remove comma in jsonpath template using bash - json

I have a JSON path template query.
oc get event jsonpath='{range .items[*]},{#.name}{","}{#.message}{","}{#.evenname}'{"\n"}{end}'> /tmp/test.csv
i'm redirecting it to csv.
name1,message of event one,eventname1
name2,message of,event,two,eventname2
name3,message of event three,eventname3
name4,message of, event four,eventname4
getting comma in a message from above output , i want to replace the comma with space for the second column(message) in the bash script.
Anyone has any thoughts on how can achieve this.
Expected result
name1,message of event one,eventname1
name2,message of event two,eventname2
name3,message of event three,eventname3
name4,message of event four,eventname4

Assuming you can change the field delimiter to a character known to not exist in the data (eg, |), you would now be generating:
name1|message of event one|eventname1
name2|message of,event,two|eventname2
name3|message of event three|eventname3
name4|message of, event four|eventname4
From here we can use sed to a) remove/replace , with <space> and then b) replace | with ,:
$ sed 's/[ ]*,[ ]*/,/g;s/,/ /g;s/|/,/g'
NOTE: the s/[ ]*,[ ]*/g is needed to address the additional requirement of stripping out repeating spaces (as would occur in line #4 if we replace , with <space>)
When applied to the data this generates:
name1,message of event one,eventname1
name2,message of event two,eventname2
name3,message of event three,eventname3
name4,message of event four,eventname4
Another option using awk (for OP's current data using the , as the field delimiter):
awk -F',' ' # input field delimiter = ","
{ x=$1"," # start new string as field #1 + ","
sep="" # initial separator = "" for fields 2 to (NF-1)
for (i=2;i<NF;i++) { # loop through fields 2 to (NF-1)
gsub(/^[ ]+|[ ]+$/,"",$i) # trim leading/trailing spaces
x=x sep $i # append current field to x along with sep
sep=" " # use " " as separator for rest of fields
}
printf "%s,%s\n", x, $NF # print "x" plus "," plus the last field (NF)
}'
When applied to the data this generates:
name1,message of event one,eventname1
name2,message of event two,eventname2
name3,message of event three,eventname3
name4,message of event four,eventname4

Related

CSV Column Insertion via awk

I am trying to insert a column in front of the first column in a comma separated value file (CSV). At first blush, awk seems to be the way to go but, I'm struggling with how to move down the new column.
CSV File
A,B,C,D,E,F
1,2,3,4,5,6
2,3,4,5,6,7
3,4,5,6,7,8
4,5,6,7,8,9
Attempted Code
awk 'BEGIN{FS=OFS=","}{$1=$1 OFS (FNR<1 ? $1 "0\nA\n2\nC" : "col")}1'
Result
A,col,B,C,D,E,F
1,col,2,3,4,5,6
2,col,3,4,5,6,7
3,col,4,5,6,7,8
4,col,5,6,7,8,9
Expected Result
col,A,B,C,D,E,F
0,1,2,3,4,5,6
A,2,3,4,5,6,7
2,3,4,5,6,7,8
C,4,5,6,7,8,9
This can be easily done using paste + printf:
paste -d, <(printf "col\n0\nA\n2\nC\n") file
col,A,B,C,D,E,F
0,1,2,3,4,5,6
A,2,3,4,5,6,7
2,3,4,5,6,7,8
C,4,5,6,7,8,9
<(...) is process substitution available in bash. For other shells use a pipeline like this:
printf "col\n0\nA\n2\nC\n" | paste -d, - file
With awk only you could try following solution, written and tested with shown samples.
awk -v value="$(echo -e "col\n0\nA\n2\nC")" '
BEGIN{
FS=OFS=","
num=split(value,arr,ORS)
for(i=1;i<=num;i++){
newVal[i]=arr[i]
}
}
{
$1=arr[FNR] OFS $1
}
1
' Input_file
Explanation:
First of all creating awk variable named value whose value is echo(shell command)'s output. NOTE: using -e option with echo will make sure that \n aren't getting treated as literal characters.
Then in BEGIN section of awk program, setting FS and OFS as , here for all line of Input_file.
Using split function on value variable into array named arr with delimiter of ORS(new line).
Then traversing through for loop till value of num(total values posted by echo command).
Then creating array named newVal with index of i(1,2,3 and so on) and its value is array arr value.
In main awk program, setting first field's value to array arr value and $1 and printing the line then.

