Need help in solving an expression in shell - unix

How to calculate X power Y in the unix shell script where value of Y is being supplied by a loop eg. Y=1 to 5. It means I would like to calculte (X^Y)

In bash you could do:
$ for i in {1..5}; do printf "$((2 ** $i))\n"; doneprintf "$((2 ** $i))\n"; done
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32
Many shells however do not support raise to power operation and in
such situations you need to use bc:
$ for i in $(seq 5); do printf "%s\n" "$(echo "2 ^ $i" | bc)"; done
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16
32

Related

How to Remove Code Specific Code Lines using Unix

Could someone please help/advise how could I removed the first 4 line and the last 2 line of codes in my 3 JavaScript files using the Shell Script?
I tried using this guide: UNIX - delete specific lines but it will only work for the first 4 lines. All 3 Javascript files have different set of line of codes.
set -vx
lines2del="(1,2,3,4)"
sedCmds=${lines2del//,/d;}
sedCmds=${sedCmds/(/}
sedCmds=${sedCmds/)/}
sedCmds=${sedCmds}d
sed -i "$sedCmds" file
Any inputs are highly appreciated. Thanks
This might work for you (GNU sed):
sed -i '1,4d;N;$d;P;D' file
This deletes the lines 1 to 4 and then prints all other lines except the last two which it also deletes.
Add the following to your lines2del:
$(($(cat file | wc -l)-2)) // third last line
$(($(cat file | wc -l)-1)) // second last line
$(cat file | wc -l) // last line
$ seq 10 | tail -n +5 | head -n -2
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$ seq 10 | awk '{p3=p2; p2=p1; p1=$0} NR>6{print p3}'
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$ seq 10 | awk '{p[NR%6]=$0} NR>6{print p[(NR-2)%6]}'
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$ seq 10 | awk -v b=4 -v a=2 'BEGIN{t=b+a} {p[NR%t]=$0} NR>t{print p[(NR-a)%t]}'
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$ seq 10 | awk -v b=3 -v a=5 'BEGIN{t=b+a} {p[NR%t]=$0} NR>t{print p[(NR-a)%t]}'
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5

Bash. Hex to ascii. Is possible without xxd or perl?

