to insert line breaks in a file whenever a comma is encountered-Shell script - shell

I need to write a shell script to re-format a file by inserting line breaks. The condition is that a line break should be inserted when we encounter comma in the file.
For example, if the file delimiter.txt contains:
this, is a file, that should, be added, with a, line break, when we find, a comma.
The output should be:
this
is a file
that should
be added
with a
line break
when we find a
a comma.
Can this be done grep or awk?

Using GNU sed:
sed 's/, /\n/g' your.file
Output:
this
is a file
that should
be added
with a
line break
when we find a
a comma.
Note: the syntax above will work only on system that have the \n as line delimiter as Linux and the most UNIXes.
If you need a portal solution in a a script then use the following expression that uses a literal new line instead of \n:
sed 's/,[[:space:]]/\
/g' your.file
Thanks #EdMorten for this advice.

This is what tr is for
$ tr ',' '\n' <<< 'this, is a file, that should, be added, with a, line break, when we find, a comma.'
this
is a file
that should
be added
with a
line break
when we find
a comma.
Or if you must use awk:
awk '{gsub(", ", "\n", $0)}1' delimiter.txt

Solution using awk:
awk 1 RS=", " file
this
is a file
that should
be added
with a
line break
when we find
a comma.

Here's the solution using perl:
perl -pe 's#,#\n#g'
Here's a sample of it working properly on OpenBSD or OS X:
% echo 'a,b,c,d,e' | perl -pe 's#,#\n#g'
a
b
c
d
e
%
E.g., unlike the sed solutions earlier, this perl works everywhere, because the same search/replace snippet wouldn't work with the BSD sed on OpenBSD or OS X:
% echo 'a,b,c,d,e' | sed -E 's#,#\n#g'
anbncndne
%

Related

Read in a file AS a single line [duplicate]