transform multiline text into csv with awk sed and grep

I run a shell command that returns a list of repeated values like this (note the indentation):
Name: vm346
cpu 1 (12%) 6150m (76%)
memory 1130Mi (7%) 1130Mi (7%)
Name: vm847
cpu 6 (75%) 30150m (376%)
memory 12980Mi (87%) 12980Mi (87%)
Name: vm848
cpu 3500m (43%) 17150m (214%)
memory 6216Mi (41%) 6216Mi (41%)
I am trying to transform that data like this (in csv):
vm346,1,(12%),6150m,(76%),1130Mi,(7%),1130Mi,(7%)
vm847,6,(75%),30150m,(376%),12980Mi,(87%),12980Mi,(87%)
vm848,3500m,(43%),17150m,(214%),6216Mi,(41%),6216Mi,(41%)
The problem is that any given dataset like the one above is always on more than one line.
when I pipe that into it awk it drives me mad because even if I use:
BEGIN{ FS="\n" }
to try and stitch the data together in one line, it doesn't work. No matter what I do, awk keeps the name value as a separated line above everything else.
I am sorry I haven't much code to share but I have been spinning my wheels with this for a few hours now and I am running out of ideas...
I can solve this in Perl:
perl -ane 'print join ",", #F[1 .. $#F]; print $F[0] eq "memory" ? "\n" : ","'
It should be easy to translate it to awk if you need it.
How does it work?
-a splits each line on whitespace into the #F array
-n reads the input line by line and runs the code specified after -e for each line
We print all the elements but the first one separated by commas (see join)
We then look at the first column, if it's memory, we are at the last line of the block, so we print a newline, otherwise we print a comma
With AWK, one option is to set RS to "Name: ", and ignore the first record with NR > 1, e.g.
awk -v RS="Name: " 'BEGIN{OFS=","} NR > 1 {print $1, $3, $4, $5, $6, $8, $9, $10, $11}' file
#> vm346,1,(12%),6150m,(76%),1130Mi,(7%),1130Mi,(7%)
#> vm847,6,(75%),30150m,(376%),12980Mi,(87%),12980Mi,(87%)
#> vm848,3500m,(43%),17150m,(214%),6216Mi,(41%),6216Mi,(41%)
awk '{$1=""}1' | paste -sd' \n' - | awk '{$1=$1}1' OFS=,
Get rid of the first column. Join every three rows. Same idea with sed:
sed 's/^ *[^ ]* *//' | paste -sd' \n' - | sed 's/ */,/g'
Something else:
awk '
$1=="Name:" {
sep=ors
ors=ORS
} {
for (i=2;i<=NF;++i) {
printf "%s%s",sep,$i
sep=OFS
}
} END {printf "%s",ors}'
Or if you want to print an ORS based on the first field being "memory" (note that this program may end without printing a terminating ORS):
awk '{for (i=2;i<=NF;++i) printf "%s%s",$i,(i==NF && $1=="memory" ? ORS : OFS)}'
something else else:
awk -v OFS=, '
index($0,$1)==1 {
OFS=ors
ors=ORS
} {
$1=""
printf "%s",$0
OFS=ofs
} END {printf "%s",ors} BEGIN {ofs=OFS}'
This might work for you (GNU sed):
sed -nE '/^ +\S+ +/{s///;H;$!d};x;/./s/\s+/,/gp;x;s/^\S+ +//;h' file
In overview the sed program processes indented lines, already gathered lines (except in the case that the current line is the first line of the file) and non-indented lines.
Turn off implicit printing and enable extended regexp's. (-nE).
If the current line is indented, remove the indent, the first field and any following spaces, append the result to the hold space and if it is not the last line, delete it.
Otherwise, check the hold space for gathered lines and if found, replace one or more whitespaces by commas and print the result. Then prep the current line by removing the first field and any following spaces and replace the hold space with the result.
The solution seems logically back-to-front, but programming in this style avoids having to check for end-of-file multiple times and invoking labels and gotos.
N.B. This solution will work for any number of indented lines.
Here is a ruby to do that:
ruby -e '
s=$<.read
s.scan(/^([^ \t]+:)([\s\S]+?)(?=^\1|\z)/m). # parse blocks
map(&:last). # get data part
# parse and join the data fields:
map{|block| block.split(/\n[ \t]+[^ \t]+[ \t]+/)}.
map{|lines| lines.map(&:strip).join(" ").split().join(",")}.
each{|l| puts "#{l}"}
' file
vm346,1,(12%),6150m,(76%),1130Mi,(7%),1130Mi,(7%)
vm847,6,(75%),30150m,(376%),12980Mi,(87%),12980Mi,(87%)
vm848,3500m,(43%),17150m,(214%),6216Mi,(41%),6216Mi,(41%)
The advantage is that this is not dependent on the number of lines or the number of fields. It is parsing data that is in blocks of the form:
START: ([ \t]+[data_with_no_space])*\n
l1 ([ \t]+[data_with_no_space])*\n
...
START:
...
Works this way:
Parse the blocks with THIS REGEX;
Save an array of the data elements;
Join the sub arrays and then split into data fields;
Join(',') to make a csv.

