I have two questions. The first is that sometimes when I am coding in Unix and I input a command and I do it wrong I get a new line without my prompt and no matter what I type, nothing happens until I exit out and re-enter. Does anyone know why this is happening?
Secondly,
I have a file that consists of: filename space data
I need to get the data, I heard that I should use awk or sed but I am not sure how to do it. Any help is welcome.
Dennis has already answered your first question well. (Note: please put only one question in at a time!)
For your second question, it can be done much more simply.
awk '{ print $2 }' yourfile
By default, awk uses space as its column delimiter, so this simply tells awk to print out the second column. If you want the output sent to a new file, then just do this:
awk '{ print $2 }' yourfile > newfile
First question:
Enter echo " and it will happen. Unix supports multi-line commands.
Example:
echo "
is a multi-line
command"
Type in " and enter to terminate.
Second question:
Here's a link to a nice AWK tutorial: Awk - A Tutorial and Introduction
Basicaly, you use
awk '{ print "echo " $2 }' filename | sh
for example, to echo all the data.
$2 accesses the second chunk of information of each line (chunks are seperates by spaces).
print "echo " $2 will cause awk to output echo data.
Last, you pipe to sh to execute the command of awk's output.
Related
I currently have the following line of code:
ls /some/dir/prefix* | sed -e 's/prefix.//' | tr '\n' ' '
Which does achieve what I want it to do:
Get list of files starting with prefix
Remove path and prefix from each string
Remove newlines and replace with spaces for later processing.
For example:
/some/dir/prefix.hello
/some/dir/prefix.world
Should become
hello world
But I feel like there's a nicer way of doing this. Is there a better way to do this in one line?
Here is a two-liner using just built-ins that does it:
fnames=(some/dir/prefix*)
echo "${fnames[#]##*.}"
And here's how this works:
fnames=(some/dir/prefix*) creates an array with all the files starting with prefix and avoids all the problems that come with parsing ls
echo "${fnames[#]##*.}" is a combination of two parameter expansions: ${fnames[#]} prints all array elements, and the ##*. part removes the longest match of anything that ends with . from each array element, leaving just the extension
If you're hell-bent on a one-liner, just join the two commands with &&.
passing ls output to external programs is not recommended, following bash solution may help you here.
for file in prefix*; do echo ${file##*.}; done
Adding a non-one liner form of solution too now.
for file in prefix*
do
echo ${file##*.}
done
Here is a very simple Awk one-liner to achieve this :
awk -F. '{$0=FILENAME; printf $NF" "; nextfile}' /some/dir/prefix*
It essentially does the following :
-F.: Set the field separator FS to a .. This way $NF represents the extension.
$0=FILENAME: Ignore the current record and set it to FILENAME, reparse everything this way.
print $NF; nextfile : print the extension and go to the next file.
The problem with this is that the file still reads a record of the current file. If that file is empty this will fail.
To make this work with empty files, you could use the gawk extension BEGINFILE
awk -F. 'BEGINFILE{$0=FILENAME; printf $NF" "; nextfile}' /some/dir/prefix*
Or you can loop over all the arguments :
awk -F. 'BEGIN{for(i in ARGV){$0=ARGV[i]; printf $NF" "};exit}' /some/dir/prefix*
One approach with awk:
ls /some/dir/prefix* | awk -F"." '{printf "%s ", $2} END {print ""}'
It might qualify as being "nicer" because there's only one command the output is piped through?!
I'm on a Mac, and I want to find a field in a CSV file adjacent to a search string
This is going to be a single file with a hard path; here's a sample of it:
84:a5:7e:6c:a6:b0, AP-ATC-151g84
84:a5:7e:6c:a6:b1, AP-A88-131g84
84:a5:7e:73:10:32, AP-AG7-133g56
84:a5:7e:73:10:30, AP-ADC-152g81
84:a5:7e:73:10:31, AP-D78-152e80
so if my search string is "84:a5:7e:73:10:32"
I want to get returned "AP-AG7-133g56"
I had been working within an Applescript, but maybe a shell script will do.
