using a wildcard in awk - awk
Using awk, I want to print all lines that have a string in the first column that starts with 22_
I tried the following, but obviously * does not work as a wildcard in awk:
awk '$1=="22_*" {print $0}' input > output
Is this possible in awk?
Let's start with a test file:
$ cat >file
22_something keep
23_other omit
To keep only lines that start with 22_:
$ awk '/^22_/' file
22_something keep
Alternatively, if you prefer to reference the first field explicitly, we could use:
$ awk '$1 ~ /^22_/' file
22_something keep
Note that we don't have to write {print $0} after the condition because that is exactly the default action that awk associates with a condition.
At the start of a regular expressions, ^ matches the beginning of a line. Thus, if you want 22_ to occur at the start of a line or the start of a field, you want to write ^22_.
In the condition $1 ~ /^22_/, note that the operator is ~. That operator tells awk to check if the preceding string, $1, matches the regular expression ^22_.
Chosen answer does not answer how to use a wildcard in awk, which is achieved using .* (instead of *):
awk '$1=="22_.*" {print $0}' input > output
Related
awk command works, but not in openwrt's awk
Works here: 'awk.js.org/`
but not in openwrt's awk, which returns the error message:
awk: bad regex '^(server=|address=)[': Missing ']'
Hello everyone!
I'm trying to use an awk command I wrote which is:
'!/^(server=|address=)[/][[:alnum:]][[:alnum:]-.]+([/]|[/]#)$|^#|^\s*$/ {count++}; END {print count+0}'
Which counts invalid lines in a dns blocklist (oisd in this case):
Input would be eg:
server=/0--foodwarez.da.ru/anyaddress.1.1.1
serverspellerror=/0-000.store/
server=/0-24bpautomentes.hu/
server=/0-29.com/
server=/0-day.us/
server=/0.0.0remote.cryptopool.eu/
server=/0.0mail6.xmrminingpro.com/
server=/0.0xun.cryptopool.space/
Output for this should be "2" since there are two lines that don't match the criteria (correctly formed address, comments, or blank lines).
I've tried formatting the command every which way with [], but can't find anything that works. Does anyone have an idea what format/syntax/option needs adjusting?
Thanks!
To portably include - in a bracket expression it has to be the first or last character, otherwise it means a range, and \s is shorthand for [[:space:]] in only some awks. This will work in any POSIX awk:
$ awk '!/^(server=|address=)[/][[:alnum:]][[:alnum:].-]+([/]|[/]#)$|^#|^[[:space:]]*$/ {count++}; END {print count+0}' file
2
Per #tripleee's comment below if your awk is broken such that a / inside a bracket expression isn't treated as literal then you may need this instead:
$ awk '!/^(server=|address=)\/[[:alnum:]][[:alnum:].-]+(\/|\/#)$|^#|^[[:space:]]*$/ {count++}; END {print count+0}' file
2
but get a new awk, e.g. GNU awk, as who knows what other surprises the one you're using may have in store for you!
'!/^(server=|address=)[/][[:alnum:]][[:alnum:]-.]+([/]|[/]#)$|^#|^\s*$/ {count++}; END {print count+0}'
- has special meaning inside [ and ], it is used to denote range e.g. [A-Z] means uppercase ASCII letter, use \ escape sequence to make it literal dash, let file.txt content be
server=/0--foodwarez.da.ru/anyaddress.1.1.1
serverspellerror=/0-000.store/
server=/0-24bpautomentes.hu/
server=/0-29.com/
server=/0-day.us/
server=/0.0.0remote.cryptopool.eu/
server=/0.0mail6.xmrminingpro.com/
server=/0.0xun.cryptopool.space/
then
awk '!/^(server=|address=)[/][[:alnum:]][[:alnum:]\-.]+([/]|[/]#)$|^#|^\s*$/ {count++}; END {print count+0}' file.txt
gives output
2
You might also consider replacing \s using [[:space:]] in order to main consistency.
(tested in GNU Awk 5.0.1)
Can I delete a field in awk?
This is test.txt:
0x01,0xDF,0x93,0x65,0xF8
0x01,0xB0,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0xB2,0x00,0x76
If I run
awk -F, 'BEGIN{OFS=","}{$2="";print $0}' test.txt
the result is:
0x01,,0x93,0x65,0xF8
0x01,,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,,0x00,0x76
The $2 wasn't deleted, it just became empty.
