Field spearator to used if they are not escaped using awk - unix

i have once question, suppose i am using "=" as fiels seperator, in this case if my string contain for example
abc=def\=jkl
so if i use = as fields seperator, it will split into 3 as
abc def\ jkl
but as i have escaped 2nd "=" , my output should be as
abc def\=jkl
Can anyone please provide me any suggestion , if i can achieve this.
Thanks in advance

I find it simplest to just convert the offending string to some other string or character that doesn't appear in your input records (I tend to use RS if it's not a regexp* since that cannot appear within a record, or the awk builtin SUBSEP otherwise since if that appears in your input you have other problems) and then process as normal other than converting back within each field when necessary, e.g.:
$ cat file
abc=def\=jkl
$ awk -F= '{
gsub(/\\=/,RS)
for (i=1; i<=NF; i++) {
gsub(RS,"\\=",$i)
print i":"$i
}
}' file
1:abc
2:def\=jkl
* The issue with using RS if it is an RE (i.e. multiple characters) is that the gsub(RS...) within the loop could match a string that didn't get resolved to a record separator initially, e.g.
$ echo "aa" | gawk -v RS='a$' '{gsub(RS,"foo",$1); print "$1=<"$1">"}'
$1=<afoo>
When the RS is a single character, e.g. the default newline, that cannot happen so it's safe to use.

If it is like the example in your question, it could be done.
awk doesn't support look-around regex. So it would be a bit difficult to get what you want by setting FS.
If I were you, I would do some preprocessing, to make the data easier to be handled by awk. Or you could read the line, and using other functions by awk, e.g. gensub() to remove those = s you don't want to have in result, and split... But I guess you want to achieve the goal by playing field separator, so I just don't give those solutions.
However it could be done by FPAT variable.
awk -vFPAT='\\w*(\\\\=)?\\w*' '...' file
this will work for your example. I am not sure if it will work for your real data.
let's make an example, to split this string: "abc=def\=jkl=foo\=bar=baz"
kent$ echo "abc=def\=jkl=foo\=bar=baz"|awk -vFPAT='\\w*(\\\\=)?\\w*' '{for(i=1;i<=NF;i++)print $i}'
abc
def\=jkl
foo\=bar
baz
I think you want that result, don't you?
my awk version:
kent$ awk --version|head -1
GNU Awk 4.0.2

Related

Unix Text Processing - how to remove part of a file name from the results?