I just need the proper syntax for opening the file and having awk search it. Again, I'm weak conceptually on how shell commands run, how they must be executed, etc
This errors, gives me ("command not found"):
set the_file to "/Users/Paw/Desktop/AP-Decoder 3.app/Contents/Resources/BSSIDtable.csv"
set the_val to "70:56:81:cb:a2:dc"
do shell script "'awk $1 ~ the_val {print $2} the_file'"
Thank you for coddling me...
This is a relatively simple:
awk '$1 == "70:56:81:cb:a2:dc," {print "The answer is "$2}' 'BSSIDtable.csv'
(the "The answer is " text can be omitted if you only wish to see only the data, but this shows you how to get more user-friendly output if desired).
The comma is included since awk uses white space for separators so the comma becomes part of column 1.
If the thing you're looking for is in a shell variable, you can use -v to provide that to awk as an awk variable:
lookfor="70:56:81:cb:a2:dc,"
awk -v mac=$lookfor '$1 == mac {print "The answer is "$2}' 'BSSIDtable.csv'
As an aside, your AppleScript solution is probably not working because the $1/$2 are being interpreted as shell variable rather than awk variables. If you insist on using AppleScript, you will have to figure out how to construct a shell command that quotes the awk commands correctly.
My advice is to just use the shell directly, the number of people proficient in that almost certainly far outnumber those proficient in AppleScript :-)
if sed is available (normaly on mac, event if not tagged in OP)
simple but read all the file
sed -n 's/84:a5:7e:73:10:32,[[:blank:]]*//p' YourFile
quit after first occurence (so average of 50% faster on huge file)
sed -n -e '/84:a5:7e:73:10:32,[[:blank:]]*/!b' -e 's///p;q' YourFile
awk
awk '/^84:a5:7e:73:10:32/ {print $2}'
# OR using a variable for batch interaction
awk -v Src='84:a5:7e:73:10:32' '$1 == Src {print $2}'
# OR assuming that case is unknow
awk -v Src='84:a5:7e:73:10:32' 'BEGIN{IGNORECASE=1} $1 == Src {print $2}'
by default it take $0 as compare test if a regex is present, just add the ^ to take first field content
I have those lines in my text file :
msg_wdraw[] = "whatever a sentence here,"
"This is the second part of this text1 ."
msg_sp2million[] = "whatever a sentence here,"
"This is the second part of this text2."
I need the sentence between msg_sp2million and the period "." and print them out.
i.e ("whatever a sentence here,"
"This is the second part of this text2.")
I tried this : sed -n "/msg_sp2million/,/./p" filename.txt
However, this sed command also returns me the value of msg_wdraw (the first variable)
I also tried awk, grep, other sed..... but failed eventually.
How can I fix this problem ? And Why this returns me not only the value of msg_sp2million and also the value of msg_wdraw ?
Please help # ~ #
Maybe something like this:
awk '/msg_sp2million/{ split($0,a,"="); print a[length(a)]; getline; print}' file.txt
Match regexp, print what comes after the =, get next line, and print that too.
Returns:
"whatever a sentence here,"
"This is the second part of this text2."
Using simple awk command:
awk -F '= *' -v RS='.' -v ORS='."\n' '$1 ~ /msg_sp2million/ {sub(/" *\n */, "\" ", $2);
print $2}' file
"whatever a sentence here," "This is the second part of this text2."
I'm unable to add my solution (a POSIX-compliant derivative of qwwqwwq's solution, referred as qww below) as a comment. So, qww's solution works, but ONLY in GNU awk from a certain version onward (apparently 3.1.5, see also http://awk.freeshell.org/AwkFeatureComparison).
Tip: Try
awk -W posix '/msg_sp2million/{ split($0,a,"="); print a[length(a)]; getline; print}' file.txt
in a non-GNU environment and you will 99% sure get an error message, e. g. about using an array in a scalar context.