I hope, when printing $0, that the result is:
0x01,0x93,0x65,0xF8
0x01,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0x00,0x76
All the existing solutions are good though this is actually a tailor made job for cut: cut -d, -f 1,3- file 0x01,0x93,0x65,0xF8 0x01,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0 0x01,0x00,0x76 If you want to remove 3rd field then use: cut -d, -f 1,2,4- file To remove 4th field use: cut -d, -f 1-3,5- file
I believe simplest would be to use sub function to replace first occurrence of continuous ,,(which are getting created after you made 2nd field NULL) with single ,. But this assumes that you don't have any commas in between field values.
awk 'BEGIN{FS=OFS=","}{$2="";sub(/,,/,",");print $0}' Input_file
2nd solution: OR you could use match function to catch regex from first comma to next comma's occurrence and get before and after line of matched string.
awk '
match($0,/,[^,]*,/){
print substr($0,1,RSTART-1)","substr($0,RSTART+RLENGTH)
}' Input_file
It's a bit heavy-handed, but this moves each field after field 2 down a place, and then changes NF so the unwanted field is not present:
$ awk -F, -v OFS=, '{ for (i = 2; i < NF; i++) $i = $(i+1); NF--; print }' test.txt
0x01,0x93,0x65,0xF8
0x01,0x01,0x03,0x02,0x00,0x64,0x06,0x01
0x01,0x00,0x76
$
Tested with both GNU Awk 4.1.3 and BSD Awk ("awk version 20070501" on macOS Mojave 10.14.6 — don't ask; it frustrates me too, but sometimes employers are not very good at forward thinking). Setting NF may or may not work on older versions of Awk — I was a little surprised it did work, but the surprise was a pleasant one, for a change.
If Awk is not an absolute requirement, and the input is indeed as trivial as in your example, sed might be a simpler solution.
sed 's/,[^,]*//' test.txt
This is especially elegant if you want to remove the second field. A more generic approach to remove, the nth field would require you to put in a regex which matches the first n - 1 followed by the nth, then replace that with just the the first n - 1.
So for n = 4 you'd have
sed 's/\([^,]*,[^,]*,[^,]*,\)[^,]*,/\1/' test.txt
or more generally, if your sed dialect understands braces for specifying repetitions
sed 's/\(\([^,]*,\)\{3\}\)[^,]*,/\1/' test.txt
Some sed dialects allow you to lose all those pesky backslashes with an option like -r or -E but again, this is not universally supported or portable.
In case it's not obvious, [^,] matches a single character which is not (newline or) comma; and \1 recalls the text from first parenthesized match (back reference; \2 recalls the second, etc).
Also, this is completely unsuitable for escaped or quoted fields (though I'm not saying it can't be done). Every comma acts as a field separator, no matter what.
With GNU sed you can add a number modifier to substitute nth match of non-comma characters followed by comma: sed -E 's/[^,]*,//2' file
Using awk in a regex-free way, with the option to choose which line will be deleted:
awk '{ col = 2; n = split($0,arr,","); line = ""; for (i = 1; i <= n; i++) line = line ( i == col ? "" : ( line == "" ? "" : "," ) arr[i] ); print line }' test.txt
Step by step:
{
col = 2 # defines which column will be deleted
n = split($0,arr,",") # each line is split into an array
# n is the number of elements in the array
line = "" # this will be the new line
for (i = 1; i <= n; i++) # roaming through all elements in the array
line = line ( i == col ? "" : ( line == "" ? "" : "," ) arr[i] )
# appends a comma (except if line is still empty)
# and the current array element to the line (except when on the selected column)
print line # prints line
}
Another solution:
You can just pipe the output to another sed and squeeze the delimiters.
$ awk -F, 'BEGIN{OFS=","}{$2=""}1 ' edward.txt | sed 's/,,/,/g'
0x01,0x93,0x65,0xF8
0x01,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0x00,0x76
$
Commenting on the first solution of #RavinderSingh13 using sub() function:
awk 'BEGIN{FS=OFS=","}{$2="";sub(/,,/,",");print $0}' Input_file
The gnu-awk manual: https://www.gnu.org/software/gawk/manual/html_node/Changing-Fields.html
It is important to note that making an assignment to an existing field changes the value of $0 but does not change the value of NF, even when you assign the empty string to a field." (4.4 Changing the Contents of a Field)
So, following the first solution of RavinderSingh13 but without using, in this case,sub() "The field is still there; it just has an empty value, delimited by the two colons":
awk 'BEGIN {FS=OFS=","} {$2="";print $0}' file
0x01,,0x93,0x65,0xF8
0x01,,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,,0x00,0x76
My solution:
awk -F, '
{
regex = "^"$1","$2
sub(regex, $1, $0);
print $0;
}'
or one line code:
awk -F, '{regex="^"$1","$2;sub(regex, $1, $0);print $0;}' test.txt
I found that OFS="," was not necessary
I would do it following way, let file.txt content be:
0x01,0xDF,0x93,0x65,0xF8
0x01,0xB0,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0xB2,0x00,0x76
then
awk 'BEGIN{FS=",";OFS=""}{for(i=2;i<=NF;i+=1){$i="," $i};$2="";print}' file.txt
output
0x01,0x93,0x65,0xF8
0x01,0x01,0x03,0x02,0x00,0x64,0x06,0x01,0xB0
0x01,0x00,0x76
Explanation: I set OFS to nothing (empty string), then for 2nd and following column I add , at start. Finally I set what is now comma and value to nothing. Keep in mind this solution would need rework if you wish to remove 1st column.
