Is it possible to use grep for the space character - bash

I have a text file (example.txt) like this:
100 this is a string
50 word
10
(Note that there are trailing space characters on the last line.)
When I do the following in my shell script:
cat example.txt | sed '1!d' | awk '{for (i=2; i < NF; i++) printf $i " "; print $NF}' - returns this is a string
cat example.txt | sed '2!d' | awk '{for (i=2; i < NF; i++) printf $i " "; print $NF}' - returns word
cat example.txt | sed '3!d' | awk '{for (i=2; i < NF; i++) printf $i " "; print $NF}' - returns 10 (incorrect, should be a space character instead)
Is there any method to use grep in bash to return the result I am looking for?

Is there any method to use grep in bash to return the result I am looking for?
Well, grep can match space characters. You have to quote them to avoid the shell interpreting them as delimiters. But grep will output either the whole line or the part of it that matches, depending on the options given to it, and I don't think that will satisfy your output requirement.
It looks like your input format may employ fixed field widths, or at least a fixed-width first field, and that you're trying to remove that first field. In that case, why not use sed? For example,
cat example.txt | sed 's/^....//'
will remove the first four characters from each line. You can also spell that
sed 's/^....//' example.txt
. If you want instead to cut a variable-length head of the line consisting of decimal digits up to the first space then that would be
sed 's/^[0-9]* //' example.txt
Note that although that's what you said in comments you want, it will produce different output than your awk example in the case of your second input line -- it will output a leading space:
word
Note also that your awk-based approach will replace multiple adjacent whitespace in the retained part of your lines with single spaces. That behavior could be obtained from sed, too, but I'm inclined to think that it's not actually wanted.

Related

Remove first two lines, last two lines and space from file and add quotes on each line and replace newline with commas in shell script

