I am trying to parse "/boot/grub/grubenv" but really not very good at regexp.
Suppose the content of /boot/grub/grubenv is:
saved_entry=1
I want to output the number "1", like below. I am currently using "awk", but open to other tools.
$ awk '/^(saved_entry=)([0-9]+)/ {print $2}' /boot/grub/grubenv
But obviously not working, thanks for the help.
Specify a field separator with -F option:
awk -F= '/^saved_entry=/ {print $2}' /boot/grub/grubenv
$1, $2, .. here represents fields (separated by =), not a backreferences to captured groups.
If you want to match things probably best to use match!
This will work even if there are more fields after and does not need you to change the field separator(incase you are doing any other stuff with the data).
The only drawback of this method is that it will only match the left-most match of the record, so if the data appears twice in the same record(line) then it will only match the first one it finds.
awk 'match($0,/^(saved_entry=)([0-9]+)/,a){print a[2]}' file
Example
input
saved_entry=1 blah blah more stuff
output
1
Explanation
Matches the regex in $0(the record) and then stores anything in brackets as separate array elements.
From the example, there would be these outputs
a[0] is saved_entry=1
a[1] is saved_entry=
a[2] is 1
Related
I would like to first take out of the string in the first column parenthesis which I can do with:
awk -F"[()]" '{print $2}'
Then, concatenate it with the second column to create a URL with the following format:
"https://ftp.drupal.org/files/projects/"[firstcolumn stripped out of parenthesis]-[secondcolumn].tar.gz
With input like:
Admin Toolbar (admin_toolbar) 8.x-2.5
Entity Embed (entity_embed) 8.x-1.2
Views Reference Field (viewsreference) 8.x-2.0-beta2
Webform (webform) 8.x-5.28
Data from the first line would create this URL:
https://ftp.drupal.org/files/projects/admin_toolbar-8.x-2.5.tar.gz
Something like
sed 's!^[^(]*(\([^)]*\))[[:space:]]*\(.*\)!https://ftp.drupal.org/files/projects/\1-\2.tar.gz!' input.txt
If a file a has your input, you can try this:
$ awk -F'[()]' '
{
split($3,parts," *")
printf "https://ftp.drupal.org/files/projects/%s-%s.tar.gz\n", $2, parts[2]
}' a
https://ftp.drupal.org/files/projects/admin_toolbar-8.x-2.5.tar.gz
https://ftp.drupal.org/files/projects/entity_embed-8.x-1.2.tar.gz
https://ftp.drupal.org/files/projects/viewsreference-8.x-2.0-beta2.tar.gz
https://ftp.drupal.org/files/projects/webform-8.x-5.28.tar.gz
The trick is to split the third field ($3). Based on your field separator ( -F'[()]'), the third field contains everything after the right paren. So, split can be used to get rid of all the spaces. I probably should have searched for an awk "trim" equivalent.
In the example data, the second last column seems to contain the part with the parenthesis that you are interested in, and the value of the last column.
If that is always the case, you can remove the parenthesis from the second last column, and concat the hyphen and the last column.
awk '{
gsub(/[()]/, "", $(NF-1))
printf "https://ftp.drupal.org/files/projects/%s-%s.tar.gz%s", $(NF-1), $NF, ORS
}' file
Output
https://ftp.drupal.org/files/projects/admin_toolbar-8.x-2.5.tar.gz
https://ftp.drupal.org/files/projects/entity_embed-8.x-1.2.tar.gz
https://ftp.drupal.org/files/projects/viewsreference-8.x-2.0-beta2.tar.gz
https://ftp.drupal.org/files/projects/webform-8.x-5.28.tar.gz
Another option with a regex and gnu awk, using match and 2 capture groups to capture what is between the parenthesis and the next field.
awk 'match($0, /^[^()]*\(([^()]+)\)\s+(\S+)/, ary) {
printf "https://ftp.drupal.org/files/projects/%s-%s.tar.gz%s", ary[1], ary[2], ORS
}' file
This might work for you (GNU sed):
sed 's#.*(#https://ftp.drupal.org/files/projects/#;s/)\s*/-/;s/\s*$/.tar.gz/' file
Pattern match, replacing the unwanted parts by the required strings.
N.B. The use of the # as a delimiter for the substitution command to avoid inserting back slashes into the literal replacement.