I'm searching through text files using grep and sed commands and I also want the file names displayed before my results. However, I'm trying to remove part of the file name when it is displayed.
The file names are formatted like this: aja_EPL_1999_03_01.txt
I want to have only the date without the beginning letters and without the .txt extension.
I've been searching for an answer and it seems like it's possible to do that with a sed or a grep command by using something like this to look forward and back and extract between _ and .txt:
(?<=_)\d+(?=\.)
But I must be doing something wrong, because it hasn't worked for me and I possibly have to add something as well, so that it doesn't extract only the first number, but the whole date. Thanks in advance.
Edit: Adding also the working command I've used just in case. I imagine whatever command is needed would have to go at the beginning?
sed '/^$/d' *.txt | grep -P '(^([A-ZÖÄÜÕŠŽ].*)?[Pp][Aa][Ll]{2}.*[^\.]$)' *.txt --colour -A 1
The results look like this:
aja_EPL_1999_03_02.txt:PALLILENNUD : korraga üritavad ümbermaailmalendu kaks meeskonda
A desired output would be this:
1999_03_02:PALLILENNUD : korraga üritavad ümbermaailmalendu kaks meeskonda
First off, you might want to think about your regular expression. While the one you have you say works, I wonder if it could be simplified. You told us:
(^([A-ZÖÄÜÕŠŽ].*)?[Pp][Aa][Ll]{2}.*[^\.]$)
It looks to me as if this is intended to match lines that start with a case insensitive "PALL", possibly preceded by any number of other characters that start with a capital letter, and that lines must not end in a backslash or a dot. So valid lines might be any of:
PALLILENNUD : korraga üritavad etc etc
Õlu on kena. Do I have appalling speling?
Peeter Pall is a limnologist at EMU!
If you'd care to narrow down this description a little and perhaps provide some examples of lines that should be matched or skipped, we may be able to do better. For instance, your outer parentheses are probably unnecessary.
Now, let's clarify what your pipe isn't doing.
sed '/^$/d' *.txt
This reads all your .txt files as an input stream, deletes any empty lines, and prints the output to stdout.
grep -P 'regex' *.txt --otheroptions
This reads all your .txt files, and prints any lines that match regex. It does not read stdin.
So .. in the command line you're using right now, your sed command is utterly ignored, as sed's output is not being read by grep. You COULD instruct grep to read from both files and stdin:
$ echo "hello" > x.txt
$ echo "world" | grep "o" x.txt -
x.txt:hello
(standard input):world
But that's not what you're doing.
By default, when grep reads from multiple files, it will precede each match with the name of the file from whence that match originated. That's also what you're seeing in my example above -- two inputs, one x.txt and the other - a.k.a. stdin, separated by a colon from the match they supplied.
While grep does include the most minuscule capability for filtering (with -o, or GNU grep's \K with optional Perl compatible RE), it does NOT provide you with any options for formatting the filename. Since you can'd do anything with the output of grep, you're limited to either parsing the output you've got, or using some other tool.
Parsing is easy, if your filenames are predictably structured as they seem to be from the two examples you've provided.
For this, we can ignore that these lines contain a file and data. For the purpose of the filter, they are a stream which follows a pattern. It looks like you want to strip off all characters from the beginning of each line up to and not including the first digit. You can do this by piping through sed:
sed 's/^[^0-9]*//'
Or you can achieve the same effect by using grep's minimal filtering to return every match starting from the first digit:
grep -o '[0-9].*'
If this kind of pipe-fitting is not to your liking, you may want to replace your entire grep with something in awk that combines functionality:
$ awk '
/[\.]$/ {next} # skip lines ending in backslash or dot
/^([A-ZÖÄÜÕŠŽ].*)?PALL/ { # lines to match
f=FILENAME
sub(/^[^0-9]*/,"",f) # strip unwanted part of filename, like sed
printf "%s:%s\n", f, $0
getline # simulate the "-A 1" from grep
printf "%s:%s\n", f, $0
}' *.txt
Note that I haven't tested this, because I don't have your data to work with.
Also, awk doesn't include any of the fancy terminal-dependent colourization that GNU grep provides through the --colour option.