I have to input.txt file which needs to be formatted by shell script with following condition
remove first two lines and
last two lines
remove all spaces in each
lines(each line have two spaces at
beginning and one space at end)
Each line should be within single
quotes(' ')
At last replace newline($) with
commas.
(original)
input.txt
sql
--------
Abce
Bca
Efr
-------
Row (3)
Desired output file
output.txt
'Abce','Bca','Efr'
I have tried using following commands
Sed -i 1,2d input.txt > input.txt
Sed "$(( $(wc -l <input.txt) -2+1)), $ d" Input.txt > input.txt
Sed ':a;N;$!ba;s/\n/, /g' input.txt > output.txt
But i get blank output.txt
Would you please try the following:
mapfile -t ary < <(tail -n +3 input.txt | head -n -2 | sed -E "s/^[[:blank:]]*/'/; s/[[:blank:]]*$/'/")
(IFS=,; echo "${ary[*]}")
tail -n +3 outputs lines after the 3rd line, inclusive.
head -n -2 outputs lines excluding the last 2 lines.
sed -E "s/^[[:blank:]]*/'/" removes leading whitespaces and prepends
a single quote.
Similarly the sed command "s/[[:blank:]]*$/'/" removes trailing
whitespaces and appends a single quote.
The syntax <(command ..) is a process substitution and the
output of the commands within the parentheses is fed to the mapfile
via the redirect.
mapfile -t ary reads lines from the standard input into the array
variable named ary.
echo "${ary[*]}" expands to a single string with the contents of
the array ary separated by the value of IFS, which is just assigned
to a comma.
The assignment of IFS and the array expansion are enclosed with
parentheses to be executed in the subshell. This prevents the IFS
to be modified in the current process.
With your shown samples, please try following awk program. Written and tested in GNU awk, should work with any version.
awk -v s1="'" -v lines="$(wc -l < Input_file)" '
BEGIN{ OFS="," }
FNR==(lines-1) {
print val
exit
}
FNR>2{
sub(/^[[:space:]]+/,"")
val=(val?val OFS:"") (s1 $0 s1)
}
' Input_file
Explanation: Adding detailed explanation for above code, this is only for explanation purposes.
awk -v s1="'" -v lines="$(wc -l < Input_file)" ' ##Starting awk program, setting s1 variable to ' and creating lines which has total number of lines in it, using wc -l command on Input_file file.
BEGIN{ OFS="," } ##Setting OFS to comma in BEGIN section of this program.
FNR==(lines-1) { ##Checking condition if its 2nd last line of Input_file.
print val ##Then printing val here.
exit ##exiting from program from here.
}
FNR>2{ ##Checking condition if FNR is greater than 2 then do following.
sub(/^[[:space:]]+/,"") ##Substituting initial spaces with NULL here.
val=(val?val OFS:"") (s1 $0 s1) ##Creating val which has ' current line ' in it and keep adding it in val.
}
' Input_file ##Mentioning Input_file name here.
If you know the input is small enough to fit in memory:
$ awk '
NR>4 { gsub(/^ *| *$/,"\047",p2); out=out sep p2; sep="," }
{ p2=p1; p1=$0 }
END { print out }
' input.txt
'Abce','Bca','Efr'
Otherwise:
$ awk '
NR>4 { gsub(/^ *| *$/,"\047",p2); printf "%s%s", sep, p2; sep="," }
{ p2=p1; p1=$0 }
END { print "" }
' input.txt
'Abce','Bca','Efr'
Either script will work using any awk in any shell on every Unix box.
This might work for you (GNU sed):
sed -E '1,2d;$!H;$!d;x;s/^\s*(.*)\s*$/'\''\1'\''/mg;s/\n[^\n]*$//;y/\n/,/' file
Delete the first two lines.
Append each line to the hold space, except for the last (this means the second from last line will still be present - see later).
Delete all lines except for the last.
Swap to the hold space.
Remove all spaces either side of the words on each line and surround those words by single quotes.
Remove the last line and its newline.
Replace all newlines by commas.
The first sed -i overwrites input.txt with an empty file. You can't write output back to the file you are reading, and sed -i does not produce any output anyway.
The minimal fix is to take out the -i and string together the commands into a pipeline; but of course, sed allows you to combine the commands into a single script.
len=$(wc -l <input.txt)
sed -e '1,2d' -e "$((len - 3))"',$d' \
-e ':a' \
-e 's/^ \(.*\) $/'"'\\1'/" \
-e N -e '$!ba' -e 's/\n/, /g' input.txt >output.txt
(Untested; if your sed does not allow multiple -e options, needs refactoring to use a single string with semicolons or newlines between the commands.)
This is hard to write and debug and brittle because of the ways you have to combine the quoting features of the shell with the requirements of sed and this particular script, but also more inherently because sed is a terse and obscure language.
A much more legible and maintainable solution is to switch to Awk, which allows you to express the logic in more human terms, and avoid having to pull in support from the shell for simple tasks like arithmetic and string formatting.
awk 'FNR > 2 { sub(/^ /, ""); sub(/ $/, "");
a[++i] = sprintf("\047%s\047,", $0); }
END { for(j=1; j < i-1; ++j) printf "%s", a[j] }' input.txt >output.txt
This literally replaces all newlines with commas; perhaps you would in fact like to print a newline instead of the comma on the last line?
awk 'FNR > 2 { sub(/^ /, ""); sub(/ $/, "");
a[++i] = sprintf("%s\047%s\047", sep, $0); sep="," }
END { for(j=1; j < i-1; ++j) printf "%s", a[j]; printf "\n" }' input.txt >output.txt
If the input file is really large, you might want to refactor this to not keep all the lines in memory. The array a collects the formatted output and we print all its elements except the last two in the END block.
sed -E '
/^-+$/,/^-+$/!d
//d
s/^[[:space:]]*|[[:space:]]*$/'\''/g
' input.txt |
paste -sd ,
This uses a trick that doesn't work on all sed implementations, to print the lines between two patterns (the dashes in this case), excluding those patterns.
On the plus side if the ---- pattern is at a different line number, it still works. Down side is it breaks, if that pattern (a line containing only dashes) occurs an odd number of times (ie. not in pairs, that wrap the lines you want).
Then sub line start and end (including white space) with single quotes.
Finally pipe to paste to sub the new lines with commas, excluding a trailing comma.
Using sed
$ sed "1,2d; /-/,$ d; s/\s\+//;s/.*/'&'/" input_file | sed -z 's/\n/,/g;s/,$/\n/'
'Abce','Bca','Efr'
I'll post a sed solution which is rather light.
sed '$d' input.txt | sed "\$d; 1,2d; s/^\s*\|\s*$/'/g" | paste -sd ',' > output.txt
$d Remove last line with first sed
\$d Remove the last line. $ escaped with backslash as we are within double-quotes.
1,2d Remove the first two lines.
s/^\s*\|\s*$/'/g Replace all leading and trailing whitespace with single quotes.
Use paste to concatenate to a single, comma delimited strings.
If we know that the relevant lines always start with two spaces, then it can even be simplified further.
sed -n "s/\s*$/'/; s/^ /'/p" input.txt | paste -sd ',' > output.txt
-n suppress printing lines unless told to
s/\s*$/'/ replace trailing whitespace with single quotes
s/^ /'/p replace two leading spaces and print lines that match
paste to concat
Then an awk solution:
awk -v i=1 -v q=\' 'FNR>2 {
gsub(/^[[:space:]]*|[[:space:]]*$/, q)
a[i++]=$0
} END {
for(i=1; i<=length(a)-3; i++)
printf "%s,", a[i]
print a[i++]
}' input.txt > output.txt
-v i=1 create an awk variable starting at one
-v q=\' create an awk variable for the single quote character
FNR>2 { ... tells it to only process line 3+
gsub(/^[[:space:]]*|[[:space:]]*$/, q) substitute leading and trailing whitespace with single quotes
a[i++]=$0 add line to array
END { ... Process the rest after reaching end of file
for(i=1; i<=length(a)-3; i++) take the length of the array but subtract three -- representing the last three lines
printf "%s,", a[i] print all but last three entries comma delimited
print a[i++] print next entry and complete the script (skipping the last two entries)
Not a one liner but works
sed "s/^ */\'/;s/\$/\',/;1,2d;N;\$!P;\$!D;\$d" | sed ' H;1h;$!d;x;s/\n//g;s/,$//'
Explanation:
s/^ */\'/;s/\$/\',/ ---> Adds single quotes and comma
N;$!P;$!D;$d ---> Deletes last two lines
H;1h;$!d;x;s/\n//g;s/,$//' ---> Loads entire file and merge all lines and remove last comma

