unix bash - extract a line from file - bash

need your help!!! I tried looking for this but to no avail.
How can I achieve the following using bash?
I've a flat file called "cube.mdl" that contains:
[...]
bla bla bla bla lots of lines above
Cube 8007841 "BILA_" MdcFile "BILA_CO_PM_MKT_BR_CUBE.mdc"
bla bla bla more lines below
[...]
I need to open that file, look for the word "MdcFile" and get the string that follows between quotes, which would be BILA_CO_PM_MKT_BR_CUBE.mdc
I know AWK or grep are powerful enough to do this in one line, but I couldn't find an example that could help me do it on my own.
Thanks in advance!
JMA

You can use:
grep -o -P "MdcFile.*" cube.mdl | awk -F\" '{ print $2 }'
This will use grep's regex to only return MdcFile and everything after it in the current line. Then, awk will use the " as a delimiter and print only the second word - which would be your "in-quotes" word(s), returned without the quotes of course.
The option -o, --only-matching specifies to return only the text matching that matches and the -P, --perl-regexp specifies that the pattern is a Perl-Regex pattern. It appears that some versions of grep do not contain these options. The OP's version is a version that does not include them, but the following appears to work for him instead:
grep "MdcFile.*" cube.mdl | awk -F\" '{ print $2 }'

grep MdcFile cube.mdl | awk '{print $5}'
would do it, assuming there's no spaces in any of those bits to throw off the position count.

This might do it.
sed -n '/MdcFile / s/.*MdcFile "\(\[^"\]\+\)".*/\1/;/MdcFile / p' INPUTFILE

Use awk for the whole thing with " as record separator:
awk -v RS='"' '/MdcFile/ { getline; print }' cube.mdl

Related

print text between pipes

I have a text file with ids as below
ref|XP_029641976.1|
ref|XP_014779594.1|
gb|KOF78315.1|
I wish to print only the text between pipes and I've tried this sed -n '/|/,/|/p and also tried substituting but they don't work. The string in front of the first pipe varies. Any ideas?
Thank you
You can get text between pipes with sed regexp like this
#!/bin/bash
echo "ref|XP_029641976.1|
ref|XP_014779594.1|
gb|KOF78315.1|" | sed -E "s:.*\|(.*?)\|:\1:g"
As others have suggested, awk is a good candidate for this:
awk -F'|' '{ print $2 }' file
Specify the field delimiter with -F and then print the second pipe delimited field with $2