I'm developing a script on which I have a hex string 31323334353637383930313233 and I want to transform it into ASCII. Desired output is 1234567890123.
I already have it working using:
echo "31323334353637383930313233" | xxd -r -p
or
echo "31323334353637383930313233" | perl -pe 's/(..)/chr(hex($1))/ge'
But the point is try to use the minimum possible requirements for the script. I want it working in suse, fedora, debian, ubuntu, arch, etc... It seems the xxd command is included in vim package. I'm wondering if there is a way to achieve this using only awk or any internal Linux tool which is going to be present by default in all Linux systems.
Found this script here:
#!/bin/bash
function hex2string () {
I=0
while [ $I -lt ${#1} ];
do
echo -en "\x"${1:$I:2}
let "I += 2"
done
}
hex2string "31323334353637383930313233"
echo
You may change the line hex2string "31323334353637383930313233" so that it takes the hex value from parameters, that is:
#!/bin/bash
function hex2string () {
I=0
while [ $I -lt ${#1} ];
do
echo -en "\x"${1:$I:2}
let "I += 2"
done
}
hex2string "$1"
echo
So when executed as:
./hexstring.sh 31323334353637383930313233
It will provide the desired ascii output.
NOTE: Can't test if it works in all Linux systems.
Using gawk, from HEX to ASCII
$ gawk '{
gsub(/../,"0x& ");
for(i=1;i<=NF;i++)
printf("%c", strtonum($i));
print ""
}' <<<"31323334353637383930313233"
1234567890123
Using any awk
$ cat hex2asc_anyawk.awk
BEGIN{
split("0 1 2 3 4 5 6 7 8 9 A B C D E F", d, / /)
for(i in d)Decimal[d[i]]=i-1
}
function hex2dec(hex, h,i,j,dec)
{
hex = toupper(hex);
i = length(hex);
while(i)
{
dec += Decimal[substr(hex,i,1)] * 16 ^ j++
i--
}
return dec;
}
{
gsub(/../,"& ");
for(i=1;i<=NF;i++)
printf("%d",hex2dec($i));
print ""
}
Execution
$ awk -f hex2asc_anyawk.awk <<<"31323334353637383930313233"
1234567890123
Explanation
Steps :
Get the decimal equivalent of hex from table.
Multiply every digit with 16 power of digit location.
Sum all the multipliers.
Example :
BEGIN{
# Here we created decimal conversion array, like above table
split("0 1 2 3 4 5 6 7 8 9 A B C D E F", d, / /)
for(i in d)Decimal[d[i]]=i-1
}
function hex2dec(hex, h,i,j,dec)
{
hex = toupper(hex); # uppercase conversion if any A,B,C,D,E,F
i = length(hex); # length of hex string
while(i)
{
# dec var where sum is stored
# substr(hex,i,1) gives 1 char from RHS
# multiply by 16 power of digit location
dec += Decimal[substr(hex,i,1)] * 16 ^ j++
i-- # decrement by 1
}
return dec;
}
{
# it modifies record
# suppose if given string is 31323334353637383930313233
# after gsub it becomes 31 32 33 34 35 36 37 38 39 30 31 32 33
# thus re-evaluate the fields
gsub(/../,"& ");
# loop through fields , NF gives no of fields
for(i=1;i<=NF;i++)
# convert from hex to decimal
# and print equivalent ASCII value
printf("%c",hex2dec($i));
# print newline char
print ""
}
Meaning of dec += Decimal[substr(hex,i,1)] * 16 ^ j++
dec += Decimal[substr(hex,i,1)] * 16 ^ j++
^ ^ ^
| | |
| | 2.Multiply every digit with 16 power of digit location.
| |
| 1.Gives decimal equivalent of hex
|
|
3. Sum all the multipliers
here's a special cheating trick for u - due to ingenuity of how they originally mapped decimal digits to bytes, their hex are all x3[0-9],
so therefore, if u already know they would decode out to digits and nothing else, here's a fast shortcut :
echo "31323334353637383930313233" |
mawk 'gsub("..","_&") + gsub("_3",_)^_'
1234567890123
if it's already URL-percent-encoded, then it's even simpler :
echo '%31%32%33%34%35%36%37%38%39%30%31%32%33' |
mawk 'gsub("%3",_)^_'
or
gawk ++NF FS='%3' OFS=
1234567890123
This specialized approach can handle hex of absolutely any arbitrary size, even for awks that don't have built-in support for bigints
TL;DR : don't "do math" when none is needed
Alternate (g)awk solution:
echo "31323334353637383930313233" | awk 'RT{printf "%c", strtonum("0x"RT)}' RS='[0-9]{2}'

Misunderstanding While loop [duplicate]

This question already has answers here:
Why should there be spaces around '[' and ']' in Bash?
(5 answers)
Closed 6 years ago.
Having trouble getting this to exit the loop and perhaps I don't understand the while loop well enough to get it to stop. This is what I'm typing in my command line. I expect it to stop at 10.
c=1; while : [[$c -le 10]]; do df -Th; echo "$c"; date; c=`expr $((c+=1))`; sleep 6; done
I've already read article but it didn't seem to help: bash while loop won't stop itself
Final code:
c=1; while [[ $c -le 10 ]]; do df -Th; echo "$c"; date; c=$((c+=1)); sleep 6; done
The final code required removing the : and adding between the [[$c and 10]]
In bash, spaces are important. Replace:
: [[$c -le 10]]
With
[[ $c -le 10 ]]
What went wrong
When bash encounters
: [[$c -le 10]]
it executes the command : with three arguments, [[$c, -le, and 10]]. This is not what you want. First, : is the no-op command: regardless of its arguments, it does nothing. Second, if you want to run the [[ test command, it needs, as shown above, spaces around it.
Alternatives
There are several ways to loop something 10 times in bash. For one:
$ for c in {1..10}; do echo "$c"; done
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For another:
$ for ((c=1;c<=10;c++)); do echo "$c"; done
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UNIX (AIX) Command Help - Sed & Awk