Remove double quotes if delimiter value is not present in data

An input file is given, each line of which contains quotes for each column and carriage return/ new line character.
If the line contains new lines it has be appended with in the same
line which is inside the quotes i.e for example line 1
Removing of double quotes for each column if the delimiter(,) is
not present.
Removing of Carriage Return characters i.e(^M)
To exemplify, given the following input file
"name","address","age"^M
"ram","abcd,^M
def","10"^M
"abhi","xyz","25"^M
"ad","ram,John","35"^M
I would like to obtain the following output by means of a sed/perl/awk script/oneliner.
name,address,age
ram,"abcd,def",10
abhi,xyz,25
ad,"ram,John",35
Solutions which i have tired it so far
For appending with previous line
sed '/^[^"]*"[^"]*$/{N;s/\n//}' sample.txt
for replacing control-m characters
perl -pne 's/\\r//g' sample.txt
But i didn't achieve final output what i required below
Use a library to parse CSV files. Apart from always wanting to use a library for that here you also have very specific reasons, with embedded newlines and delimiters.
In Perl a good library is Text::CSV (which wraps Text::CSV_XS if installed). A basic example
use warnings;
use strict;
use feature 'say';
use Text::CSV;
my $file = shift or die "Usage: $0 file.csv\n";
my $csv = Text::CSV->new({ binary => 1, auto_diag => 1 });
open my $fh, '<', $file or die "Can't open $file: $!";
while (my $row = $csv->getline($fh)) {
s/\n+//g for #$row;
$csv->say(\*STDOUT, $row);
}
Comments
The binary option in the constructor is what handles newlines embedded in data
Once a line is read into the array reference $row I remove newlines in each field with a simplistic regex. By all means please improve this as/if needed
The pruning of $row works as follows. In a foreach loop each element is really aliased by the loop variable, so if that gets changed the array changes. I used default where elements are aliased by $_, which the regex changes so $row changes.
I like this compact shortcut because it has such a distinct look that I can tell from across the room that an array is being changed in place; so I consider it a sort-of-an-idiom. But if it is in fact confusing please by all means write out a full and proper loop
The processed output is printed to STDOUT. Or, open an output file and pass that filehandle to say (or to print in older module versions) so the output goes directly to that file
The above prints, for the sample input provided in the question
name,address,age
ram,"abcd,def",10
abhi,xyz,25
ad,"ram,John",35
This might work for you (GNU sed):
sed ':a;/[^"]$/{N;s/\n//;ba};s/"\([^",]*\)"/\1/g' file
The solution is in two parts:
Join broken lines to make whole ones.
Remove double quotes surrounding fields that do not contain commas.
If the current line does not end with double quotes, append the next line, remove the newline and repeat. Otherwise: remove double quotes surrounding fields that do not contain double quotes or commas.
N.B. Supposes that fields do not contain quoted double quotes. If that is the case, the condition for the first step would need to be amended and double quotes within fields would need to catered for.
FPAT is the way to go using gnu awk, it handles comma separated files.
remove ^m
clean lines
remove qutes
.
dos2unix sample.txt
awk '{printf "%s"(/,$/?"":"\n"),$0}' sample.txt > tmp && mv tmp sample.txt
"name","address","age"
"ram","abcd,def","10"
"abhi","xyz","25"
"ad","ram,John","35"
awk -v FPAT="([^,]+)|(\"[^\"]+\")" -v OFS=, '{for (i=1;i<=NF;i++) if($i!~",") $i=substr($i,2,length($i)-2)}1' sample.txt
name,address,age
ram,"abcd,def",10
abhi,xyz,25
ad,"ram,John",35
All in one go:
dos2unix sample.txt && awk '{printf "%s"(/,$/?"":"\n"),$0}' sample.txt | awk -v FPAT="([^,]+)|(\"[^\"]+\")" -v OFS=, '{for (i=1;i<=NF;i++) if($i!~",") $i=substr($i,2,length($i)-2)}1'
Normally you set Filed Separator FS or F to tell how filed are separated. FPAT="([^,]+)|(\"[^\"]+\")" FPAT tells how the filed looks like using a regex. This regex is complicated and often used with CSV.
(i=1;i<=NF;i++) loop through on by one field on the line.
if($i!~",") if it does not contain comma, then
$i=substr($i,2,length($i)-2) remove first and last character, the "
If a field for some reason do not contain ", this is more robust:
awk -v FPAT="([^,]+)|(\"[^\"]+\")" -v OFS=, '{for (i=1;i<=NF;i++) if($i!~",") {n=split($i,a,"\"");$i=(n>1?a[2]:$i)}}1' file
It will not do any thing to a field not contains double quote.
With perl, please try the following:
perl -e '
while (<>) {
s/\r$//; # remove trailing CR code
$str .= $_;
}
while ($str =~ /("(("")|[^"])*"\n?)|((^|(?<=,))[^,]*((?=,)|\n))/g) {
$_ = $&;
if (/,/) { # the element contains ","
s/\n//g; # then remove newline(s) if any
} else { # otherwise remove surrounding double quotes
s/^"//s; s/"$//s;
}
push(#ary, $_);
if (/\n$/) { # newline terminates the element
print join(",", #ary);
#ary = ();
}
}' sample.txt
Output:
name,address,age
ram,"abcd,def",10
abhi,xyz,25
ad,"ram,John",35

