | cut -d\| -f5-)
$ echo $(( ${#string} - ${#newstring} ))
132

What is a unix command for deleting the first N characters of a line?

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

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