How can I replace a newline ("\n") with a space ("") using the sed command?
I unsuccessfully tried:
sed 's#\n# #g' file
sed 's#^$# #g' file
How do I fix it?
sed is intended to be used on line-based input. Although it can do what you need.
A better option here is to use the tr command as follows:
tr '\n' ' ' < input_filename
or remove the newline characters entirely:
tr -d '\n' < input.txt > output.txt
or if you have the GNU version (with its long options)
tr --delete '\n' < input.txt > output.txt
Use this solution with GNU sed:
sed ':a;N;$!ba;s/\n/ /g' file
This will read the whole file in a loop (':a;N;$!ba), then replaces the newline(s) with a space (s/\n/ /g). Additional substitutions can be simply appended if needed.
Explanation:
sed starts by reading the first line excluding the newline into the pattern space.
Create a label via :a.
Append a newline and next line to the pattern space via N.
If we are before the last line, branch to the created label $!ba ($! means not to do it on the last line. This is necessary to avoid executing N again, which would terminate the script if there is no more input!).
Finally the substitution replaces every newline with a space on the pattern space (which is the whole file).
Here is cross-platform compatible syntax which works with BSD and OS X's sed (as per #Benjie comment):
sed -e ':a' -e 'N' -e '$!ba' -e 's/\n/ /g' file
As you can see, using sed for this otherwise simple problem is problematic. For a simpler and adequate solution see this answer.
Fast answer
sed ':a;N;$!ba;s/\n/ /g' file
:a create a label 'a'
N append the next line to the pattern space
$! if not the last line, ba branch (go to) label 'a'
s substitute, /\n/ regex for new line, / / by a space, /g global match (as many times as it can)
sed will loop through step 1 to 3 until it reach the last line, getting all lines fit in the pattern space where sed will substitute all \n characters
Alternatives
All alternatives, unlike sed will not need to reach the last line to begin the process
with bash, slow
while read line; do printf "%s" "$line "; done < file
with perl, sed-like speed
perl -p -e 's/\n/ /' file
with tr, faster than sed, can replace by one character only
tr '\n' ' ' < file
with paste, tr-like speed, can replace by one character only
paste -s -d ' ' file
with awk, tr-like speed
awk 1 ORS=' ' file
Other alternative like "echo $(< file)" is slow, works only on small files and needs to process the whole file to begin the process.
Long answer from the sed FAQ 5.10
5.10. Why can't I match or delete a newline using the \n escape
sequence? Why can't I match 2 or more lines using \n?
The \n will never match the newline at the end-of-line because the
newline is always stripped off before the line is placed into the
pattern space. To get 2 or more lines into the pattern space, use
the 'N' command or something similar (such as 'H;...;g;').
Sed works like this: sed reads one line at a time, chops off the
terminating newline, puts what is left into the pattern space where
the sed script can address or change it, and when the pattern space
is printed, appends a newline to stdout (or to a file). If the
pattern space is entirely or partially deleted with 'd' or 'D', the
newline is not added in such cases. Thus, scripts like
sed 's/\n//' file # to delete newlines from each line
sed 's/\n/foo\n/' file # to add a word to the end of each line
will NEVER work, because the trailing newline is removed before
the line is put into the pattern space. To perform the above tasks,
use one of these scripts instead:
tr -d '\n' < file # use tr to delete newlines
sed ':a;N;$!ba;s/\n//g' file # GNU sed to delete newlines
sed 's/$/ foo/' file # add "foo" to end of each line
Since versions of sed other than GNU sed have limits to the size of
the pattern buffer, the Unix 'tr' utility is to be preferred here.
If the last line of the file contains a newline, GNU sed will add
that newline to the output but delete all others, whereas tr will
delete all newlines.
To match a block of two or more lines, there are 3 basic choices:
(1) use the 'N' command to add the Next line to the pattern space;
(2) use the 'H' command at least twice to append the current line
to the Hold space, and then retrieve the lines from the hold space
with x, g, or G; or (3) use address ranges (see section 3.3, above)
to match lines between two specified addresses.
Choices (1) and (2) will put an \n into the pattern space, where it
can be addressed as desired ('s/ABC\nXYZ/alphabet/g'). One example
of using 'N' to delete a block of lines appears in section 4.13
("How do I delete a block of specific consecutive lines?"). This
example can be modified by changing the delete command to something
else, like 'p' (print), 'i' (insert), 'c' (change), 'a' (append),
or 's' (substitute).
Choice (3) will not put an \n into the pattern space, but it does
match a block of consecutive lines, so it may be that you don't
even need the \n to find what you're looking for. Since GNU sed
version 3.02.80 now supports this syntax:
sed '/start/,+4d' # to delete "start" plus the next 4 lines,
in addition to the traditional '/from here/,/to there/{...}' range
addresses, it may be possible to avoid the use of \n entirely.
A shorter awk alternative:
awk 1 ORS=' '
Explanation
An awk program is built up of rules which consist of conditional code-blocks, i.e.:
condition { code-block }
If the code-block is omitted, the default is used: { print $0 }. Thus, the 1 is interpreted as a true condition and print $0 is executed for each line.