How to print keys from all key-value pairs

Text file looks like this:
key11=val1|key12=val2|key13=val3
key21=val1|key22=val2|key23=val3
How can I extract keys so that:
key11|key12|key13
key21|key22|key23
I have tried unsuccessfully :
awk '{ gsub(/[^[|]=]+=/,"") }1' file.txt
gives back the actual data:
key11=val1|key12=val2|key13=val3
key21=val1|key22=val2|key23=val3
Since you tagged bash
while IFS='=|' read -ra words; do
n=${#words[#]}
for ((i=1; i<n; i+=2)); do
unset words[i]
done
( IFS='|'; echo "${words[*]}" )
done < file
gawk
This can be done by awk, by setting FS and OFS :
kent$ awk -F'=[^|]*' -v OFS="" '$1=$1' file
key11|key12|key13
key21|key22|key23
or safer: awk -F.... '{$1=$1}1' file
substitution (by sed for example):
kent$ sed 's/=[^|]*//g' file
key11|key12|key13
key21|key22|key23
Here's one solution
echo "key11=val1|key12=val2|key13=val3" \
| awk -F'[=|]' '{
for (i=1;i<=NF;i+=2){
printf("%s%s", $i, (i<(NF-1))?"|":"")
}
print""
}'
output
key11|key12|key13
It should also work by passing in the filename as an argument to awk, i.e.
awk -F'[=|]' '{for (i=1;i<=NF;i+=2){printf("%s%s", $i, (i<(NF-1))?"|":"") }print""}' file1 [file_more_as_will_fit]
Discussion
We use a multiple character value for FS (FieldSeperator) so each = and | char mark the beginning of a new field.
-F'[=|]'
Because we know we want to start with field1 for output and skip every other field, we use
for (i=1;i<=NF;i+=2)
printf formats the output as defined by the format string '%s%s' . There area a zillion options available for printf format strs, but you only need the value for $i (the looping value that generates the key) and whether to print a | char or not.
printf("%s%s", $i ...)
And we use awk's ternary operator, which evaluates what element number is being processed (i<..). As long as it is not the 2nd to last field, the | char is emitted.
(i<(NF-1))?"|":""
IHTH
sed
I did this with sed:
sed -r 's/([[:alnum:]]*)=[[:alnum:]]*/\1/g' < file.txt
tested here and got:
key11|key12|key13
key21|key22|key23
s/<pattern>/<subst>/ means "replace <pattern> by <subst>", and with the g in the end it will do it for every pattern found in the line.
The [[:alnum:]]* is equivalent to [0-9a-zA-Z]*, and means any number of letters or digits.
The first pattern between parentesis will correspond to \1 in the substitution, the second \2 and so on.
So, it will match every "key=value" and replace it by "key".
awk -F'[=|]' '{print $1,$3,$5}' OFS="|" file
key11|key12|key13
key21|key22|key23