Need help parsing a file via UNIX commands

I have a file that has lines that look like this
LINEID1:FIELD1=ABCD,&FIELD2-0&FIELD3-1&FIELD4-0&FIELD9-0;
LINEID2:FIELD1=ABCD,&FIELD5-1&FIELD6-0;
LINEID3:FIELD1=ABCD,&FIELD7-0&FIELD8-0;
LINEID1:FIELD1=XYZ,&FIELD2-0&FIELD3-1&FIELD9-0
LINEID3:FIELD1=XYZ,&FIELD7-0&FIELD8-0;
LINEID1:FIELD1=PQRS,&FIELD3-1&FIELD4-0&FIELD9-0;
LINEID2:FIELD1=PQRS,&FIELD5-1&FIELD6-0;
LINEID3:FIELD1=PQRS,&FIELD7-0&FIELD8-0;
I'm interested in only the lines that begin with LINEID1 and only some elements (FIELD1, FIELD2, FIELD4 and FIELD9) from that line. The output should look like this (no & signs.can replace with |)
FIELD1=ABCD|FIELD2-0|FIELD4-0|FIELD9-0;
FIELD1=XYZ|FIELD2-0|FIELD9-0;
FIELD1=PQRS|FIELD4-0|FIELD9-0;
If additional information is required, do let me know, I'll post them in edits. Thanks!!
This is not exactly what you asked for, but no-one else is answering and it is pretty close for you to get started with!
awk -F'[&:]' '/^LINEID1:/{print $2,$3,$5,$6}' OFS='|' file
Output
FIELD1=ABCD,|FIELD2-0|FIELD4-0|FIELD9-0;
FIELD1=XYZ,|FIELD2-0|FIELD9-0|
FIELD1=PQRS,|FIELD3-1|FIELD9-0;|
The -F sets the Input Field Separator to colon or ampersand. Then it looks for lines starting LINEID1: and prints the fields you need. The OFS sets the Output Field Separator to the pipe symbol |.
Pure awk:
awk -F ":" ' /LINEID1[^0-9]/{gsub(/FIELD[^1249]+[-=][A-Z0-9]+/,"",$2); gsub(/,*&+/,"|",$2); print $2} ' file
Updated to give proper formatting and to omit LINEID11, etc...
Output:
FIELD1=ABCD|FIELD2-0|FIELD4-0|FIELD9-0;
FIELD1=XYZ|FIELD2-0|FIELD9-0
FIELD1=PQRS|FIELD4-0|FIELD9-0;
Explanation:
awk -F ":" - split lines into LHS ($1) and RHS ($2) since output only requires RHS
/LINEID1[^0-9]/ - return only lines that match LINEID1 and also ignores LINEID11, LINEID100 etc...
gsub(/FIELD[^1249]+[-=][A-Z0-9]+/,"",$2) - remove all fields that aren't 1, 4 or 9 on the RHS
gsub(/,*&+/,"|",$2) - clean up the leftover delimiters on the RHS
To select rows from data with Unix command lines, use grep, awk, perl, python, or ruby (in increasing order of power & possible complexity).
To select columns from data, use cut, awk, or one of the previously mentioned scripting languages.
First, let's get only the lines with LINEID1 (assuming the input is in a file called input).
grep '^LINEID1' input
will output all the lines beginning with LINEID1.
Next, extract the columns we care about:
grep '^LINEID1' input | # extract lines with LINEID1 in them
cut -d: -f2 | # extract column 2 (after ':')
tr ',&' '\n\n' | # turn ',' and '&' into newlines
egrep 'FIELD[1249]' | # extract only fields FIELD1, FIELD2, FIELD4, FIELD9
tr '\n' '|' | # turn newlines into '|'
sed -e $'s/\\|\\(FIELD1\\)/\\\n\\1/g' -e 's/\|$//'
The last line inserts newlines in front of the FIELD1 lines, and removes any trailing '|'.
That last sed pattern is a little more challenging because sed doesn't like literal newlines in its replacement patterns. To put a literal newline, a bash escape needs to be used, which then requires escapes throughout that string.
Here's the output from the above command:
FIELD1=ABCD|FIELD2-0|FIELD4-0|FIELD9-0;
FIELD1=XYZ|FIELD2-0|FIELD9-0
FIELD1=PQRS|FIELD4-0|FIELD9-0;
This command took only a couple of minutes to cobble up.
Even so, it's bordering on the complexity threshold where I would shift to perl or ruby because of their excellent string processing.
The same script in ruby might look like:
#!/usr/bin/env ruby
#
while line = gets do
if line.chomp =~ /^LINEID1:(.*)$/
f1, others = $1.split(',')
fields = others.split('&').map {|f| f if f =~ /FIELD[1249]/}.compact
puts [f1, fields].flatten.join("|")
end
end
Run this script on the same input file and the same output as above will occur:
$ ./parse-fields.rb < input
FIELD1=ABCD|FIELD2-0|FIELD4-0|FIELD9-0;
FIELD1=XYZ|FIELD2-0|FIELD9-0
FIELD1=PQRS|FIELD4-0|FIELD9-0;

Unix: formatting data with the CLI

What I would like is a way to pipe the output of a program into another program that will format each line into a MySQL insert. Does Unix provide an easy way of doing this?
you should be looking at printf command in bash
http://wiki.bash-hackers.org/commands/builtin/printf
I found this does what I need it to do
echo "something from file" | awk '{print "something left " $0 " something right" }'
To address your comment on Gurubaran's answer and give printf some TLC:
$ printf "something left %s something right\n" $(cat test.txt)
something left Test something right
$
You could do something like:
echo "something" | while read line; do
echo "INSERT INTO table VALUES($line);"
done > file
As you might have gathered from the other answers, there are a myriad ways to do it. They've not even touched on sed, Perl, Python, and the like. The key issue is 'how many values are there in each line of data', and what is the correct way of presenting that information to MySQL. If the input lines are nicely formatted so that the data is ready for presentation, then it can be as simple as:
data-generator-script |
while read line
do
echo "INSERT INTO Table VALUES($line);"
done
or:
data-generator-script |
sed 's/.*/INSERT INTO Table(Col1, Col2, …, ColN) VALUES(&);/'
On the other hand, if the data is presented with multiple values separated by blanks, with strings needing to be enclosed in quotes, then life is messier:
data-generator-script |
awk '{printf("INSERT INTO Table(Col1, Col2, …, ColN) VALUES(%d, '%s', …, %d);\n", $1, $2, …, $N);}'
If you have to split on some character other than blank, to allow multi-word strings, you again have to deal with more complex issues. If it is something like a pipe | and it doesn't appear in the data, then you just need to add -F'|' to the awk script. If the data is in CSV format, you may be able to use the first technique, but if there are multi-line strings in the data, you need a full CSV parser tool.