When awk reads the input it splits it into records based on the value of RS (Record Separator), which by default is a newline, thus awk will by default parse the input line-wise. The splitting also involves stripping off RS from the input record.
Now, when printing a record, ORS (Output Record Separator) is appended to it, default is again a newline. So by changing ORS to a space all newlines are changed to spaces.
GNU sed has an option, -z, for null-separated records (lines). You can just call:
sed -z 's/\n/ /g'
The Perl version works the way you expected.
perl -i -p -e 's/\n//' file
As pointed out in the comments, it's worth noting that this edits in place. -i.bak will give you a backup of the original file before the replacement in case your regular expression isn't as smart as you thought.
Who needs sed? Here is the bash way:
cat test.txt | while read line; do echo -n "$line "; done
In order to replace all newlines with spaces using awk, without reading the whole file into memory:
awk '{printf "%s ", $0}' inputfile
If you want a final newline:
awk '{printf "%s ", $0} END {printf "\n"}' inputfile
You can use a character other than space:
awk '{printf "%s|", $0} END {printf "\n"}' inputfile
tr '\n' ' '
is the command.
Simple and easy to use.
Three things.
tr (or cat, etc.) is absolutely not needed. (GNU) sed and (GNU) awk, when combined, can do 99.9% of any text processing you need.
stream != line based. ed is a line-based editor. sed is not. See sed lecture for more information on the difference. Most people confuse sed to be line-based because it is, by default, not very greedy in its pattern matching for SIMPLE matches - for instance, when doing pattern searching and replacing by one or two characters, it by default only replaces on the first match it finds (unless specified otherwise by the global command). There would not even be a global command if it were line-based rather than STREAM-based, because it would evaluate only lines at a time. Try running ed; you'll notice the difference. ed is pretty useful if you want to iterate over specific lines (such as in a for-loop), but most of the times you'll just want sed.
That being said,
sed -e '{:q;N;s/\n/ /g;t q}' file
works just fine in GNU sed version 4.2.1. The above command will replace all newlines with spaces. It's ugly and a bit cumbersome to type in, but it works just fine. The {}'s can be left out, as they're only included for sanity reasons.
Why didn't I find a simple solution with awk?
awk '{printf $0}' file
printf will print the every line without newlines, if you want to separate the original lines with a space or other:
awk '{printf $0 " "}' file
The answer with the :a label ...
How can I replace a newline (\n) using sed?
... does not work in freebsd 7.2 on the command line:
( echo foo ; echo bar ) | sed ':a;N;$!ba;s/\n/ /g'
sed: 1: ":a;N;$!ba;s/\n/ /g": unused label 'a;N;$!ba;s/\n/ /g'
foo
bar
But does if you put the sed script in a file or use -e to "build" the sed script...
> (echo foo; echo bar) | sed -e :a -e N -e '$!ba' -e 's/\n/ /g'
foo bar
or ...
> cat > x.sed << eof
:a
N
$!ba
s/\n/ /g
eof
> (echo foo; echo bar) | sed -f x.sed
foo bar
Maybe the sed in OS X is similar.
Easy-to-understand Solution
I had this problem. The kicker was that I needed the solution to work on BSD's (Mac OS X) and GNU's (Linux and Cygwin) sed and tr:
$ echo 'foo
bar
baz
foo2
bar2
baz2' \
| tr '\n' '\000' \
| sed 's:\x00\x00.*:\n:g' \
| tr '\000' '\n'
Output:
foo
bar
baz
(has trailing newline)
It works on Linux, OS X, and BSD - even without UTF-8 support or with a crappy terminal.
Use tr to swap the newline with another character.
NULL (\000 or \x00) is nice because it doesn't need UTF-8 support and it's not likely to be used.
Use sed to match the NULL
Use tr to swap back extra newlines if you need them
You can use xargs:
seq 10 | xargs
or
seq 10 | xargs echo -n
cat file | xargs
for the sake of completeness
If you are unfortunate enough to have to deal with Windows line endings, you need to remove the \r and the \n:
tr '\r\n' ' ' < $input > $output
I'm not an expert, but I guess in sed you'd first need to append the next line into the pattern space, bij using "N". From the section "Multiline Pattern Space" in "Advanced sed Commands" of the book sed & awk (Dale Dougherty and Arnold Robbins; O'Reilly 1997; page 107 in the preview):
The multiline Next (N) command creates a multiline pattern space by reading a new line of input and appending it to the contents of the pattern space. The original contents of pattern space and the new input line are separated by a newline. The embedded newline character can be matched in patterns by the escape sequence "\n". In a multiline pattern space, the metacharacter "^" matches the very first character of the pattern space, and not the character(s) following any embedded newline(s). Similarly, "$" matches only the final newline in the pattern space, and not any embedded newline(s). After the Next command is executed, control is then passed to subsequent commands in the script.
From man sed:
[2addr]N
Append the next line of input to the pattern space, using an embedded newline character to separate the appended material from the original contents. Note that the current line number changes.
I've used this to search (multiple) badly formatted log files, in which the search string may be found on an "orphaned" next line.
In response to the "tr" solution above, on Windows (probably using the Gnuwin32 version of tr), the proposed solution:
tr '\n' ' ' < input
was not working for me, it'd either error or actually replace the \n w/ '' for some reason.
Using another feature of tr, the "delete" option -d did work though:
tr -d '\n' < input
or '\r\n' instead of '\n'
I used a hybrid approach to get around the newline thing by using tr to replace newlines with tabs, then replacing tabs with whatever I want. In this case, " " since I'm trying to generate HTML breaks.
echo -e "a\nb\nc\n" |tr '\n' '\t' | sed 's/\t/ <br> /g'`
You can also use this method:
sed 'x;G;1!h;s/\n/ /g;$!d'
Explanation
x - which is used to exchange the data from both space (pattern and hold).
G - which is used to append the data from hold space to pattern space.
h - which is used to copy the pattern space to hold space.
1!h - During first line won't copy pattern space to hold space due to \n is
available in pattern space.
$!d - Clear the pattern space every time before getting the next line until the
the last line.
Flow
When the first line get from the input, an exchange is made, so 1 goes to hold space and \n comes to pattern space, appending the hold space to pattern space, and a substitution is performed and deletes the pattern space.
During the second line, an exchange is made, 2 goes to hold space and 1 comes to the pattern space, G append the hold space into the pattern space, h copy the pattern to it, the substitution is made and deleted. This operation is continued until EOF is reached and prints the exact result.
Bullet-proof solution. Binary-data-safe and POSIX-compliant, but slow.
POSIX sed
requires input according to the
POSIX text file
and
POSIX line
definitions, so NULL-bytes and too long lines are not allowed and each line must end with a newline (including the last line). This makes it hard to use sed for processing arbitrary input data.
The following solution avoids sed and instead converts the input bytes to octal codes and then to bytes again, but intercepts octal code 012 (newline) and outputs the replacement string in place of it. As far as I can tell the solution is POSIX-compliant, so it should work on a wide variety of platforms.
od -A n -t o1 -v | tr ' \t' '\n\n' | grep . |
while read x; do [ "0$x" -eq 012 ] && printf '<br>\n' || printf "\\$x"; done
POSIX reference documentation:
sh,
shell command language,
od,
tr,
grep,
read,
[,
printf.
Both read, [, and printf are built-ins in at least bash, but that is probably not guaranteed by POSIX, so on some platforms it could be that each input byte will start one or more new processes, which will slow things down. Even in bash this solution only reaches about 50 kB/s, so it's not suited for large files.
Tested on Ubuntu (bash, dash, and busybox), FreeBSD, and OpenBSD.
In some situations maybe you can change RS to some other string or character. This way, \n is available for sub/gsub:
$ gawk 'BEGIN {RS="dn" } {gsub("\n"," ") ;print $0 }' file
The power of shell scripting is that if you do not know how to do it in one way you can do it in another way. And many times you have more things to take into account than make a complex solution on a simple problem.
Regarding the thing that gawk is slow... and reads the file into memory, I do not know this, but to me gawk seems to work with one line at the time and is very very fast (not that fast as some of the others, but the time to write and test also counts).
I process MB and even GB of data, and the only limit I found is line size.
Finds and replaces using allowing \n
sed -ie -z 's/Marker\n/# Marker Comment\nMarker\n/g' myfile.txt
Marker
Becomes
# Marker Comment
Marker
You could use xargs — it will replace \n with a space by default.
However, it would have problems if your input has any case of an unterminated quote, e.g. if the quote signs on a given line don't match.
On Mac OS X (using FreeBSD sed):
# replace each newline with a space
printf "a\nb\nc\nd\ne\nf" | sed -E -e :a -e '$!N; s/\n/ /g; ta'
printf "a\nb\nc\nd\ne\nf" | sed -E -e :a -e '$!N; s/\n/ /g' -e ta
To remove empty lines:
sed -n "s/^$//;t;p;"
Using Awk:
awk "BEGIN { o=\"\" } { o=o \" \" \$0 } END { print o; }"
A solution I particularly like is to append all the file in the hold space and replace all newlines at the end of file:
$ (echo foo; echo bar) | sed -n 'H;${x;s/\n//g;p;}'
foobar
However, someone said me the hold space can be finite in some sed implementations.
Replace newlines with any string, and replace the last newline too
The pure tr solutions can only replace with a single character, and the pure sed solutions don't replace the last newline of the input. The following solution fixes these problems, and seems to be safe for binary data (even with a UTF-8 locale):
printf '1\n2\n3\n' |
sed 's/%/%p/g;s/#/%a/g' | tr '\n' # | sed 's/#/<br>/g;s/%a/#/g;s/%p/%/g'
Result:
1<br>2<br>3<br>
It is sed that introduces the new-lines after "normal" substitution. First, it trims the new-line char, then it processes according to your instructions, then it introduces a new-line.
Using sed you can replace "the end" of a line (not the new-line char) after being trimmed, with a string of your choice, for each input line; but, sed will output different lines. For example, suppose you wanted to replace the "end of line" with "===" (more general than a replacing with a single space):
PROMPT~$ cat <<EOF |sed 's/$/===/g'
first line
second line
3rd line
EOF
first line===
second line===
3rd line===
PROMPT~$
To replace the new-line char with the string, you can, inefficiently though, use tr , as pointed before, to replace the newline-chars with a "special char" and then use sed to replace that special char with the string you want.
For example:
PROMPT~$ cat <<EOF | tr '\n' $'\x01'|sed -e 's/\x01/===/g'
first line
second line
3rd line
EOF
first line===second line===3rd line===PROMPT~$

