How to set FS to eof? - bash

I want to read whole file not per lines. How to change field separator to eof-symbol?
I do:
awk "^[0-9]+∆DD$1[PS].*$" $(ls -tr)
$1 - param (some integer), .* - message that I want to print. There is a problem: message can contains \n. In that way this code prints only first line of file. How can I scan whole file not per lines?
Can I do this using awk, sed, grep? Script must have length <= 60 characters (include spaces).


Assuming you mean record separator, not field separator, with GNU awk you'd do:
gawk -v RS='^$' '{ print "<" $0 ">" }' file
Replace the print with whatever you really want to do and update your question with some sample input and expected output if you want help with that part too.
The portable way to do this, by the way, is to build up the record line by line and then process it in the END section:
awk '{rec = rec (NR>1?RS:"") $0} END{ print "<" rec ">" }' file
using nf = split(rec,flds) to create fields if necessary.

Related

Separating onto a new line based on a delimiter

I have some rows in my file that look like this
ENSG00000003096:E4.2|E5.1
ENSG00000035115:E14.2|E15.1
ENSG00000140987:E5.2|ENSG00000140987:E6.1
ENSG00000154358:E46.1|E47.1
I would like to separate them onto a new line based on the delimiter "|" , such that it becomes
ENSG00000003096:E4.2
ENSG00000003096:E5.1
ENSG00000035115:E14.2
ENSG00000035115:E15.1
ENSG00000140987:E5.2
ENSG00000140987:E6.1
ENSG00000154358:E46.1
ENSG00000154358:E47.1

With input data as advised in your question, this seems to work with gnu awk:
awk -F: -v RS="[|]|\n" 'NF==1{print p FS $0;next}NF!=1{p=$1}1' file1
#Output
ENSG00000003096:E4.2
ENSG00000003096:E5.1
ENSG00000035115:E14.2
ENSG00000035115:E15.1
ENSG00000140987:E5.2
ENSG00000140987:E6.1
ENSG00000154358:E46.1
ENSG00000154358:E47.1
Logic:
| or \n are used as record separator RS
: is used as field separator FS
If a line has more than one fields then keep the first field in a variable p
if a line has only one field then print previous $1 = variable p and the line $0

You may mean something like
awk 'BEGIN{FS=":"}{ split($2, fields, "|"); print $1 ":" fields[1]; print $1 ":" fields[2]; }' my_file.txt

Append and replace using awk/sed

I have this file:
2016,05,P,0002 ,CJGLOPSD8
00,BBF,BBDFTP999,051000100,GBP, , -2705248.00
00,BBF,BBDFTP999,059999998,GBP, , -3479679.38
00,BBF,BBDFTP999,061505141,GBP, , -0.40
00,BBF,BBDFTP999,061505142,GBP, , 6207621.00
00,BBF,BBDFTP999,061505405,GBP, , -0.16
00,BBF,BBDFTP999,061552000,GBP, , -0.24
00,BBF,BBDFTP999,061559010,GBP, , -0.44
00,BBF,BBDFTP999,062108021,GBP, , -0.34
00,BBF,BBDFTP999,063502007,GBP, , -0.28
I want to programmatically (in unix, or informatica if possible) grab the first two fields in the top row, concatenate them, append them to the end of each line and remove that first row.
Like so:
00,BBF,BBDFTP999,051000100,GBP,,-2705248.00,201605
00,BBF,BBDFTP999,059999998,GBP,,-3479679.38,201605
00,BBF,BBDFTP999,061505141,GBP,,-0.40,201605
00,BBF,BBDFTP999,061505142,GBP,,6207621.00,201605
00,BBF,BBDFTP999,061505405,GBP,,-0.16,201605
00,BBF,BBDFTP999,061552000,GBP,,-0.24,201605
00,BBF,BBDFTP999,061559010,GBP,,-0.44,201605
00,BBF,BBDFTP999,062108021,GBP,,-0.34,201605
00,BBF,BBDFTP999,063502007,GBP,,-0.28,201605
This is my current attempt:
awk -vvar1=`cat OF\ OPSDOWN8.CSV | head -1 | cut -d',' -f1` -vvar2=`cat OF\ OPSDOWN8.CSV | head -1 | cut -d',' -f2` 'BEGIN {FS=OFS=","} {print $0, var 1var2}' OF\ OPSDOWN8.CSV> OF_OPSDOWN8.csv
Any pointers? I've tried looking around the forum but can only find answers to part of my question.
Thanks for your help.

