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
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
I have a file containing many SQL statements and need to add escape characters, using SED, for single quotes withing the SQL statements. Consider the following:
INSERT INTO MYTABLE VALUES (1,'some text','Drink at O'Briens');
In the above we need to escape the single quote in O'Briens. Using regex I can find the string using [a-zA-Z ]'[a-zA-Z ].
So this will find the 3 characters of interest, however if I do the following sed command:
sed -i "s/[a-zA-Z ]'[a-zA-Z ]/''/g" file.sql
This, however, removes the O and the B so I end up with:
INSERT INTO MYTABLE VALUES (1,'some text','Drink at ''riens');
How do I isolate/reference the O and the B so the string becomes:
INSERT INTO MYTABLE VALUES (1,'some text','Drink at O''Briens');
Use capture groups to copy parts of the input to the result.
sed -r -i "s/([a-zA-Z ])'([a-zA-Z ])/\1''\2/g" file.sql
You could do this in awk. Simple explanation would be, perform substitution on last field of line, where substitute ' with 2 instances of ' and print the line then.
awk '{sub(/\047/,"&&",$NF)} 1' Input_file
Above code will only print the lines in output, in case you want to perform inplace save then try following.
awk '{sub(/\047/,"&&",$NF)} 1' Input_file > temp && mv temp Input_file
How can I use sed to add a dynamic prefix to each number in an integer list?
For example:
I have a string "A-1,2,3,4,5", I want to transform it to string "A-1,A-2,A-3,A-4,A-5" - which means I want to add prefix of first integer i.e. "A-" to each number of the list.
If I have string like "B-1,20,300" then I want to transform it to string "B-1,B-20,B-300".
I am not able to use RegEx Capturing Groups because for global match they do not retain their value in subsequent matches.
When it comes to looping constructs in sed, I like to use newlines as markers for the places I have yet to process. This makes matching much simpler, and I know they're not in the input because my input is a text line.
For example:
$ echo A-1,2,3,4,5 | sed 's/,/\n/g;:a s/^\([^0-9]*\)\([^\n]*\)\n/\1\2,\1/; ta'
A-1,A-2,A-3,A-4,A-5
This works as follows:
s/,/\n/g # replace all commas with newlines (insert markers)
:a # label for looping
s/^\([^0-9]*\)\([^\n]*\)\n/\1\2,\1/ # replace the next marker with a comma followed
# by the prefix
ta # loop unless there's nothing more to do.
The approach is similar to #potong's, but I find the regex much more readable -- \([^0-9]*\) captures the prefix, \([^\n]*\) captures everything up to the next marker (i.e. everything that's already been processed), and then it's just a matter of reassembling it in the substitution.
Don't use sed, just use the other standard UNIX text manipulation tool, awk:
$ echo 'A-1,2,3,4,5' | awk '{p=substr($0,1,2); gsub(/,/,"&"p)}1'
A-1,A-2,A-3,A-4,A-5
$ echo 'B-1,20,300' | awk '{p=substr($0,1,2); gsub(/,/,"&"p)}1'
B-1,B-20,B-300
This might work for you (GNU sed):
sed -E ':a;s/^((([^-]+-)[^,]+,)+)([0-9])/\1\3\4/;ta' file
Uses pattern matching and a loop to replace a number following a comma by the first column prefix and that number.
Assuming this is for shell scripting, you can do so with 2 seds:
set string = "A1,2,3,4,5"
set prefix = `echo $string | sed 's/^\([A-Z]\).*/\1/'`
echo $string | sed 's/,\([0-9]\)/,'$prefix'-\1/g'
Output is
A1,A-2,A-3,A-4,A-5
With
set string = "B-1,20,300"
Output is
B-1,B-20,B-300
Could you please try following(if ok with awk).
awk '
BEGIN{
FS=OFS=","
}
{
for(i=1;i<=NF;i++){
if($i !~ /^A/&&$i !~ /\"A/){
$i="A-"$i
}
}
}
1' Input_file
if your data in 'd' file, tried on gnu sed:
sed -E 'h;s/^(\w-).+/\1/;x;G;:s s/,([0-9]+)(.*\n(.+))/,\3\1\2/;ts; s/\n.+//' d
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 $3}' file
-0.185
0.185
FS=" {3,4}" here we pass a regex as FS value. This regex get parsed and set the Field Separator value to three or four spaces. In regex {min,max} called range quantifier which repeats the previous token from min to max times.
I'm trying to automagically remove all lines from a text file that contains a letter "T" that is not immediately followed by a "H". I've been using grep and sending the output to another file, but I can't come up with the magic regex that will help me do this.
I don't mind using awk, sed, or some other linux tool if grep isn't the right tool to be using.
That should do it:
grep -v 'T[^H]'
-v : print lines not matching
[^H]: matches any character but H
You can do:
grep -v 'T[^H]' input
-v is the inverse match option of grep it does not list the lines that match the pattern.
The regex used is T[^H] which matches any lines that as a T followed by any character other than a H.
Read lines from file exclude EMPTY Lines and Lines starting with #
grep -v '^$\|^#' folderlist.txt
folderlist.txt
# This is list of folders
folder1/test
folder2
# This is comment
folder3
folder4/backup
folder5/backup
Results will be:
folder1/test
folder2
folder3
folder4/backup
folder5/backup
Adding 2 awk solutions to the mix here.
1st solution(simpler solution): With simple awk and any version of awk.
awk '!/T/ || /TH/' Input_file
Checking 2 conditions:
If a line doesn't contain T OR
If a line contains TH then:
If any of above condition is TRUE then print that line simply.
2nd solution(GNU awk specific): Using GNU awk using match function where mentioning regex (T)(.|$) and using match function's array creation capability.
awk '
!/T/{
print
next
}
match($0,/(T)(.|$)/,arr) && arr[1]=="T" && arr[2]=="H"
' Input_file
Explanation: firstly checking if a line doesn't have T then print that simply. Then using match function of awk to match T followed by any character OR end of the line. Since these are getting stored into 2 capturing groups so checking if array arr's 1st element is T and 2nd element is H then print that line.