ls sort alphabetically
char - ascii is 45
char 9 ascii is 57
so file - sort before 9
but why file 999-a sort after 9999a
$ ls -1
-
9
9999a
999-a
So, I basically want to store sorted array/data into another array and use that data to print something else?
Even when I want to have the footer, sorted data is printed after the footer.
printf (" %-25s %-20s %d\n", employee_name[working_employee_id[y]], title[employee_name[working_employee_id[y]]], salary[employee_name[working_employee_id[y]]]) | "sort -nr -k2"
I want to print other things after the execution of this line instead of letting sort to print at the end
You need to close() the pipe at the end of your input before printing anything else if you want to make sure the command you're piping to finishes displaying all its output before your footer text.
Example:
$ paste <(seq 10 | shuf) <(seq 10 | shuf) |
awk '{ printf "%d\t%d\t%d\n", $1, $2, $1 + $2 | "sort -k1,1n" }
END { close("sort -k1,1n"); print "a\tb\tc" }'
1 8 9
2 3 5
3 6 9
4 4 8
5 2 7
6 10 16
7 9 16
8 1 9
9 7 16
10 5 15
a b c
How to print from 3rd column to till last columns using awk command in unix, if there are 'n' columns in a file. I am getting with cut command but I need awk command. I am trying to do with awk -F " " '{ for{i=3;i<=NF;i++) print $i}', I am getting the output but it is not in the correct format. Can anyone suggest me the proper command.
Combining Ed Morton's answers in:
Print all but the first three columns
delete a column with awk or sed
We get something like this:
awk '{sub(/^(\S+\s*){2}/,""); sub(/(\s*\S+){2}$/,"")}1'
# ^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^
# remove 2 first cols remove 2 last cols
Which you can adapt to your exact needs in terms of columns.
See an example given this input:
$ paste -d ' ' <(seq 5) <(seq 2 6) <(seq 3 7) <(seq 4 8) <(seq 5 9)
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
Let's just print the 3rd column:
$ awk '{sub(/^(\S+\s*){2}/,""); sub(/(\s*\S+){2}$/,"")}1' <(paste -d ' ' <(seq 5) <(seq 2 6) <(seq 3 7) <(seq 4 8) <(seq 5 9))
3
4
5
6
7
Your attempt was close but appears that it would print each and every column on a new line.
To correct this we create a variable called 'line' and initialize it to an empty string. The first time we are in the loop we just add the column to 'line'. From that point on we will append to 'line' with the field separator and the next column. Finally, we print 'line'. This will happen for each line in the file.
awk '{line="";for(i=3;i<=NF;i++) if(i==3) line=$i; else line=line FS $i; print line}'
In this example I assume to use awk's default field separator. Also any lines that are less than three will print blank lines.
Assuming your fields are space-separated then with GNU awk for gensub():
$ cat file
a b c d e f
g h i j k l
$ awk '{print gensub(/(\S\s){2}/,"",1)}' file
c d e f
i j k l
In general to print from, say, field 3 to field 5 if they are blank separated using GNU awk again with gensub():
$ awk '{print gensub(/(\S\s){2}((\S\s){2}\S).*/,"\\2",1)}' file
c d e
i j k
or the 3rd arg to match():
$ awk 'match($0,/(\S\s){2}((\S\s){2}\S)/,a){print a[2]}' file
c d e
i j k
or in general if they are separated by any single character:
$ awk '{print gensub("([^"FS"]"FS"){2}(([^"FS"]"FS"){2}[^"FS"]).*","\\2",1)}' file
c d e
i j k
$ awk 'match($0,"([^"FS"]"FS"){2}(([^"FS"]"FS"){2}[^"FS"])",a){print a[2]}' file
c d e
i j k
If the fields are separated by a string instead of a single-character but the RS is a single character then you should temporarily change FS to RS (which by definition you KNOW can't be present in the record) so you can negate it in the bracket expressions:
$ cat file
aSOMESTRINGbSOMESTRINGcSOMESTRINGdSOMESTRINGeSOMESTRINGf
gSOMESTRINGhSOMESTRINGiSOMESTRINGjSOMESTRINGkSOMESTRINGl
$ awk -F'SOMESTRING' '{gsub(FS,RS)} match($0,"([^"RS"]"RS"){2}(([^"RS"]"RS"){2}[^"RS"])",a){gsub(RS,FS,a[2]); print a[2]}' file
cSOMESTRINGdSOMESTRINGe
iSOMESTRINGjSOMESTRINGk
If both the FS and the RS are multi-char then there's various options but the simplest is to use the NUL character or some other character you know can't appear in your input file instead of RS as the temporary replacement FS:
$ awk -F'SOMESTRING' '{gsub(FS,"\0")} match($0,/([^\0]\0){2}(([^\0]\0){2}[^\0])/,a){gsub("\0",FS,a[2]); print a[2]}' file
cSOMESTRINGdSOMESTRINGe
iSOMESTRINGjSOMESTRINGk
Obviously change FS to OFS in the final gsub()s above if desired.