Ignore comma after backslash in a line in a text file using awk or sed

I have a text file containing several lines of the following format:
name,list_of_subjects,list_of_sports,school
Eg1: john,science\,social,football,florence_school
Eg2: james,painting,tennis\,ping_pong\,chess,highmount_school
I need to parse the text file and print the output of fields ignoring the escaped commas. Here those will be fields 2 or 3 like this:
science, social
tennis, ping_pong, chess
I do not know how to ignore escaped characters. How can I do it with awk or sed in terminal?
Substitute \, with a character that your records do not contain normally (e.g. \n), and restore it before printing. For example:
$ awk -F',' 'NR>1{ if(gsub(/\\,/,"\n")) gsub(/\n/,",",$2); print $2 }' file
science,social
painting
Since first gsub is performed on the whole record (i.e $0), awk is forced to recompute fields. But the second one is performed on only second field (i.e $2), so it will not affect other fields. See: Changing Fields.
To be able to extract multiple fields with properly escaped commas you need to gsub \ns in all fields with a for loop as in the following example:
$ awk 'BEGIN{ FS=OFS="," } NR>1{ if(gsub(/\\,/,"\n")) for(i=1;i<=NF;++i) gsub(/\n/,"\\,",$i); print $2,$3 }' file
science\,social,football
painting,tennis\,ping_pong\,chess
See also: What's the most robust way to efficiently parse CSV using awk?.
You could replace the \, sequences by another character that won't appear in your text, split the text around the remaining commas then replace the chosen character by commas :
sed $'s/\\\,/\31/g' input | awk -F, '{ printf "Name: %s\nSubjects : %s\nSports: %s\nSchool: %s\n\n", $1, $2, $3, $4 }' | tr $'\31' ','
In this case using the ASCII control char "Unit Separator" \31 which I'm pretty sure your input won't contain.
You can try it here.
Why awk and sed when bash with coreutils is just enough:
# Sorry my cat. Using `cat` as input pipe
cat <<EOF |
name,list_of_subjects,list_of_sports,school
Eg1: john,science\,social,football,florence_school
Eg2: james,painting,tennis\,ping_pong\,chess,highmount_school
EOF
# remove first line!
tail -n+2 |
# substitute `\,` by an unreadable character:
sed 's/\\\,/\xff/g' |
# read the comma separated list
while IFS=, read -r name list_of_subjects list_of_sports school; do
# read the \xff separated list into an array
IFS=$'\xff' read -r -d '' -a list_of_subjects < <(printf "%s" "$list_of_subjects")
# read the \xff separated list into an array
IFS=$'\xff' read -r -d '' -a list_of_sports < <(printf "%s" "$list_of_sports")
echo "list_of_subjects : ${list_of_subjects[#]}"
echo "list_of_sports : ${list_of_sports[#]}"
done
will output:
list_of_subjects : science social
list_of_sports : football
list_of_subjects : painting
list_of_sports : tennis ping_pong chess
Note that this will be most probably slower then solution using awk.
Note that the principle of operation is the same as in other answers - substitute \, string by some other unique character and then use that character to iterate over the second and third field elemetns.
This might work for you (GNU sed):
sed -E 's/\\,/\n/g;y/,\n/\n,/;s/^[^,]*$//Mg;s/\n//g;/^$/d' file
Replace quoted commas by newlines and then revert newlines to commas and commas to newlines. Remove all lines that do not contain a comma. Delete empty lines.
Using Perl. Change the \, to some control char say \x01 and then replace it again with ,
$ cat laxman.txt
john,science\,social,football,florence_school
james,painting,tennis\,ping_pong\,chess,highmount_school
$ perl -ne ' s/\\,/\x01/g and print ' laxman.txt | perl -F, -lane ' for(#F) { if( /\x01/ ) { s/\x01/,/g ; print } } '
science,social
tennis,ping_pong,chess
You can perhaps join columns with a function.
function joincol(col, i) {
$col=$col FS $(col+1)
for (i=col+1; i<NF; i++) {
$i=$(i+1)
}
NF--
}
This might get used thusly:
{
for (col=1; col<=NF; col++) {
if ($col ~ /\\$/) {
joincol(col)
}
}
}
Note that decrementing NF is undefined behaviour in POSIX. It may delete the last field, or it may not, and still be POSIX compliant. This works for me in BSDawk and Gawk. YMMV. May contain nuts.
Use gawk's FPAT:
awk -v FPAT='(\\\\.|[^,\\\\]*)+' '{print $3}' file
#list_of_sports
#football
#tennis\,ping_pong\,chess
then use gnusub to replace the backslashes:
awk -v FPAT='(\\\\.|[^,\\\\]*)+' '{print gensub("\\\\", "", "g", $3)}' file
#list_of_sports
#football
#tennis,ping_pong,chess

How can I find unique characters per line of input?