Use this awk:
awk 'BEGIN{FS=OFS=","} NR==1{val=$1$2;next} {gsub(/ */,"");print $0,val}' file
Explanation:
BEGIN{FS=OFS=","} - This block will set FS (Field Separator) and OFS (Output Field Separator) as ,.
NR==1 - Working with line number 1. Here, $1 and $2 denotes field number.
print $0,val - Printing $0 (whole line) and stored value from val.

I would use the following awk command:
awk 'NR==1{d=$1$2;next}{$(NF+1)=d;gsub(/[[:space:]]/,"")}1' FS=, OFS=, file
Explanation:
NR==1{d=$1$2;next} applies on line 1 and set's a variable d(ate) to the value of the first and the second field. The variable is being used when processing the remaining lines. next tells awk to go ahead with the next line right away without processing further instructions on this line.
{$(NF+1)=d;gsub(/[[:space:]]/,"")}1 appends a new field to the line (NF is the number of fields, assigning d to $(NF+1) effectively adds a field. gsub() is used to removing spaces. 1 at the end always evaluates to true and makes awk print the modified line.
FS=, is a command line argument. It set's the input field delimiter to ,.
OFS=, is a command line argument. It set's the output field delimiter to ,.
Output:
00,BBF,BBDFTP999,051000100,GBP,,-2705248.00,201605
00,BBF,BBDFTP999,059999998,GBP,,-3479679.38,201605
00,BBF,BBDFTP999,061505141,GBP,,-0.40,201605
00,BBF,BBDFTP999,061505142,GBP,,6207621.00,201605
00,BBF,BBDFTP999,061505405,GBP,,-0.16,201605
00,BBF,BBDFTP999,061552000,GBP,,-0.24,201605
00,BBF,BBDFTP999,061559010,GBP,,-0.44,201605
00,BBF,BBDFTP999,062108021,GBP,,-0.34,201605
00,BBF,BBDFTP999,063502007,GBP,,-0.28,201605

With sed :
sed '1{s/\([^,]*\),\([^,]*\),.*/\1\2/;h;d};/.*/G;s/\n/,/;s/ //g' file
in ERE mode :
sed -r '1{s/([^,]*),([^,]*),.*/\1\2/;h;d};/.*/G;s/\n/,/;s/ //g' file
Output :
00,BBF,BBDFTP999,051000100,GBP,,-2705248.00,201605
00,BBF,BBDFTP999,059999998,GBP,,-3479679.38,201605
00,BBF,BBDFTP999,061505141,GBP,,-0.40,201605
00,BBF,BBDFTP999,061505142,GBP,,6207621.00,201605
00,BBF,BBDFTP999,061505405,GBP,,-0.16,201605
00,BBF,BBDFTP999,061552000,GBP,,-0.24,201605
00,BBF,BBDFTP999,061559010,GBP,,-0.44,201605
00,BBF,BBDFTP999,062108021,GBP,,-0.34,201605
00,BBF,BBDFTP999,063502007,GBP,,-0.28,201605

This might work for you (GNU sed):
sed '1s/,//;1s/,.*//;1h;1d;s/ //g;G;s/\n/,/' file
For the first line only: remove the first comma, remove from the next comma to the end of the line, store the amended line in the hold space (HS) and then delete the current line (the d abruptly ends processing). For subsequent lines: remove all spaces, append the HS and replace the newline (from the G command) with a comma.
Or if you prefer:
sed '1{s/,//;s/,.*//;h;d};s/ //g;G;s/\n/,/' file

If you want to use Informatica for this, use two Source Qualifiers. Read the file twice - just one line in one SQ (filter out the rest) and in the second SQ read the whole file except the first line (skip header). Join the two on dummy port and you're done.

Awk, Shell Scripting

I have a file which has the following form:
#id|firstName|lastName|gender|birthday|creationDate|locationIP|browserUsed
111|Arkas|Sarkas|male|1995-09-11|2010-03-17T13:32:10.447+0000|192.248.2.123|Midori
Every field is separated with "|". I am writing a shell script and my goal is to remove the "-" from the fifth field (birthday), in order to make comparisons as if they were numbers.
For example i want the fifth field to be like |19950911|
The only solution I have reached so far, deletes all the "-" from each line which is not what I want using sed.
i would be extremely grateful if you show me a solution to my problem using awk.