Extract string between two patterns (inclusive) while conserving the format

I have a file in the following format
cat test.txt
id1,PPLLTOMaaaaaaaaaaaJACK
id2,PPLRTOMbbbbbbbbbbbJACK
id3,PPLRTOMcccccccccccJACK
I am trying to identify and print the string between TOM and JACK including these two strings, while maintaining the first column FS=,
Desired output:
id1,TOMaaaaaaaaaaaJACK
id2,TOMbbbbbbbbbbbJACK
id3,TOMcccccccccccJACK
So far I have tried gsub:
awk -F"," 'gsub(/.*TOM|JACK.*/,"",$2) && !_[$0]++' test.txt > out.txt
and have the following output
id1 aaaaaaaaaaa
id2 bbbbbbbbbbb
id3 ccccccccccc
As you can see I am getting close but not able to include TOM and JACK patterns in my output. Plus I am also losing the original FS. What am I doing wrong?
Any help will be appreciated.
You are changing a field ($2) which causes awk to reconstruct the record using the value of OFS as the field separator and so in this case changing the commas to spaces.
Never use _ as a variable name - using a name with no meaning is just slightly better than using a name with the wrong meaning, just pick a name that means something which, in this case is seen but idk what you are trying to do when using that in this context.
gsub() and sub() do not support capture groups so you either need to use match()+substr():
$ awk 'BEGIN{FS=OFS=","} match($2,/TOM.*JACK/){$2=substr($2,RSTART,RLENGTH)} 1' file
id1,TOMaaaaaaaaaaaJACK
id2,TOMbbbbbbbbbbbJACK
id3,TOMcccccccccccJACK
or use GNU awk for the 3rd arg to match()
$ gawk 'BEGIN{FS=OFS=","} match($2,/TOM.*JACK/,a){$2=a[0]} 1' file
id1,TOMaaaaaaaaaaaJACK
id2,TOMbbbbbbbbbbbJACK
id3,TOMcccccccccccJACK
or for gensub():
$ gawk 'BEGIN{FS=OFS=","} {$2=gensub(/.*(TOM.*JACK).*/,"\\1","",$2)} 1' file
id1,TOMaaaaaaaaaaaJACK
id2,TOMbbbbbbbbbbbJACK
id3,TOMcccccccccccJACK
The main difference between the match() and gensub() solutions is how they would behave if TOM appeared twice on the line:
$ cat file
id1,PPLLfooTOMbarTOMaaaaaaaaaaaJACK
id2,PPLRTOMbbbbbbbbbbbJACKfooJACKbar
id3,PPLRfooTOMbarTOMcccccccccccJACKfooJACKbar
$
$ awk 'BEGIN{FS=OFS=","} match($2,/TOM.*JACK/,a){$2=a[0]} 1' file
id1,TOMbarTOMaaaaaaaaaaaJACK
id2,TOMbbbbbbbbbbbJACKfooJACK
id3,TOMbarTOMcccccccccccJACKfooJACK
$
$ awk 'BEGIN{FS=OFS=","} {$2=gensub(/.*(TOM.*JACK).*/,"\\1","",$2)} 1' file
id1,TOMaaaaaaaaaaaJACK
id2,TOMbbbbbbbbbbbJACKfooJACK
id3,TOMcccccccccccJACKfooJACK
and just to show one way of stopping at the first instead of the last JACK on the line:
$ awk 'BEGIN{FS=OFS=","} match($2,/TOM.*JACK/,a){$2=gensub(/(JACK).*/,"\\1","",a[0])} 1' file
id1,TOMbarTOMaaaaaaaaaaaJACK
id2,TOMbbbbbbbbbbbJACK
id3,TOMbarTOMcccccccccccJACK
Use capture groups to save the parts of the line you want to keep. Here's how to do it with sed
sed 's/^\([^,]*,\).*\(TOM.*JACK\).*/\1\2/' <test.txt > out.txt
Do you mean to do the following?
$ cat test.txt
id1,PPLLTOMaaaaaaaaaaaJACKABCD
id2,PPLRTOMbbbbbbbbbbbJACKDFCC
id3,PPLRTOMcccccccccccJACKSDER
$ cat test.txt | sed -e 's/,.*TOM/,TOM/g' | sed -e 's/JACK.*/JACK/g'
id1,TOMaaaaaaaaaaaJACK
id2,TOMbbbbbbbbbbbJACK
id3,TOMcccccccccccJACK
$
This should work as long as the TOM and JACK do not repeat themselves.
sed 's/\(.*,\).*\(TOM.*JACK\).*/\1\2/' <oldfile >newfile
Output:
id1,TOMaaaaaaaaaaaJACK
id2,TOMbbbbbbbbbbbJACK
id3,TOMcccccccccccJACK

