Please suggest perfect shell script command to remove last two '||' delimiter separated columns from the file.(Lets assume below example)
File Name: abc.dat
"a1"||"a2"||"a3"||"a4"
"b1"||"b2"||"b3"||"b4"
"c1"||"c2"||"c3"||"c4"
output should be like :
"a1"||"a2"
"b1"||"b2"
"c1"||"c2"
I tried below cut and awk command but not worked:
awk -F '||' '{print $1$2}' ${file} >> ${file}
cut -d'||' -f2 --complement ${file} >> ${file} (not working as cut: the delimiter must be a single character)
With your shown samples, please try following. Make field separator as ||(escaping it to treat literal character) along with setting OFS to || too. Then print 1st and 2nd fields for each line of Input_file.
awk -F'\\|\\|' -v OFS="||" '{print $1,$2}' Input_file
Once you are happy with results of above command, also to make changes into Input_file itself try following.
awk -F'\\|\\|' -v OFS="||" '{print $1,$2}' Input_file > temp && mv temp Input_file
2nd solution: Using GNU grep try following.
grep -oP '^.*?\|\|"[^"]*' Input_file
Rather than assuming || is the delimiter, assume that | is the delimiter and the second field is empty.
$ cut -d'|' -f1-3 <<EOF
> "a1"||"a2"||"a3"||"a4"
> "b1"||"b2"||"b3"||"b4"
> "c1"||"c2"||"c3"||"c4"
> EOF
"a1"||"a2"
"b1"||"b2"
"c1"||"c2"
(This assumes that || was chosen for some aesthetic reason, rather than to allow for single pipes in each field.)
You may use:
awk '{sub(/(\|{2}[^|]*){2}$/, "")} 1' file
"a1"||"a2"
"b1"||"b2"
"c1"||"c2"
Or if you just want to remove last 2 columns without caring how many columns are there in total use:
awk -F '\\|{2}' -v OFS='||' '{
$NF = $(NF-1) = ""
sub(/([|]{2})*$/, "")
} 1' file
Related
I am trying to use awk to select/remove data based on cell entries in a CSV file.
How do I chain Awk commands to build up complex searches like I have done with grep? I plan to use Awk to select rows based on matching criteria in cells in multiple columns, not just the first column as in this example.
Test data
123,line1
123a,line2
abc,line3
G-123,line4
G-123a,line5
Separate Awk statements with intermediate files
awk '$1 !~ /^[[:digit:]]/ {print $0}' file.txt > output1.txt
awk '$1 !~ /^G-[[:digit:]]/ {print $0}' output1.txt > output2.txt
mv output2.txt output.txt
cat output.txt
Chained or multi-line grep version (I think limited to first column only)
grep -v \
-e "^[[:digit:]]" \
-e "^G-[[:digit:]]" \
file.txt > output.txt
cat output.txt
How can I rewrite the Awk command to avoid the intermediate files?
Generally, in awk there are boolean operators available (it's better than grep! :) )
awk '/match1/ || /match2/' file
awk '(/match1/ || /match2/ ) && /match3/' file
and so on ...
In your example you could use something like:
awk -F, '$1 ~ /^[[:digit:]]/ || $1 ~ /G-[[:digit:]]/' input >> output
Note: This is just an example of how to use boolean operators. Also the regular expression itself could have been used here to express the alternative match:
awk -F, '$1 ~ /^(G-)?[[:digit:]]/' input >> ouput
In your awk commands and example, awk regards file.txt as having only one field because you have not defined FS, so the default whitespace field separator is used.
With that said, you can easily AND your two pattern matches together like this:
awk '($1 !~ /^[[:digit:]]/) && ($1 !~ /^G-[[:digit:]]/) {print $0}' file.txt
To make awk use comma as a field separator, you can define it in a BEGIN block. In this example, the output should be just line3
awk 'BEGIN {FS=","} ($1 !~ /^[[:digit:]]/) && ($1 !~ /^G-[[:digit:]]/) {print $2}' file.txt
I would suggest the literal translation of that grep command in awk is
awk '
/^[[:digit:]]/ {next}
/^G-[[:digit:]]/ {next}
{print}
' file.txt
But you have several examples of how to write it more concisely.
You can use
awk '$1 !~ /^(G-)?[[:digit:]]/' file.txt > output.txt
The awk tries to find in Field 1:
^ - start of string
(G-)? - an optional G- char sequence (note the regex flavor in awk is POSIX ERE, so (...) denotes a capturing group and ? denotes a one or zero times quantifier)
[[:digit:]] - a digit.
If the match is found, the record (=line) is not printed. Else, the line is printed.
to stick to your question, I would use:
awk '$1 !~ /^[[:digit:]]/ && $1 !~ /G-[[:digit:]]/' file.txt > output.txt
But I like the #Wiktor Stribiżew REGEX approach!
With your shown samples, this could be also done in grep in a single regexp, we need not to chain the different regex, adding this solution in case you/anyone need it; could be helpful.
grep -v -E '^(G-)?[[:digit:]]' Input_file
Explanation: Simple explanation would be, using grep's -v option to omit lines which are matching the mentioned pattern. Then using -E option of it to enable ERE(extended regular expressions). In main program using regex ^(G-)?[[:digit:]] to match if line starts from G- OR digit then don't print that line.
cat file1.txt | awk -F '{print $1 "|~|" $2 "|~|" $3}' > file2.txt
I am using above command to filter first three columns from file1 and put into file.
