awk to parse field by using period and output unique digits - awk

I am trying to use awk to parse $2 on using the first . in the string and output the digits with the header row above it. The current output is close but both commands seem to taking $1 as well. Do I need to specify something in the command to only prints the digits in $2, it seems close. Thank you :).
file
R_2016_09_20_12_47
IonXpress_007 16-0001.xxx.xxx
IonXpress_008 16-0002.xxx.xxx
IonXpress_009 16-0003.xxx.xxx
R_2016_09_20_12_46
IonXpress_007 16-0004.xxx.xxx
IonXpress_008 16-0005.xxx.xxx
IonXpress_009 16-0006.xxx.xxx
desired output
R_2016_09_20_12_47
16-0001
16-0002
16-0003
R_2016_09_20_12_46
16-0004
16-0005
16-0006
awk
awk -F. '{print $1}' file
cut
cut -d'.' -f1 file
current output
R_2016_09_20_12_47
IonXpress_007 16-0001
IonXpress_008 16-0002
IonXpress_009 16-0003
R_2016_09_20_12_46
IonXpress_001 16-0004
IonXpress_002 16-0005
IonXpress_003 16-0006

Try this :
% awk -F'[ .]' '{print $2 ? $2 : $1}' file
R_2016_09_20_12_47
16-0001
16-0002
16-0003
R_2016_09_20_12_46
16-0004
16-0005
16-0006
NOTE
i take space and . as separators
i use ternary operator to make a condition on $2

Related

How to extract string from a file in bash

I have a file called DB_create.sql which has this line
CREATE DATABASE testrepo;
I want to extract only testrepo from this. So I've tried
cat DB_create.sql | awk '{print $3}'
This gives me testrepo;
I need only testrepo. How do I get this ?
With your shown samples, please try following.
awk -F'[ ;]' '{print $(NF-1)}' DB_create.sql
OR
awk -F'[ ;]' '{print $3}' DB_create.sql
OR without setting any field separators try:
awk '{sub(/;$/,"");print $3}' DB_create.sql
Simple explanation would be: making field separator as space OR semi colon and then printing 2nd last field($NF-1) which is required by OP here. Also you need not to use cat command with awk because awk can read Input_file by itself.
Using gnu awk, you can set record separator as ; + line break:
awk -v RS=';\r?\n' '{print $3}' file.sql
testrepo
Or using any POSIX awk, just do a call to sub to strip trailing ;:
awk '{sub(/;$/, "", $3); print $3}' file.sql
testrepo
You can use
awk -F'[;[:space:]]+' '{print $3}' DB_create.sql
where the field separator is set to a [;[:space:]]+ regex that matches one or more occurrences of ; or/and whitespace chars. Then, Field 3 will contain the string you need without the semi-colon.
More pattern details:
[ - start of a bracket expression
; - a ; char
[:space:] - any whitespace char
] - end of the bracket expression
+ - a POSIX ERE one or more occurrences quantifier.
See the online demo.
Use your own code but adding the function sub():
cat DB_create.sql | awk '{sub(/;$/, "",$3);print $3}'
Although it's better not using cat. Here you can see why: Comparison of cat pipe awk operation to awk command on a file
So better this way:
awk '{sub(/;$/, "",$3);print $3}' file

awk command to read a key value pair from a file

I have a file input.txt which stores information in KEY:VALUE form. I'm trying to read GOOGLE_URL from this input.txt which prints only http because the seperator is :. What is the problem with my grep command and how should I print the entire URL.
SCRIPT
$> cat script.sh
#!/bin/bash
URL=`grep -e '\bGOOGLE_URL\b' input.txt | awk -F: '{print $2}'`
printf " $URL \n"
INPUT_FILE
$> cat input.txt
GOOGLE_URL:https://www.google.com/
OUTPUT
https
DESIRED_OUTPUT
https://www.google.com/
Since there are multiple : in your input, getting $2 will not work in awk because it will just give you 2nd field. You actually need an equivalent of cut -d: -f2- but you also need to check key name that comes before first :.
This awk should work for you:
awk -F: '$1 == "GOOGLE_URL" {sub(/^[^:]+:/, ""); print}' input.txt
https://www.google.com/
Or this non-regex awk approach that allows you to pass key name from command line:
awk -F: -v k='GOOGLE_URL' '$1==k{print substr($0, length(k FS)+1)}' input.txt
Or using gnu-grep:
grep -oP '^GOOGLE_URL:\K.+' input.txt
https://www.google.com/
Could you please try following, written and tested with shown samples in GNU awk. This will look for string GOOGLE_URL and will catch further either http or https value from url, in case you need only https then change http[s]? to https in following solution please.
awk '/^GOOGLE_URL:/{match($0,/http[s]?:\/\/.*/);print substr($0,RSTART,RLENGTH)}' Input_file
Explanation: Adding detailed explanation for above.
awk ' ##Starting awk program from here.
/^GOOGLE_URL:/{ ##Checking condition if line starts from GOOGLE_URL: then do following.
match($0,/http[s]?:\/\/.*/) ##Using match function to match http[s](s optional) : till last of line here.
print substr($0,RSTART,RLENGTH) ##Printing sub string of matched value from above function.
}
' Input_file ##Mentioning Input_file name here.
2nd solution: In case you need anything coming after first : then try following.
awk '/^GOOGLE_URL:/{match($0,/:.*/);print substr($0,RSTART+1,RLENGTH-1)}' Input_file
Take your pick:
$ sed -n 's/^GOOGLE_URL://p' file
https://www.google.com/
$ awk 'sub(/^GOOGLE_URL:/,"")' file
https://www.google.com/
The above will work using any sed or awk in any shell on every UNIX box.
I would use GNU AWK following way for that task:
Let file.txt content be:
EXAMPLE_URL:http://www.example.com/
GOOGLE_URL:https://www.google.com/
KEY:GOOGLE_URL:
Then:
awk 'BEGIN{FS="^GOOGLE_URL:"}{if(NF==2){print $2}}' file.txt
will output:
https://www.google.com/
Explanation: GNU AWK FS might be pattern, so I set it to GOOGLE_URL: anchored (^) to begin of line, so GOOGLE_URL: in middle/end will not be seperator (consider 3rd line of input). With this FS there might be either 1 or 2 fields in each line - latter is case only if line starts with GOOGLE_URL: so I check number of fields (NF) and if this is second case I print 2nd field ($2) as first record in this case is empty.
(tested in gawk 4.2.1)
Yet another awk alternative:
gawk -F'(^[^:]*:)' '/^GOOGLE_URL:/{ print $2 }' infile