Add blank column using awk or sed

I have a file with the following structure (comma delimited)
116,1,89458180,17,FFFF,0403254F98
I want to add a blank column on the 4th field such that it becomes
116,1,89458180,,17,FFFF,0403254F98
Any inputs as to how to do this using awk or sed if possible ?
thank you
Assuming that none of the fields contain embedded commas, you can restate the task as replacing the third comma with two commas. This is just:
sed 's/,/,,/3'
With the example line from the file:
$ echo "116,1,89458180,17,FFFF,0403254F98" | sed 's/,/,,/3'
116,1,89458180,,17,FFFF,0403254F98
You can use this awk,
awk -F, '$4="," $4' OFS=, yourfile
(OR)
awk -F, '$4=FS$4' OFS=, yourfile
If you want to add 6th and 8th field,
awk -F, '{$4=FS$4; $1=FS$1; $6=FS$6}1' OFS=, yourfile
Through awk
$ echo '116,1,89458180,17,FFFF,0403254F98' | awk -F, -v OFS="," '{print $1,$2,$3,","$4,$5,$6}'
116,1,89458180,,17,FFFF,0403254F98
It prints a , after third field(delimited) by ,
Through GNU sed
$ echo 116,1,89458180,17,FFFF,0403254F98| sed -r 's/^([^,]*,[^,]*,[^,]*)(.*)$/\1,\2/'
116,1,89458180,,17,FFFF,0403254F98
It captures all the characters upto the third command and stored it into a group. Characters including the third , upto the last are stored into another group. In the replacement part, we just add an , between these two captured groups.
Through Basic sed,
Through Basic sed
$ echo 116,1,89458180,17,FFFF,0403254F98| sed 's/^\([^,]*,[^,]*,[^,]*\)\(.*\)$/\1,\2/'
116,1,89458180,,17,FFFF,0403254F98
echo 116,1,89458180,17,FFFF,0403254F98|awk -F',' '{print $1","$2","$3",,"$4","$5","$6}'
Non-awk
t="116,1,89458180,17,FFFF,0403254F98"
echo $(echo $t|cut -d, -f1-3),,$(echo $t|cut -d, -f4-)
You can use bellow awk command to achieve that.Replace the $3 with what ever the column that you want to make it blank.
awk -F, '{$3="" FS $3;}1' OFS=, filename
sed -e 's/\([^,]*,\)\{4\}/&,/' YourFile
replace the sequence of 4 [content (non comma) than comma ] by itself followed by a comma

How to extract specific lines from a file in bash?

I want to extract the string from a line which starts with a specific pattern from a file in shell script.
For example: I want the strings from lines that start with hello:
hi to_RAm
hello to_Hari
hello to_kumar
bye to_lilly
output should be
to_Hari
to_kumar
Can anyone help me?
sed is the most appropriate tool:
sed -n 's/^hello //p'
Use grep:
grep ^hello file | awk '{print $2}'
^ is to match lines that starts with "hello". This is assuming you want to print the second word.
If you want to print all words except the first then:
grep ^hello file | awk '{$1=""; print $0}'
You could use GNU grep's perl-compatible regexes and use a lookbehind:
grep -oP '(?<=hello ).*'

How do I print a field from a pipe-separated file?

I have a file with fields separated by pipe characters and I want to print only the second field. This attempt fails:
$ cat file | awk -F| '{print $2}'
awk: syntax error near line 1
awk: bailing out near line 1
bash: {print $2}: command not found
Is there a way to do this?
Or just use one command:
cut -d '|' -f FIELDNUMBER
The key point here is that the pipe character (|) must be escaped to the shell. Use "\|" or "'|'" to protect it from shell interpertation and allow it to be passed to awk on the command line.
Reading the comments I see that the original poster presents a simplified version of the original problem which involved filtering file before selecting and printing the fields. A pass through grep was used and the result piped into awk for field selection. That accounts for the wholly unnecessary cat file that appears in the question (it replaces the grep <pattern> file).
Fine, that will work. However, awk is largely a pattern matching tool on its own, and can be trusted to find and work on the matching lines without needing to invoke grep. Use something like:
awk -F\| '/<pattern>/{print $2;}{next;}' file
The /<pattern>/ bit tells awk to perform the action that follows on lines that match <pattern>.
The lost-looking {next;} is a default action skipping to the next line in the input. It does not seem to be necessary, but I have this habit from long ago...
The pipe character needs to be escaped so that the shell doesn't interpret it. A simple solution:
$ awk -F\| '{print $2}' file
Another choice would be to quote the character:
$ awk -F'|' '{print $2}' file
Another way using awk
awk 'BEGIN { FS = "|" } ; { print $2 }'
And 'file' contains no pipe symbols, so it prints nothing. You should either use 'cat file' or simply list the file after the awk program.

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