match duplicate string before a specified delimiter - awk

cat test.txt
serverabc.test.net
serverabc.qa.net
serverabc01.test.net
serverstag.staging.net
serverstag.test.net
here i need to match the duplicate strings just before the delimiter '.'
So the expected output would be like below. because string "serverabc" and "serverstag" found to be duplicates. Please help.
serverabc.test.net
serverabc.qa.net
serverstag.staging.net
serverstag.test.net


awk to the rescue!
$ awk -F\. '{c[$1]++; a[$1]=a[$1]?a[$1]RS$0:$0}
END{for(k in c) if(c[k]>1) print a[k]}' file
serverabc.test.net
serverabc.qa.net
serverstag.staging.net
serverstag.test.net


If it is not going to be used allot I would probably just do something like this:
cut -f1 -d\. foo.txt | sort |uniq -c | grep -v " 1 " | cut -c 9-|sed 's/\(.*\)/^\1\\./' > dup.host
grep -f dup.host foo.txt
serverabc.test.net
serverabc.qa.net
serverstag.staging.net
serverstag.test.net

Related

Find the second word delimited by space or comma then insert strings before and after

I have a file containing TABLE schema.table and want to put strings around it to make a command like MARK string REJECT
the file contains many lines
TABLE SCHEMA.MYTAB, etc. etc....
or
TABLE SCHEMA.MYTAB , etc. etc....
The result is
MARK SCHEMA.MYTAB REJECT
..etc
I have
grep TABLE dirx/myfile.txt | awk -F, '{print $1}' | awk '{print $2}' | sed -e 's/^/MARK /' |sed -e 's/$/ REJECT/'
It works, but can this be tidier? I think I can combine the awk and sed into single commands but not sure how.

Maybe:
awk '/^TABLE/ {gsub(/,.*$/, ""); print "MARK " $2 " REJECT"}' dirx/myfile.txt

printf usage with awk when printing multiple columns

I am trying this below command:
cat dcl1serrfip_check.csv | grep -Fi 'BANK0_F5_WRDAT_P0[0]' | grep -i setup | grep 'L2H' | grep highv | grep -i low | awk -F ',' -v dev="0.861" -v rc="1.105" -v inte="0.872" '{ print ($10+$11)-(($12+$13)-($14))","($10*dev)+($11*rc)-(($12*dev)+($13*rc)-($14*inte))}'
This gives below output:
-6.93889e-18,0.000288
I want this output to be formatted to 4 decimal places. How to do it? The desired output would be
-0.0000,0.0002

You need, %0.4f or %.4f
To Test use :
awk 'BEGIN{ printf("%0.4f\n", -6.93889e-18) }'
So it becomes:
printf("%0.4f,%0.4f\n", ($10+$11)-(($12+$13)-($14)), ($10*dev)+($11*rc)-(($12*dev)+($13*rc)-($14*inte)) )
Actually you can rewrite your command in awk itself, no need of so many grep and cat combination

Add text to beginning of awk result

Good day all,
I am running below command:
netstat -an | awk '/:25/{ print $4 }' | sed 's/:25//' | paste -sd ',' -
which produces
192.168.2.22,127.0.0.1
I would like to amend the result to something like below (to be parsed as a csv by an application)
Manuallyaddedtext 192.168.2.22,127.0.0.1
Many thanks

echo -n "Mytext " ; netstat...

Delete multiple strings/characters in a file

I have a curl output generated similar below, Im working on a SED/AWK script to eliminate unwanted strings.
File
{id":"54bef907-d17e-4633-88be-49fa738b092d","name":"AA","description","name":"AAxxxxxx","enabled":true}
{id":"20000000000000000000000000000000","name":"BB","description","name":"BBxxxxxx","enabled":true}
{id":"542ndf07-d19e-2233-87gf-49fa738b092d","name":"AA","description","name":"CCxxxxxx","enabled":true}
{id":"20000000000000000000000000000000","name":"BB","description","name":"DDxxxxxx","enabled":true}
......
I like to modify this file and retain similar below,
AA AAxxxxxx
BB BBxxxxxx
AA CCxxxxxx
BB DDxxxxxx
AA n.....
BB n.....
Is there a way I could remove word/commas/semicolons in-between so I can only retain these values?

Try this awk
curl your_command | awk -F\" '{print $(NF-9),$(NF-3)}'
Or:
curl your_command | awk -F\" '{print $7,$13}'
A semantic approach ussing perl:
curl your_command | perl -lane '/"name":"(\w+)".*"name":"(\w+)"/;print $1." ".$2'
For any number of name ocurrences:
curl your_command | perl -lane 'printf $_." " for ( $_ =~ /"name":"(\w+)"/g);print ""'

This might work for you (GNU sed):
sed -r 's/.*("name":")([^"]*)".*\1([^"]*)".*/\2 \3/p;d' file
This extracts the fields following the two name keys and prints them if successful.
Alternatively, on simply pattern matching:
sed -r 's/.*:.*:"([^"]*)".*:"([^"]*)".*:.*/\1 \2/p;d' file

In this particular case, you could do
awk -F ":|," '{print $4,$7}' file2 |tr -d '"'
and get
AA AAxxxxxx
BB BBxxxxxx
AA CCxxxxxx
BB DDxxxxxx
Here, the field separator is either : or ,, we print the fourth and seventh field (because all lines have the entries in these two fields) and finally, we use tr to delete the " because you don't want to have it.

AWK to search for a for a string and print full text where string occurs

I have a document containing several lines of text.
Example(not actual):
*Prepare 42 Locked delete from table where type='test' and user_id='099'and number='+66719919*
I want to be able to search for user_id where ever it occurs in the document (which does not follow a pattern) and have the output as:
user_id=009
OR
009
Please how do I achieve this using awk?
Thanks.

awk '/user_id/{for(i=1;i<=NF;i++){if($i~/user_id/){split($i,a,"=");print a[2]}}}' your_file
tested:
> echo "*Prepare type='test' and user_id='099' and number='+66719919*" | awk '/user_id/{for(i=1;i<=NF;i++){if($i~/user_id/){split($i,a,"=");print a[2]}}}'
'099'
another one:
> echo "*Prepare type='test' and user_id='099' and number='+66719919*" | awk '/user_id/{for(i=1;i<=NF;i++){if($i~/user_id/){ print $i}}}'
user_id='099'

You could also use grep:
grep -o "user_id='\?[0-9]*'\?"
Append tr to remove the quotes:
grep -o "user_id='\?[0-9]*'\?" | tr -d \'