How to replace string by an escape character plus string in unix

How can I convert a one line like below:
794170|VWSD|AAA|e|h|i|j|STRING1|794170|VWSD|BBB|q|w|e|r|STRING2|794170|VWSD|CCC|z|x|c|v|STRING3|...and so on
to a linefeed-delimted,
Expected Output:
794170|VWSD|AAA|e|h|i|j|STRING1|
794170|VWSD|BBB|q|w|e|r|STRING2|
794170|VWSD|CCC|z|x|c|v|STRING3|
and so on.
BTW I'n not a unix expert and just want steps or simple commands to resolve. Appreciate your help.
I assume you have your string in a file with name "x", then you can do this.
I use the character ":" to represent the carriage return that 'sed' adds to your string. Choose something else if ":" occurs in your string. Then "tr" changes ":" to carriage return. The output is as you desire except that there is an extra carriage return at the beginning.
cat x | sed 's/794170/:794170/g' | tr ':' "\n"
You can use the fold command:
$ fold -w32 file
794170|VWSD|AAA|e|h|i|j|STRING1|
794170|VWSD|BBB|q|w|e|r|STRING2|
794170|VWSD|CCC|z|x|c|v|STRING3|
I don't think you can do it with a simple command. There are several options for creating scripts that can split lines more or less arbitrarily. Any Unix will have the awk utility available. On most systems you will also find Python and Perl. My guess is that a Perl or Python script is the easiest way to split lines like the one you gave.
This would be one way to do it in Python
inline = "794170|VWSD|AAA|e|h|i|j|STRING1|794170|VWSD|BBB|q|w|e|r|STRING2|794170|VWSD|CCC|z|x|c|v|STRING3|"
splits = ['794170' + s for s in inline.split('794170')]
for s in splits[1:]:
print s
794170|VWSD|AAA|e|h|i|j|STRING1|
794170|VWSD|BBB|q|w|e|r|STRING2|
794170|VWSD|CCC|z|x|c|v|STRING3|

How do you split a file base on a token?

Let's say you got a file containing texts (from 1 to N) separated by a $
How can a slit the file so the end result is N files?
text1 with newlines $
text2 $etc... $
textN
I'm thinking something with awk or sed but is there any available unix app that already perform that kind of task?
awk 'BEGIN{RS="$"; ORS=""} { textNumber++; print $0 > "text"textNumber".out" }' fileName
Thank to Bill Karwin for the idea.
Edit : Add the ORS="" to avoid printing a newline at the end of each files.
Maybe split -p pattern?
Hmm. That may not be exactly what you want. It doesn't split a line, it only starts a new file when it sees the pattern. And it seems to be supported only on BSD-related systems.
You could use something like:
awk 'BEGIN {RS = "$"} { ... }'
edit: You might find some inspiration for the { ... } part here:
http://www.gnu.org/manual/gawk/html_node/Split-Program.html
edit: Thanks to comment from dmckee, but csplit also seems to copy the whole line on which the pattern occurs.
If I'm reading this right, the UNIX cut command can be used for this.
cut -d $ -f 1- filename
I might have the syntax slightly off, but that should tell cut that you're using $ separated fields and to return fields 1 through the end.
You may need to escape the $.
awk -vRS="$" '{ print $0 > "text"t++".out" }' ORS="" file
using split command we can split using strings.
but csplit command will allow you to slit files basing on regular expressions as well.

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