Is there any way to extract the unique characters of each line?
I know I can find the unique lines of a file using
sort -u file
I would like to determine the unique characters of each line (something like sort -u for each line).
To clarify: given this input:
111223234213
111111111111
123123123213
121212122212
I would like to get this output:
1234
1
123
12
Using sed
sed ':;s/\(.\)\(.*\)\1/\1\2/;t' file
Basically what it does is capture a character and check if it appears anywhere else on the line. It also captures all the characters between these.
Then it replaces all of that including the second occurence with just first occurence and then what was inbetween.
t is test and jumps to the : label if the previous command was successful. Then this repeats until the s/// command fails meaning only unique characters remain.
; just separates commands.
1234
1
123
12
Keeps order as well.
It doesn't get things in the original order, but this awk one-liner seems to work:
awk '{for(i=1;i<=length($0);i++){a[substr($0,i,1)]=1} for(i in a){printf("%s",i)} print "";delete a}' input.txt
Split apart for easier reading, it could be stand-alone like this:
#!/usr/bin/awk -f
{
# Step through the line, assigning each character as a key.
# Repeated keys overwrite each other.
for(i=1;i<=length($0);i++) {
a[substr($0,i,1)]=1;
}
# Print items in the array.
for(i in a) {
printf("%s",i);
}
# Print a newline after we've gone through our items.
print "";
# Get ready for the next line.
delete a;
}
Of course, the same concept can be implemented pretty easily in pure bash as well:
#!/usr/bin/env bash
while read s; do
declare -A a
while [ -n "$s" ]; do
a[${s:0:1}]=1
s=${s:1}
done
printf "%s" "${!a[#]}"
echo ""
unset a
done < input.txt
Note that this depends on bash 4, due to the associative array. And this one does get things in the original order, because bash does a better job of keeping array keys in order than awk.
And I think you've got a solution using sed from Jose, though it has a bunch of extra pipe-fitting involved. :)
The last tool you mentioned was grep. I'm pretty sure you can't do this in traditional grep, but perhaps some brave soul might be able to construct a perl-regexp variant (i.e. grep -P) using -o and lookarounds. They'd need more coffee than is in me right now though.
One way using perl:
perl -F -lane 'print do { my %seen; grep { !$seen{$_}++ } #F }' file
Results:
1234
1
123
12
Another solution,
while read line; do
grep -o . <<< $line | sort -u | paste -s -d '\0' -;
done < file
grep -o . convert 'row line' to 'column line'
sort -u sort letters and remove repetead letters
paste -s -d '\0' - convert 'column line' to 'row line'
- as a filename argument to paste to tell it to use standard input.
This awk should work:
awk -F '' '{delete a; for(i=1; i<=NF; i++) a[$i]; for (j in a) printf "%s", j; print ""}' file
1234
1
123
12
Here:
-F '' will break the record char by char giving us single character in $1, $2 etc.
Note: For non-gnu awk use:
awk 'BEGIN{FS=""} {delete a; for(i=1; i<=NF; i++) a[$i];
for (j in a) printf "%s", j; print ""}' file
This might work for you (GNU sed):
sed 's/\B/\n/g;s/.*/echo "&"|sort -u/e;s/\n//g' file
Split each line into a series of lines. Unique sort those lines. Combine the result back into a single line.
Unique and sorted alternative to the others, using sed and gnu tools:
sed 's/\(.\)/\1\n/g' file | sort | uniq
which produces one character per line; If you want those on one line, just do:
sed 's/\(.\)/\1\n/g' file | sort | uniq | sed ':a;N;$!ba;s/\n//g;'
This has the advantage of showing the characters in sorted order, rather than order of appearance.