How do I remove commas only from inside double quotes for every line in a comma delimited csv?

I have a comma delimited CSV file that encapsulates the fields in double quote that I am attempting to operate on in bash. I would like to remove commas from inside the double quoted field for each line. I've looked at other solutions for the question asked here, and they revolved around using external libraries for CSV parsing, which isn't an option for my limited environment where the majority of the work is being done in awk and sed.
"A","B","C D","E, F","G"
desired output
"A","B","C D","E F","G"
With sed, to remove all commas followed by one non quote character and commas not preceded by one non quote character:
sed 's/,*\([^"]\)/\1/g;s/\([^"]\),*/\1/g' file
Edit:
Added * quantifier to match subsequent commas.
Easy with Perl's Text::CSV_XS module:
perl -MText::CSV_XS=csv -we 'csv(
in => shift,
always_quote => 1,
on_in => sub { tr/,//d for #{ $_[1] } }
);' -- file.csv
in specifies the input, shift just takes one from the command line arguments
always_quote adds quotes even to fields that don't need them
on_in introduces code to run on each line, in this case, it iterates over all the cells in the row and removes commas using the transliteration operator tr.
With GNU awk and FPAT:
$ awk '
BEGIN {
FPAT = "([^,]+)|(\"[^\"]+\")" # field definition
OFS="," # output field separator
}
{
for(i=1;i<=NF;i++) # loop all fields
gsub(/,/,"",$i)} # replace all commas in fields
1' file # output
"A","B","C D","E F","G"
I like ruby for CSV one-liners:
ruby -rcsv -ne '
CSV.parse($_) {|row|
puts row.map {|field| field.delete(",")}
.to_csv(:force_quotes => true)
}
'