The above solution could be ameliorated into:
sed -E 's#.*\((.*)\)\s*(\S*).*#https://ftp.drupal.org/files/projects/\1-\2.tar.gz#' file
The following is what I have written that would allow me to display only the phone numbers
in the file. I have posted the sample data below as well.
As I understand (read from left to right):
Using awk command delimited by "," if the first char is an Int and then an int preceded by [-,:] and then an int preceded by [-,:]. Show the 3rd column.
I used "www.regexpal.com" to validate my expression. I want to learn more and an explanation would be great not just the answer.
GNU bash, version 4.4.12(1)-release (x86_64-pc-linux-gnu)
awk -F "," '/^(\d)+([-,:*]\d+)+([-,:*]\d+)*$/ {print $3}' bashuser.csv
bashuser.csv
Jordon,New York,630-150,7234
Jaremy,New York,630-250-7768
Jordon,New York,630*150*7745
Jaremy,New York,630-150-7432
Jordon,New York,630-230,7790
Expected Output:
6301507234
6302507768
....
You could just remove all non int
awk '{gsub(/[^[:digit:]]/, "")}1' file.csv
gsub remove all match
[^[:digit:]] the ^ everything but what is next to it, which is an int [[:digit:]], if you remove the ^ the reverse will happen.
"" means remove or delete in awk inside the gsub statement.
1 means print all, a shortcut for print
In sed
sed 's/[^[:digit:]]*//g' file.csv
Since your desired output always appears to start on field #3, you can simplify your regrex considerably using the following:
awk -F '[*,-]' '{print $3$4$5}'
Proof of concept
$ awk -F '[*,-]' '{print $3$4$5}' < ./bashuser.csv
6301507234
6302507768
6301507745
6301507432
6302307790
Explanation
-F '[*,-]': Use a character class to set the field separators to * OR , OR -.
print $3$4$5: Concatenate the 3rd through 5th fields.
awk is not very suitable because the comma occurs not only as a separator of records, better results will give sed:
sed 's/[^,]\+,[^,]\+,//;s/[^0-9]//g;' bashuser.csv
first part s/[^,]\+,[^,]\+,// removes first two records
second part //;s/[^0-9]//g removes all remaining non-numeric characters
sorry for the nth simple question on regexp but I'm not able to get what I need without a what seems to me a too complicated solution. I'm parsing a file containing sequence of only 3 letters A,E,D as in
AADDEEDDA
EEEEEEEE
AEEEDEEA
AEEEDDAAA
and I'd like to identify only those that start with E and ends in D with only one change in the sequence as for example in
EDDDDDDDD
EEEDDDDDD
EEEEEEEED
I'm fighting with the proper regexp to do that. Here my last attempt
echo "1,AAEDDEED,1\n2,EEEEDDDD,2\n3,EDEDEDED" | gawk -F, '{if($2 ~ /^E[(ED){1,1}]*D$/ && $2 !~ /^E[(ED){2,}]*D$/) print $0}'
which does not work. Any help?
Thanks in advance.
If i understand correctly your request a simple
awk '/^E+D+$/' file.input
will do the trick.
UPDATE: if the line format contains pre/post numbers (with post optional) as showed later in the example, this can be a possible pure regex adaptation (alternative to the use of field switch-F,):
awk '/^[0-9]+,E+D+(,[0-9]+)?$/' input.test
First of all, you need the regular expression:
^E+[^ED]*D+$
This matches one or more Es at the beginning, zero or more characters that are neither E nor D in the middle, and one or more Ds at the end.
Then your AWK program will look like
$2 ~ /^E+[^ED]*D+$/
$2 refers to the 2nd field of the current record, ~ is the regex matching operator, and /s delimit a regular expression. Together, these components form what is known in AWK jargon as a "pattern", which amounts to a boolean filter for input records. Note that there is no "action" (a series of statements in {s) specified here. That's because when no action is specified, AWK assumes that the action should be { print $0 }, which prints the entire line.
If I understand you correct you want to match patterns that starts with at least one E and then continues with at least one D until the end.
echo "1,AAEDDEED,1\n2,EEEEDDDD,2\n3,EDEDEDED" | gawk -F, '{if($2 ~ /^E+D+$) print $0}'
I have a rather large chart to parse. Each column is separated by either 4 spaces or by 3 spaces and a hyphen (since the numbers in the chart can be negative).
cat DATA.txt | awk "{ print match($0,/\s\s/) }"
does nothing but print a slew of 0's. I'm trying to understand AWK and when to escape, etc, but I'm not getting the hang of it. Help is appreciated.