But only getting the column names and not the column data.
How to do that?
|~| - is the delimiter.
file1.txt has values as :
a|~|b|~|c|~|d|~|e
1|~|2|~|3|~|4|~|5
11|~|22|~|33|~|44|~|55
111|~|222|~|333|~|444|~|555
my expedted output is :
a|~|b|~|c
1|~|2|~|3
11|~|22|~|33
111|~|222|~|333
With your shown samples, please try following awk code. You need to set field separator to |~| and remove starting space from lines, then print the lines.
awk -F'\\|~\\|' -v OFS='|~|' '{sub(/^[[:blank:]]+/,"");print $1,$2,$3}' Input_file
In case you want to keep spaces(which was in initial post before edit) then try following:
awk -F'\\|~\\|' -v OFS='|~|' '{print $1,$2,$3}' Input_file
NOTE: Had a chat with user in room and got to know why this code was not working for user because of gunzip -c file was being used wrongly, its output was being saved into a variable on which user was running awk program, so correcting that command generated right file and awk program ran fine on it. Adding this as a reference for future readers.
One approach would be:
awk -v FS="," -v OFS="|~|" '{gsub(/[|][~][|]/,","); sub(/^\s*/,""); print $1,$2,$3}' file1.txt
The approach simply replaces all "|~|" with a "," setting the output file separator to "|~|". All leading whitespace is trimmed with sub().
Example Use/Output
With your data in file1.txt, you would have:
$ awk -v FS="," -v OFS="|~|" '{gsub(/[|][~][|]/,","); sub(/^\s*/,""); print $1,$2,$3}' file1.txt
a|~|b|~|c
1|~|2|~|3
11|~|22|~|33
111|~|222|~|333
Let me know if this is what you intended. You can simply redirect, e.g. > file2.txt to write to the second file.
For such cases, my bash+awk script rcut comes in handy:
rcut -Fd'|~|' -f-3 ip.txt
The -F option enables fixed string input delimiter (which is given using the -d option). And by default, the output field separator will also be same as -d when -F is active. -f-3 is similar to cut syntax to specify first three fields.
For better speed, use hck command:
hck -Ld'|~|' -D'|~|' -f-3 ip.txt
Here, -L enables literal field separator and -D specifies output field separator.
Another benefit is that hck supports -z option to automatically handle common compressed formats based on filename extension (adding this since OP had an issue with compressed input).
Another way:
sed 's/|~|/\t/g' file1.txt | awk '{print $1"|~|"$2"|~|"$3}' > file2.txt
First replace the |~| delimiter, and use the default awk separator, then print columns what you need.
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
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
I've got a file with following records:
depots/import/HDN1YYAA_15102018.txt;1;CAB001
depots/import/HDN1YYAA_20102018.txt;2;CLI001
depots/import/HDN1YYAA_20102018.txt;32;CLI001
depots/import/HDN1YYAA_25102018.txt;1;CAB001
depots/import/HDN1YYAA_50102018.txt;1;CAB001
depots/import/HDN1YYAA_65102018.txt;1;CAB001
depots/import/HDN1YYAA_80102018.txt;2;CLI001
depots/import/HDN1YYAA_93102018.txt;2;CLI001
When I execute following oneliner awk:
cat lignes_en_erreur.txt | awk 'FS=";"{ if(NR==1){print $1}}END {}'
the output is not the expected:
depots/import/HDN1YYAA_15102018.txt;1;CAB001
While I am suppose get only the frist column:
If I run it through all the records:
cat lignes_en_erreur.txt | awk 'FS=";"{ if(NR>0){print $1}}END {}'
then it will start filtering only after the second line and I get the following output:
depots/import/HDN1YYAA_15102018.txt;1;CAB001
depots/import/HDN1YYAA_20102018.txt
depots/import/HDN1YYAA_20102018.txt
depots/import/HDN1YYAA_25102018.txt
depots/import/HDN1YYAA_50102018.txt
depots/import/HDN1YYAA_65102018.txt
depots/import/HDN1YYAA_80102018.txt
depots/import/HDN1YYAA_93102018.txt
Does anybody knows why awk is skiping the first line only.
I tried deleting first record but the behaviour is the same, it will skip the first line.
First, it should be
awk 'BEGIN{FS=";"}{ if(NR==1){print $1}}END {}' filename
You can omit the END block if it is empty:
awk 'BEGIN{FS=";"}{ if(NR==1){print $1}}' filename
You can use the -F command line argument to set the field delimiter:
awk -F';' '{if(NR==1){print $1}}' filename
Furthermore, awk programs consist of a sequence of CONDITION [{ACTIONS}] elements, you can omit the if:
awk -F';' 'NR==1 {print $1}' filename
You need to specify delimiter in either BEGIN block or as a command-line option:
awk 'BEGIN{FS=";"}{ if(NR==1){print $1}}'
awk -F ';' '{ if(NR==1){print $1}}'
cut might be better suited here, for all lines
$ cut -d';' -f1 file
to skip the first line
$ sed 1d file | cut -d';' -f1
to get the first line only
$ sed 1q file | cut -d';' -f1
however at this point it's better to switch to awk
if you have a large file and only interested in the first line, it's better to exit early
$ awk -F';' '{print $1; exit}' file