Change output field seperator on first row
I have a file like this
name|age
Bob|30
Tom|50
Cindy|10
I want the first row to have a different seperator, "^".
awk 'NR==1 { gsub("|","^")1}1' f
But I keep getting
^n^a^m^e^|^a^g^e^
Bob|30
Tom|50
Cindy|10
Desired output is
name^age
Bob|30
Tom|50
Cindy|10
Your code with gsub("|","^") doesn't have special meta character | (used for alternation in regex) escaped hence it will match every position in input.
You may use this awk without involving any regex:
awk 'BEGIN{FS=OFS="|"} FNR==1{OFS="^"; $1=$1; OFS=FS} 1' f
name^age
Bob|30
Tom|50
Cindy|10
Details:
FS="|": Sets FS as |
OFS="^": Sets OFS as ^
$1=$1: Forces awk to reformat each of the fields using OFS
You can also use sed: sed '1 s/|/^/' ip.txt 1 address for the command, which is first line here | is not special, because by default sed uses BRE, see this Q&A for BRE vs ERE differences use s/|/^/g if you can have multiple matches
Like this :
awk -F'|' 'NR==1{print $1,$2;next}1' OFS='^' file
or a mix between anubhava response and mine:
awk -F'|' 'NR==1{$1=$1}1' OFS='^' file
Could you please try following.
awk 'FNR==1{sub(/\|/,"^")} 1' Input_file
Use gsub in place of sub in case of multiple occurrences needs to be changed.
awk 'FNR==1{gsub(/\|/,"^")} 1' Input_file
Grep part of string after symbol and shuffle columns
I would like to take the number after the - sign and put is as column 2 in my matrix. I know how to grep the string but not how to print it after the text string. in: 1-967764 GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 3-425354 GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 4-376323 GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 5-221398 GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 6-180339 TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT out: GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764 GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354 GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323 GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398 TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339
awk -F'[[:space:]-]+' '{print $3,$2}' file
Seems like a simple substitution should do the job:
sed -E 's/[0-9]+-([0-9]+)[[:space:]]*(.*)/\2 \1/' file
Capture the parts you're interested in and use them in the replacement.
Alternatively, using awk:
awk 'sub(/^[0-9]+-/, "") { print $2, $1 }' file
Remove the leading digits and - from the start of the line. When this is successful, sub returns true, so the action is performed, printing the second field, followed by the first.
Using regex ( +|-) as field separator:
$ awk -F"( +|-)" '{print $3,$2}' file
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339
here is another awk
$ awk 'split($1,a,"-") {print $2,a[2]}' file
awk '{sub(/.-/,"");print $2,$1}' file
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339
Unable to match regex in string using awk
I am trying to fetch the lines in which the second part of the line contains a pattern from the first part of the line.
$ cat file.txt
String1 is a big string|big
$ awk -F'|' ' { if ($2 ~ /$1/) { print $0 } } ' file.txt
But it is not working.
I am not able to find out what is the mistake here.
Can someone please help?
Two things: No slashes, and your numbers are backwards. awk -F\| '$1~$2' file.txt
I guess what you meant is part of the string in the first part should be a part of the 2nd part.if this is what you want! then,
awk -F'|' '{n=split($1,a,' ');for(i=1,i<=n;i++){if($2~/a[i]/)print $0}}' your_file
There are surprisingly many things wrong with your command line: 1) You aren't using the awk condition/action syntax but instead needlessly embedding a condition within an action, 2) You aren't using the default awk action but instead needlessly hand-coding a print $0. 3) You have your RE operands reversed. 4) You are using RE comparison but it looks like you really want to match strings. You can fix the first 3 of the above by modifying your command to: awk -F'|' '$1~$2' file.txt but I think what you really want is "4" which would mean you need to do this instead: awk -F'|' 'index($1,$2)' file.txt