How to extract word from a string that may/may not start with a single quote

Sample string:
'kernel-rt|kernel-alt|/kernel-' 'headers|xen|firmware|tools|python|utils'
cut -d' ' -f 1 string.txt gives me
'kernel-rt|kernel-alt|/kernel-'
But how do we proceed further to get just the 'kernel' from it?
Assuming you want only the 3rd kernel (in bold) and not the others
'kernel-rt|kernel-alt|/kernel-' 'headers|xen|firmware|tools|python|utils'
Here is how you extract it using single command awk (standard Linux gawk).
input="kernel-rt|kernel-alt|/kernel-' 'headers|xen|firmware|tools|python|utils"
echo $input|awk -F"|" '{split($3,a,"-");match(a[1],"[[:alnum:]]+",b);print b[0]}'
explanation
-F"|" specify field separator is | so that only is 3rd field required
split($3,a,"-") split 3rd field by -, left part assigned to a[1]
match(a[1],"[[:alnum:]]+",b) from a[1] extract sequence of alphanumeric string into b[0]
print b[0] output the matched string.
If you want to extract kernel from 2nd or 1st fields. Change $3 to $2 or $1.
$ cat file
'kernel-rt|kernel-alt|/kernel-' 'headers|xen|firmware|tools|python|utils'
$
$ awk '{print $1}' file
'kernel-rt|kernel-alt|/kernel-'
$
$ awk '{gsub(/\047/,"",$1); print $1}' file
kernel-rt|kernel-alt|/kernel-
$
$ awk '{gsub(/\047/,""); split($1,f,/[|]/); print f[1]}' file
kernel-rt
and just to make you think...
$ awk '{gsub(/\047|\.*/,"")}1' file
kernel-rt

Extract fields from logs with awk and aggregate them for a new command

I have this kind of log:
2018-10-05 09:12:38 286 <190>1 2018-10-05T09:12:38.474640+00:00 app web - - Class uuid=uuid-number-one cp=xxx action='xxxx'
2018-10-05 10:11:23 286 <190>1 2018-10-05T10:11:23.474640+00:00 app web - - Class uuid=uuid-number-two cp=xxx action='xxxx'
I need to extract uuid and run a second query with:
./getlogs --search 'uuid-number-one OR uuid-number-two'
For the moment for the first query I do this to extract uuid:
./getlogs | grep 'uuid' | awk 'BEGIN {FS="="} { print $2 }' | cut -d' ' -f1
My three question :
I think I could get rid of grep and cut and use only awk?
How could I capture only the value of uuid. I tried awk '/uuid=\S*/{ print $1 }' or awk 'BEGIN {FS="uuid=\\S*"} { print $1 }' but it's a failure.
How could I aggregate the result and turn it into one shell variable that I can use after for the new command?
You could define two field separators:
$ awk -F['= '] '/uuid/{print $12}' file
Result:
uuid-number-one
uuid-number-two
Question 2:
The pattern part in awk just selects lines to process. It doesn't change the internal variables like $1 or NF. You need to do the replacement afterwards:
$ awk '/uuid=/{print gensub(/.*uuid=(\S*).*/, "\\1", "")}' file
Question 3:
var=$(awk -F['= '] '/uuid/{r=r","$12}END{print substr(r,2)}' file)
Implement the actual aggregation for each line (here r=r","$12).
Could you please try following(tested on shown samples and in BASH environment).
awk 'match($0,/uuid=[^ ]*/){print substr($0,RSTART+5,RLENGTH-5)}' Input_file
Solution 2nd: In case your uid is not having space in it then use following.
awk '{sub(/.*uuid=/,"");sub(/ .*/,"")} 1' Input_file
solution 3rd: using sed following may help you(considering that uid is not having any space in its values).
sed 's/\(.*uuid=\)\([^ ]*\)\(.*\)/\2/' Input_file
Solution 4th: using awk field separator method for shown samples.
awk -F'uuid=| cp' '{print $2}' Input_file
To concatenate all values into a shell variable use following.
shell_var=$(awk 'match($0,/uuid=[^ ]*/){val=val?val OFS substr($0,RSTART+5,RLENGTH-5):substr($0,RSTART+5,RLENGTH-5)} END{print val}' Input_file)

Get only part of a file name in Awk

I have tried
awk '{print FILENAME}'
And the result was full path of the file.
I want to get only the file name, example: from "test/testing.test.txt" I just want to get "testing" without ".test.txt".
Use -F to delimit by the period and print the first string before that delimiter:
awk -F'.' '{ print $1 }'
Alternatively,
ls -l | awk '{ print $9 }' | awk -F"." '{ print $1 }'
will run through the whole folder👍
(there's a fancier way to do it, but that's easy).
Use the sub and/or split functions to extract the part of FILENAME you want.