One line:
1979 1 -0.176 -0.185 -0.412 0.069 -0.129 0.297 -2.132 -0.334 -0.019
1979 1 -0.176 0.185 -0.412 0.069 -0.129 0.297 -2.132 -0.334 -0.019
I would like to get just, say, the second column. I copied the line, but I'd like to see -0.185 and 0.185.
You need to start by thinking about bash quoting, since it is bash which interprets the argument to awk which will be the awk program. Inside double-quoted strings, bash expands $0 to the name of the bash executable (or current script); that's almost certainly not what you want, since it will not be a quoted string. In fact, you almost never want to use double quotes around the awk program argument, so you should get into the habit of writing awk '...'.
Also, awk regular expressions don't understand \s (although Gnu awk will handle that as an extension). And match returns the position of the match, which I don't think you care about either.
Since by default, awk considers any sequence of whitespace a field separator, you don't really need to play any games to get the fourth column. Just use awk '{print $4}'
Why not just use this simple awk
awk '$0=$4' Data.txt
-0.185
0.185
It sets $0 to value in $4 and does the default action, print.
PS do not use cat with program that can read data itself, like awk
In case of filed 4 containing 0, you can make it more robust like:
awk '{$0=$4}1' Data.txt
If you're trying to split the input according to 3 or 4 spaces then you will get the expected output only from column 3.
$ awk -v FS=" {3,4}" '{print $3}' file
-0.185
0.185
FS=" {3,4}" here we pass a regex as FS value. This regex get parsed and set the Field Separator value to three or four spaces. In regex {min,max} called range quantifier which repeats the previous token from min to max times.
I've got some textfiles that hold names, phone numbers and region codes. One combination per line.
The syntax is always "Name Region_code number"
With any number of spaces between the 3 variables.
What I want to do is search for specific region codes, like 23 or 493, forexample.
The problem is that these numbers might appear in the longer numbers too, which might enable a return that shouldn't have been returned.
I was thinking of this sort of command:
grep '04' numbers.txt
But if I do that, a line that contains 04 in the number but not as region code will show as a result too... which is not correct.
I'm sure you are about to get buried in clever regular expressions, but I think in this case all you need to do is include one of the spaces on each side of your region code in the grep.
grep ' 04 ' numbers.txt
I'd do:
awk '$2 == "04"' < numbers.txt
and with grep:
grep -e '^[^ ]*[ ]*04[ ]*[^ ]*$' numbers.txt
If you want region codes alone, you should use:
grep "[[:space:]]04[[:space:]]"
this way it will only look for numbers on the middle column, while start or end of strings are considered word breaks.
You can even do:
function search_region_codes {
grep "[[:space:]]${1}[[:space:]]" FILE
}
replacing FILE with the name of your file,
and use
search_region_codes 04
or even
function search_region_codes {
grep "[[:space:]]${1}[[:space:]]" $2
}
and using
search_region_codes NUMBER FILE
Are you searching for an entire region code, or a region code that contains the subpattern?
If you want the whole region code, and there is at least one space on either side, then you can format the grep by adding a single space on either side of the specific region code. There are other ways to indicate word boundaries using regular expressions.
grep ' 04 ' numbers.txt
If there can be spaces in the name or phone number fields, than that solution might not work. Also, if you the pattern can be a sub-part of the region code, then awk is a better tool. This assumes that the 'name' field contains no spaces. The matching operator '==' requires that the pattern exactly match the field. This can be tricky when there is whitespace on either side of the field.
awk '$2 == "04" {print $0}' < numbers.txt
If the file has a delimiter, than can be set in awk using the '-F' argument to awk to set the field separator character. In this example, a comma is used as the field separator. In addition, the matching operator in this example is a '~' allowing the pattern to be any part of the region code (if that is applicable). The "/y" is a way to match work boundaries at the beginning and end of the expression.
awk -F , '$2 ~ /\y04\y/ {print $0}' < numbers.txt
In both examples, the {print $0} is optional, if you want the full line to be printed. However, if you want to do any formatting on the output, that can be done inside that block.
use word boundaries. not sure if this works in grep, but in other regex implementations i'd surround it with whitespace or word boundary patterns
'\s+04\s+' or '\b04\b'
Something like that