If this is a homework writing the complete script will be a disservice. Some hints: the function you should be using is gsub in awk. The fifth field is $5 and you can set the field separator by -F'|' or in BEGIN block as FS="|"
Also, line numbers are in NR variable, to skip first line for example, you can add a condition NR>1

An awk one liner:
awk 'BEGIN { FS="|" } { gsub("-","",$5); print }' infile.txt

To keep "|" as output separator, it is better to define OFS value as "|" :
... | awk 'BEGIN { FS="|"; OFS="|"} {gsub("-","",$5); print $0 }'

How to print the line number where a string appears in a file?

I have a specific word, and I would like to find out what line number in my file that word appears on.
This is happening in a c shell script.
I've been trying to play around with awk to find the line number, but so far I haven't been able to. I want to assign that line number to a variable as well.

Using grep
To look for word in file and print the line number, use the -n option to grep:
grep -n 'word' file
This prints both the line number and the line on which it matches.
Using awk
This will print the number of line on which the word word appears in the file:
awk '/word/{print NR}' file
This will print both the line number and the line on which word appears:
awk '/word/{print NR, $0}' file
You can replace word with any regular expression that you like.
How it works:
/word/
This selects lines containing word.
{print NR}
For the selected lines, this prints the line number (NR means Number of the Record). You can change this to print any information that you are interested in. Thus, {print NR, $0} would print the line number followed by the line itself, $0.
Assigning the line number to a variable
Use command substitution:
n=$(awk '/word/{print NR}' file)
Using shell variables as the pattern
Suppose that the regex that we are looking for is in the shell variable url:
awk -v x="$url" '$0~x {print NR}' file
And:
n=$(awk -v x="$url" '$0~x {print NR}' file)

Sed
You can use the sed command
sed -n '/pattern/=' file
Explanation
The -n suppresses normal output so it doesn't print the actual lines. It first matches the /pattern/, and then the = operator means print the line number. Note that this will print all lines that contains the pattern.

Use the NR Variable
Given a file containing:
foo
bar
baz
use the built-in NR variable to find the line number. For example:
$ awk '/bar/ { print NR }' /tmp/foo
2

find the line number for which the first column match RRBS
awk 'i++ {if($1~/RRBS/) print i}' ../../bak/bak.db

Cut and replace bash

I have to process a file with data organized like this
AAAAA:BB:CCC:EEEE:DDDD
FF:III:JJJ:KK:LLL
MMMM:NN:OOO:PP
etc
Columns can have different length but lines always have the same number of columns.
I want to be able to cut a specific column of a given line and change it to the value I want.
For example I'd apply my command and change the file to
AAAAA:BB:XXXX:EEEE:DDDD
FF:III:JJJ:KK:LLL
MMMM:NN:OOO:PP
I know how to select a specific line with sed and then cut the field but I have no idea on how to replace the field with the value I have.
Thanks

Here's a way to do it with awk:
Going with your example, if you wanted to replace the 3rd field of the 1st line:
awk 'BEGIN{FS=OFS=":"} {if (NR==1) {$3 = "XXXX"}; print}' input_file
Input:
AAAAA:BB:CCC:EEEE:DDDD
FF:III:JJJ:KK:LLL
MMMM:NN:OOO:PP
Output:
AAAAA:BB:XXXX:EEEE:DDDD
FF:III:JJJ:KK:LLL
MMMM:NN:OOO:PP
Explanation:
awk: invoke the awk command
'...': everything enclosed by single-quotes are instructions to awk
BEGIN{FS=OFS=":"}: Use : as delimiters for both input and output. FS stands for Field Separator. OFS stands for Output Field Separator.
if (NR==1) {$3 = "XXXX"};: If Number of Records (NR) read so far is 1, then set the 3rd field ($3) to "XXXX".
print: print the current line
input_file: name of your input file.
If instead what you are trying to accomplish is simply replace all occurrences of CCC with XXXX in your file, simply do:
sed -i 's/CCC/XXXX/g` input_file
Note that this will also replace partial matches, such as ABCCCDD -> ABXXXXDD

This might work for you (GNU sed):
sed -r 's/^(([^:]*:?){2})CCC/\1XXXX/' file
or
awk -F: -vOFS=: '$3=="CCC"{$3="XXXX"};1' file

Resources