If the FS was a regexp instead of a string and you want to retain it in the output then you need to look at GNU awk for the 4th arg for split().
If you don't mind normalizing the space, the most straightforward way is
$ awk '{$1=$2=""}1' | sed -r 's/^ +//'
in action
$ seq 11 40 | pr -6ts' ' | awk '{$1=$2=""}1' | sed -r 's/^ +//'
21 26 31 36
22 27 32 37
23 28 33 38
24 29 34 39
25 30 35 40
for the input
$ seq 11 40 | pr -6ts' '
11 16 21 26 31 36
12 17 22 27 32 37
13 18 23 28 33 38
14 19 24 29 34 39
15 20 25 30 35 40
To print from third column to till end then
cat filename|awk '{for(i=1;i<3;i++) $i="";print $0}'
I'm trying to encode a TCP header myself, but can't understand what is the right order of bits/octets in it. This is what RFC 793 says:
0 1 2 3
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Source Port | Destination Port |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Sequence Number |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
...
This means that Source Port should take first two octets and the lowest bit should be in the first octet. This means to me that in order to encode source port 180 I should start my TCP header with these two bytes:
B4 00 ...
However, all examples I can find tell me to do it the other way around:
00 B4 ...
Why?
This means that Source Port should take first two octets
Correct.
and the lowest bit should be in the first octet.
Incorrect. It doesn't mean that. It doesn't say anything about it.
All multi-byte integers in all IP headers are represented in network byte order, which is big-endian. This is specified in RFC 1700.
I am working on an NTP Client. A few other threads indicate that the a message containing "\x1b' + 47 * '\0" is sent to the NTP server, but none of these threads give an explanation about what this message actually means or why it is sent. I've tried looking at the NTP RFC but I was unable to find any information about it in there either.
"\x1b' + 47 * '\0" represents a data field of 48 bytes. 0x1B followed by 47 times
0. 48 bytes is the size of an NTP UDP packet. The first byte (0x1B) specifies LI, VN, and Mode.
RFC 5905 NTP Specification (7.3. Packet Header Variables) specifies the message header as follows:
0 1 2 3
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
|LI | VN |Mode | Stratum | Poll | Precision |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Setting the first byte of the data to 0x1B or 00 011 011 means
LI = 0 (Leap indicator)
VN = 3 (Version number)
Mode = 3 (Mode, mode 3 is client mode)
You may also use the more recent version (VN = 4). This would require the first header byte to be set to
0x23 (00 100 011).
The modes are defined as
+-------+--------------------------+
| Value | Meaning |
+-------+--------------------------+
| 0 | reserved |
| 1 | symmetric active |
| 2 | symmetric passive |
| 3 | client |
| 4 | server |
| 5 | broadcast |
| 6 | NTP control message |
| 7 | reserved for private use |
+-------+--------------------------+
Specifying Mode = 3 indicates the message as a client request message.
Sending such a packet to port 123 of an NTP server will force the server to send a reply package.