Split a big txt file to do grep - unix

I work (unix, shell scripts) with txt files that are millions field separate by pipe and not separated by \n or \r.
something like this:
field1a|field2a|field3a|field4a|field5a|field6a|[...]|field1d|field2d|field3d|field4d|field5d|field6d|[...]|field1m|field2m|field3m|field4m|field5m|field6m|[...]|field1z|field2z|field3z|field4z|field5z|field6z|
All text is in the same line.
The number of fields is fixed for every file.
(in this example I have field1=name; field2=surname; field3=mobile phone; field4=email; field5=office phone; field6=skype)
When I need to find a field (ex field2), command like grep doesn't work (in the same line).
I think that a good solution can be do a script that split every 6 field with a "\n" and after do a grep. I'm right? Thank you very much!
With awk :
$ cat a
field1a|field2a|field3a|field4a|field5a|field6a|field1d|field2d|field3d|field4d|field5d|field6d|field1m|field2m|field3m|field4m|field5m|field6m|field1z|field2z|field3z|field4z|field5z|field6z|
$ awk -F"|" '{for (i=1;i<NF;i=i+6) {for (j=0; j<6; j++) printf $(i+j)"|"; printf "\n"}}' a
field1a|field2a|field3a|field4a|field5a|field6a|
field1d|field2d|field3d|field4d|field5d|field6d|
field1m|field2m|field3m|field4m|field5m|field6m|
field1z|field2z|field3z|field4z|field5z|field6z|
Here you can easily set the length of line.
Hope this helps !
you can use sed to split the line in multiple lines:
sed 's/\(\([^|]*|\)\{6\}\)/\1\n/g' input.txt > output.txt
explanation:
we have to use heavy backslash-escaping of (){} which makes the code slightly unreadable.
but in short:
the term (([^|]*|){6}) (backslashes removed for readability) between s/ and /\1, will match:
[^|]* any character but '|', repeated multiple times
| followed by a '|'
the above is obviously one column and it is grouped together with enclosing parantheses ( and )
the entire group is repeated 6 times {6}
and this is again grouped together with enclosing parantheses ( and ), to form one full set
the rest of the term is easy to read:
replace the above (the entire dataset of 6 fields) with \1\n, the part between / and /g
\1 refers to the "first" group in the sed-expression (the "first" group that is started, so it's the entire dataset of 6 fields)
\n is the newline character
so replace the entire dataset of 6 fields by itself followed by a newline
and do so repeatedly (the trailing g)
you can use sed to convert every 6th | to a newline.
In my version of tcsh I can do:
sed 's/\(\([^|]\+|\)\{6\}\)/\1\n/g' filename
consider this:
> cat bla
a1|b2|c3|d4|
> sed 's/\(\([^|]\+|\)\{6\}\)/\1\n/g' bla
a1|b2|
c3|d4|
This is how the regex works:
[^|] is any non-| character.
[^|]\+ is a sequence of at least one non-| characters.
[^|]\+| is a sequence of at least one non-| characters followed by a |.
\([^|]\+|\) is a sequence of at least one non-| characters followed by a |, grouped together
\([^|]\+|\)\{6\} is 6 consecutive such groups.
\(\([^|]\+|\)\{6\}\) is 6 consecutive such groups, grouped together.
The replacement just takes this sequence of 6 groups and adds a newline to the end.
Here is how I would do it with awk
awk -v RS="|" '{printf $0 (NR%7?RS:"\n")}' file
field1a|field2a|field3a|field4a|field5a|field6a|[...]
field1d|field2d|field3d|field4d|field5d|field6d|[...]
field1m|field2m|field3m|field4m|field5m|field6m|[...]
field1z|field2z|field3z|field4z|field5z|field6z|
Just adjust the NR%7 to number of field you to what suites you.
What about printing the lines on blocks of six?
$ awk 'BEGIN{FS=OFS="|"} {for (i=1; i<=NF; i+=6) {print $(i), $(i+1), $(i+2), $(i+3), $(i+4), $(i+5)}}' file
field1a|field2a|field3a|field4a|field5a|field6a
field1d|field2d|field3d|field4d|field5d|field6d
field1m|field2m|field3m|field4m|field5m|field6m
field1z|field2z|field3z|field4z|field5z|field6z
Explanation
BEGIN{FS=OFS="|"} set input and output field separator as |.
{for (i=1; i<=NF; i+=6) {print $(i), $(i+1), $(i+2), $(i+3), $(i+4), $(i+5)}} loop through items on blocks of 6. Every single time, print six of them. As print end up writing a new line, then you are done.
If you want to treat the files as being in multiple lines, then make \n the field separator. For example, to get the 2nd column, just do:
tr \| \\n < input-file | sed -n 2p
To see which columns match a regex, do:
tr \| \\n < input-file | grep -n regex

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