I'm running this on an AIX 6.1.
The intended purpose of this command is to display the following information in the following format:
GetUsedRAM:GetUsedSwap:CPU_0_System:CPU_0_User:…CPU_N_System:CPU_N_User
The command is composed of several sub commands:
echo `vmstat 1 2 | tr -s ' ' ':' | cut -d':' -f4,5,14-15 | tail -1 | sed 's/\([0-9]*:[0-9]*:\)\([0-9]*:[0-9]*\)/\1/'``mpstat -a 1 1 | tr -s ' ' '|' | head -8 | tail -4 | cut -d'|' -f 25,27 | awk -F "|" '{printf "%.0f:%.0f:",$2,$1}' | sed '$s/.$//'| sed -e "s/ \{1,\}$//"| awk '{int a[10];split($1, a,":");printf("%d:%d:%d:%d:%d:%d:%d:%d",a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7])}'`
Which I'll re format for clarity:
echo \
`vmstat 1 2 |
tr -s ' ' ':' |
cut -d':' -f4,5,14-15 |
tail -1 |
sed 's/\([0-9]*:[0-9]*:\)\([0-9]*:[0-9]*\)/\1/' \
` \
`mpstat -a 1 1 |
tr -s ' ' '|' |
head -8 |
tail -4 |
cut -d'|' -f 25,27 |
awk -F "|" '{printf "%.0f:%.0f:",$2,$1}' |
sed '$s/.$//' |
sed -e "s/ \{1,\}$//" |
awk '{int a[10];split($1, a,":");printf("%d:%d:%d:%d:%d:%d:%d:%d",a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7])}' \
`
I understand all of the tr, cut, head tail, and (roughly) vmstat/mpstat commands. The first sed is where I get lost, I've tried running the command in smaller segments and not quite sure why it seems to work as a whole but not when I truncate the command before the next tr.
I'm also not so sure on the awk command although I understand the premise vaguely, as a function allowing formatted output.
Similarly, I have a vague understanding of sed being a command allowing certain strings/characters being replaced in some file.
I'm not able to make out what this specific implementation in the above case is.
Could anyone provide some clarity or direction as to exactly what is happening at each sed and awk step within the context of the entire command?
Thanks for your help.
Simplification
This two simpler commands will get the exact same output:
# GetUsedRAM:GetUsedSwap:CPU_0_System:CPU_0_User:…CPU_N_System:CPU_N_User
# Select fields 4,5 of last line, and format with :
comm1=`vmstat 1 2 |
awk '$4~/[0-9]/{avm=$4;fre=$5} END{printf "%s:%s",avm,fre}'
`
# Select fields 27 (sy) and 25 (us) for four cpu, print as decimal.
comm2=`mpstat -A 1 1 |
awk -v firstline=6 -v cpus=4 '
BEGIN{start=firstline-1; end=firstline+cpus;}
NR>start && NR<end {printf( ":%d:%d", $27,$25)}'
`
echo "${comm1}${comm2}"
Description.
Description of original commands
The whole command is the concatenation of two commands.
The first command:
The output of the vmstat is shown in this link.
The columns 4 and 5 are 'avm' and 'fre'. The output in columns 14 and 15,
seem to be 'us' (user) and 'sy' (system). And I say seem as no output
from the user is available to confirm.
The first command
`vmstat 1 2 | # Execute the command vmstat.
tr -s ' ' ':' | # convert all spaces to colon (:).
cut -d':' -f4,5,14-15 | # select fields 4,5,14,and 15
tail -1 | # select last line.
sed 's/\([0-9]*:[0-9]*:\)\([0-9]*:[0-9]*\)/\1/' \ # See below.
`
The sed command selects inside braces all digits [0-9]* before a colon
repeated twice. And then again (without the last colon). That's the whole
string in two parts: « (dd:dd:)(dd:dd) » (d means digit).
And finally, it replaces such whole string by what was selected inside
the first braces /\1/.
All this complexity just removes fields 14 and 15 as selected by cut.
A simpler command with exactly the same output is:
Select fields 4,5 of last line, and format with (:).
`vmstat 1 2 | awk '
$4~/[0-9]/{avm=$4;fre=$5} END{printf "%s:%s:",avm,fre}'
`
The second command:
The output of mpstat -A is similar to this one from Linux.
And also similar to this AIX mpstat -d output.
However, the exact output of AIX 6.1 for mpstat -a (ALL) on the computer
used could have several variations. Anyway, guided by the intended final
output desired: CPU_0_System:CPU_0_User:…CPU_N_System:CPU_N_User.
It seems that the columns to be selected should be us (user) and sy
(sys) percent of time that used the cpu for all cpu in use,
which seem to be four on the computer measured.
The manual for AIX 6.1 mpstat is here.
It has a list of all the 40 columns that are presented when the option
-a ALL is used:
CPU min maj mpcs mpcr dev soft dec ph cs ics bound rq push
S3pull S3grd S0rd S1rd S2rd S3rd S4rd S5rd S3hrd S4hrd S5hrd
sysc us sy wa id pc %ec ilcs vlcs lcs %idon %bdon %istol %bstol %nsp
us and sy are listed as the fields 27 and 28, however the command presented
by the user selects fields number 25 and 27. Close but not the same. The
only way to confirm would be to receive the output of the command from the user.
For testing I will be using the output of mpstat 5 1 from here.
# mpstat 5 1
System configuration: lcpu=4 ent=1.0 mode=Uncapped
cpu min maj mpc int cs ics rq mig lpa sysc us sy wt id pc %ec lcs
0 4940 0 1 632 685 268 0 320 100 263924 42 55 0 4 0.57 35.1 277
1 990 0 3 1387 2234 805 0 684 100 130290 28 47 0 25 0.27 16.6 649
2 3943 0 2 531 663 223 0 389 100 276520 44 54 0 3 0.57 34.9 270
3 1298 0 2 1856 2742 846 0 752 100 82141 31 40 0 29 0.22 13.4 650
ALL 11171 0 8 4406 6324 2142 0 2145 100 752875 39 51 0 10 1.63 163.1 1846
The second command
`mpstat -A 1 1 | # execute command
tr -s ' ' '|' | # replace all spaces with (|).
head -8 | # select 8 first lines.
tail -4 | # select last four lines.
cut -d'|' -f 25,27 | # select fields 25 and 27
awk -F "|" '{printf "%.0f:%.0f:",$2,$1}' | # print the fields as integers.
sed '$s/.$//' | # on the last line ($), substitute the last character (.$) by nothing.
sed -e "s/ \{1,\}$//" | # remove trailing space(s).
awk '{
int a[10];
split($1, a,":");
printf("%d:%d:%d:%d:%d:%d:%d:%d",a[0],a[1],a[2],a[3],a[4],a[5],a[6],a[7])
}' \
`
About the int: For older versions of awk, calling a function without the parentheses is equivalent to call the function on $0. int is equivalent to int($0), which is not printed, nor used. The same happens to the value of a[10].
The split sets each value of the command in a[i]. Then, all values of a[i] are printed as decimals.
The equivalent, and way simpler is:
Command #2
`mpstat -A 1 1 |
awk -v firstline=6 -v cpus=4 '
BEGIN{start=firstline-1; end=firstline+cpus;}
NR>start && NR<end {printf( ":%d:%d", $27,$25)}'
`