The following solution should also work on a HP-UX workstation:
(well, the -W posix may be omitted of course, but is always invaluable while in testing stage)
awk -W posix '/msg_sp2million/{ amount=split($0,a,"="); print a[amount]; getline; print}' file.txt
So, I have a file called 'dummy' which contains the string:
"There is 100% packet loss at node 1".
I also have a small script that I want to use to grab the percentage from this file. The script is below.
result=`grep 'packet loss' dummy` |
awk '{ first=match($0,"[0-9]+%")
last=match($0," packet loss")
s=substr($0,first,last-first)
print s}'
echo $result
I want the value of $result to basically be 100% in this case. But for some reason, it just prints out a blank string. Can anyone help me?
You would need to put the closing backtick after the end of the awk command, but it's preferable to use $() instead:
result=$( grep 'packet loss' dummy |
awk '{ first=match($0,"[0-9]+%")
last=match($0," packet loss")
s=substr($0,first,last-first)
print s}' )
echo $result
but you could just do:
result=$( grep 'packet loss' | grep -o "[0-9]\+%" )
Try
awk '{print $3}'
instead.
the solution below can be used when you don't know where the percentage numbers are( and there's no need to use awk with greps)
$ results=$(awk '/packet loss/{for(i=1;i<=NF;i++)if($i~/[0-9]+%$/)print $i}' file)
$ echo $results
100%
You could do this with bash alone using expr.
i=`expr "There is 98.76% packet loss at node 1" : '[^0-9.]*\([0-9.]*%\)[^0-9.]*'`; echo $i;
This extracts the substring matching the regex within \( \).
Here I'm assuming that the output lines you're interested in adhere strictly to your example, with the percentage value being the only variation.
With that assumption, you really don't need anything more complicated than:
awk '/packet loss/ { print $3 }' dummy
This quite literally means "print the 3rd field of any lines containing 'packet loss' in them". By default awk treats whitespace as field delimiters, which is perfect for you.
If you are doing more than simply printing the percentage, you could save the results to a shell variable using backticks, or redirect the output to a file. But your sample code simply echoes the percentages to stdout, and then exits. The one-liner does the exact same thing. No need for backticks or $() or any other shell machinations whatsoever.
NB: In my experience, piping the output of grep to awk is usually doing something that awk can do all by itself.
I have a file with fields separated by pipe characters and I want to print only the second field. This attempt fails:
$ cat file | awk -F| '{print $2}'
awk: syntax error near line 1
awk: bailing out near line 1
bash: {print $2}: command not found
Is there a way to do this?
Or just use one command:
cut -d '|' -f FIELDNUMBER
The key point here is that the pipe character (|) must be escaped to the shell. Use "\|" or "'|'" to protect it from shell interpertation and allow it to be passed to awk on the command line.
Reading the comments I see that the original poster presents a simplified version of the original problem which involved filtering file before selecting and printing the fields. A pass through grep was used and the result piped into awk for field selection. That accounts for the wholly unnecessary cat file that appears in the question (it replaces the grep <pattern> file).
Fine, that will work. However, awk is largely a pattern matching tool on its own, and can be trusted to find and work on the matching lines without needing to invoke grep. Use something like:
awk -F\| '/<pattern>/{print $2;}{next;}' file
The /<pattern>/ bit tells awk to perform the action that follows on lines that match <pattern>.
The lost-looking {next;} is a default action skipping to the next line in the input. It does not seem to be necessary, but I have this habit from long ago...
The pipe character needs to be escaped so that the shell doesn't interpret it. A simple solution:
$ awk -F\| '{print $2}' file
Another choice would be to quote the character:
$ awk -F'|' '{print $2}' file
Another way using awk
awk 'BEGIN { FS = "|" } ; { print $2 }'
And 'file' contains no pipe symbols, so it prints nothing. You should either use 'cat file' or simply list the file after the awk program.