Insert newline character at index in .bash

I'm taking an introductory course to bash at my university and am working on a little MotD script that uses a json-object grabbed from an API using curl.
I want to make absolutely certain that you understand that this is NOT an assignment, but something I'm playing around with to learn more about how to script with bash.
I've found myself stuck with what could possibly be a very simply issue; I want to insert a new line ('\n') on a specific index if the 'quote' value of my json-object is too long (in this case on index 80).
I've been following a bunch of SO threads and this is my current solution:
#!/bin/bash
json_object=$(curl -s 'http://quotes.stormconsultancy.co.uk/random.json')
quote=$(echo ${json_object} | jq .quote | sed -e 's/^"//' -e 's/"$//')
author=$(echo ${json_object} | jq .author)
count=${#quote}
echo $quote
echo $author
echo "wc: $count"
if((count > 80));
then
quote=${quote:0:80}\n${quote:80:(count - 80)}
else
echo "lower"
fi
printf "$quote"
The current output I receive from the printf is the first word of the quote, whereas if I have an echo before trying to do the string-manipulation I get the entire quote.
I'm sorry if it's not following best practice or anything, but I'm an absolute beginner using both vi and bash.
I'd be very happy with any sort of advice. :)
EDIT:
Sample output:
$ ./json.bash
You should name a variable using the same care with which you name a first-born child.
"James O. Coplien"
86
higher
You should name a variable using the same care with which you name a first-born nchild.
You can just use a single line bash command to achieve this,
string="You should name a variable using the same care with which you name a first-born child."
(( "${#string}" > 80 )) && printf "%s\n" "${string:0:80}"$'\n'"${string:80}" || printf "%s\n" "$string"
You should name a variable using the same care with which you name a first-born
child.
(and) for an input line less than 80 charaacters
string="You should name a variable using the same care"
(( "${#string}" > 80 )) && printf "%s\n" "${string:0:80}"$'\n'"${string:80}" || printf "%s\n" "$string"
You should name a variable using the same care
An explanation,
(( "${#string}" > 80 )) && printf "%s\n" "${string:0:80}"$'\n'"${string:80}" || printf "%s\n" "$string"
# The syntax is a indirect implementation of ternary operator as bash doesn't
# directly support it.
#
# (( "${#string}" > 80 )) will return a success/fail depending upon the length
# of the string variable and if it is greater than 80, the command after && is
# executed and if it fails the command after || is executed
#
# "${string:0:80}"$'\n'"${string:80}"
# A parameter expansion syntax for sub-string extraction.
#
# ${PARAMETER:OFFSET}
#
# ${PARAMETER:OFFSET:LENGTH}
#
# This one can expand only a part of a parameter's value, given a position
# to start and maybe a length. If LENGTH is omitted, the parameter will be
# expanded up to the end of the string. If LENGTH is negative, it's taken as
# a second offset into the string, counting from the end of the string.
#
# So in our example we basically extract the characters from position 0 to 80
# insert a new-line and append the rest of the string
#
# The $'\n' syntax allows to include all escape sequence characters be
# included, in this case just the new line character.
Not really in the original question, but adding some extra code to #Inian great answer to allow not to break in the middle of a word, but rather at the last white space in ${string:0:80}:
#!/usr/bin/env bash
string="You should really name a variable using the same care with which you name a first-born child."
if (( "${#string}" > 80 )); then
maxstring="${string:0:80}"
lastspace="${maxstring##*\ }"
breakat="$((${#maxstring} - ${#lastspace}))"
printf "%s\n" $"${string:0:${breakat}}"$'\n'"${string:${breakat}}"
else
printf "%s\n" "$string"
fi
maxstring=${string:0:80}:
Let's get the first 80 characters of the quote.
lastspace=${maxstring##*\ }:
Deletes longest match of *\ (white space is escaped) from front of $maxstring, ${lastspace} will be the remaining string from last white space until end of the string.
breakat="$((${#maxstring} - ${#lastspace}))":
Subtract the length of ${lastspace} with the length of ${maxstring} to get the last index of the white space from ${maxstring}. This is the index where \n will be inserted.
Example output with "hard" break at character 80:
You should really name a variable using the same care with which you name a firs
t-born child.
Example output with a "soft" break at the closest white space from character 80:
You should really name a variable using the same care with which you name a
first-born child.