How can I get a range of line every nth interval using awk, sed, or other unix command?

I know how to get a range of lines by using awk and sed.
I also do know how to print out every nth line using awk and sed.
However, I don't know how to combined the two.
For example, I have a file with 1780000 lines.
For every 17800th line, I would like to print 17800th line plus the two after that.
So if I have a file with 1780000 lines and it starts from 1 and ends at 1780000, this will print:
1
2
3
17800
17801
17802
35600
35601
35602
# ... and so on.
Does anyone know how to get a range of line every nth interval using awk, sed, or other unix command?
Using GNU sed:
sed -n '0~17800{N;N;p}' input
Meaning,
For every 17800th line: 0~17800
Read two lines: {N;N;
And print these out: p}
We can also add the first three lines:
sed -n -e '1,3p' -e '0~17800{N;N;p}' input
Using Awk, this would be simpler:
awk 'NR%17800<3 || NR==3 {print}' input
$ cat file
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$ awk '!(NR%3)' file
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$ awk -v intvl=3 -v delta=2 '!(NR%intvl){print "-----"; c=delta} c&&c--' file
-----
3
4
-----
6
7
-----
9
10
$ awk -v intvl=4 -v delta=2 '!(NR%intvl){print "-----"; c=delta} c&&c--' file
-----
4
5
-----
8
9
$ awk -v intvl=4 -v delta=3 '!(NR%intvl){print "-----"; c=delta} c&&c--' file
-----
4
5
6
-----
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10
seq -f %.0f 1780000 | awk 'NR < 4 || NR % 17800 < 3' | head
output:
1
2
3
17800
17801
17802
35600
35601
35602
53400
Explanation
The NR < 4 is for the first 3 lines because the requirement For every 17800th line, print 17800th line plus the two after that. doesn't fit the output you gave.
Here I use head for reducing the output size and you should remove it in your use case.
For GNU seq, you don't need -f %.0f.

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