How to replace '. ' with '.\n' using MacOS sed? [duplicate]

I am trying to scrub some lists into a properly formatted CSV file for database import.
My starting file, looks something like this with what is supposed to be each "line" spanning multiple lines like below
Mr. John Doe
Exclusively Stuff, 186
Caravelle Drive, Ponte Vedra
33487.
I created a sed script that cleans up the file (there's lots of "dirty" formatting like double spaces and spaces before/after commas). The problem is the Zip with the period. I would like to change that period for a new line, but I cannot get it to work.
The command that I use is:
sed -E -f scrub.sed test.txt
and the scrub.sed script is as follows:
:a
N
s|[[:space:]][[:space:]]| |g
s|,[[:space:]]|,|g
s|[[:space:]],|,|g
s|\n| |g
s|[[:space:]]([0-9]{5})\.|,FL,\1\n |g
$!ba
What I get is
Mr. John Doe,Exclusively Stuff,186 Caravelle Drive,Ponte Vedra,FL,33487n
If figured that the Zip+.(period) would be a great "delimiter" to use the substitution on and while I can find it, I can't seem to tell it to put a newline there.
Most of the things I found online are about replacing the newline with something else (usually deleting them), but not much on replacing with a newline. I did find this, but it didn't work: How to insert newline character after comma in `),(` with sed?
Is there something I am missing?
Update:
I edited my scrub.sed file putting the literal new line as instucted. It still doesn't work
:a
N
s|[[:space:]][[:space:]]| |g
s|,[[:space:]]|,|g
s|[[:space:]],|,|g
s|\n| |g
s|[[:space:]]([0-9]{5})\.|,FL,\1\
|g
$!ba
What I get is (everything on one line):
Mr. John Doe,Exclusively Stuff,186 Caravelle Drive,Ponte Vedra,FL,33487 Mrs. Jane Smith,Props and Stuff,123 Main Drive,Jacksonville,FL,336907
My expected output should be:
Mr. John Doe,Exclusively Stuff,186 Caravelle Drive,Ponte Vedra,FL,33487
Mrs. Jane Smith,Props and Stuff,123 Main Drive,Jacksonville,FL,336907
The sed on BSD does not support the \n representation of a new line (turning it into a literal n):
$ echo "123." | sed -E 's/([[:digit:]]*)\./\1\n next line/'
123n next line
GNU sed does support the \n representation:
$ echo "123." | gsed -E 's/([[:digit:]]*)\./\1\nnext line/'
123
next line
Alternatives are:
Use a single character delimiter that you then use tr translate into a new line:
$ echo "123." | sed -E 's/([[:digit:]]*)\./\1|next line/' | tr '|' '\n'
123
next line
Or use an escaped literal new line in your sed script:
$ echo "123." | sed -E 's/([[:digit:]]*)\./\1\
next line/'
123
next line
Or define a new line:
POSIX:
nl='
'
BASH / zsh / others that support ANSI C quoting:
nl=$'\n'
And then use sed with appropriate quoting and escapes to insert the literal \n:
echo "123." | sed 's/\./'"\\${nl}"'next line/'
123
next line
Or use awk:
$ echo "123." | awk '/^[[:digit:]]+\./{sub(/\./,"\nnext line")} 1'
123
next line
Or use GNU sed which supports \n
The portable way to get a newline in sed is a backslash followed by a literal newline:
$ echo 'foo' | sed 's/foo/foo\
bar/'
foo
bar
I guarantee there's a far simpler solution to your whole problem by using awk rather than sed though.
The following works on Oracle Linux, x8664:
$ echo 'foobar' | sed 's/foo/foo\n/'
foo
bar
If you need it to match more than once per line, you'll need to place a g at the end, as in:
$ echo 'foobarfoobaz' | sed 's/foo/foo\n/g'
foo
barfoo
baz
Add a line after a match.
The sed command can add a new line after a pattern match is found. The "a" command to sed tells it to add a new line after a match is found.
sed '/unix/ a "Add a new line"' file.txt
unix is great os. unix is opensource. unix is free os.
"Add a new line"
learn operating system.
unixlinux which one you choose.
"Add a new line"
Add a line before a match
The sed command can add a new line before a pattern match is found. The "i" command to sed tells it to add a new line before a match is found.
sed '/unix/ i "Add a new line"' file.txt
"Add a new line"
unix is great os. unix is opensource. unix is free os.
learn operating system.
"Add a new line"
unixlinux which one you choose.

Bash script delete a line in the file

I have a file, which has multiple lines.
For example:
a
ab#
ad.
a12fs
b
c
...
I want to use sed or awk delete the line, if the line include symbols or numbers. (For example, I want to delete: ab#, ad., a12fs.... lines)
or in another words, I just want to keep the line which include [a-z][A-Z] .
I know how to delete number line,
sed '/[0-9]/d' file.txt
but I do not know how to delete symbols lines.
Or there has any easy way to do that?
To keep blank lines:
grep '^[[:alpha:]]*$' file
sed '/[^[:alpha:]]/d' file
awk '/^[[:alpha:]]*$/' file
To remove blank lines:
grep '^[[:alpha:]]+$' file
sed -E -n '/^[[:alpha:]]+$/p' file
awk '/^[[:alpha:]]+$/' file
grep works well too and is even simpler: just do the reverse: keep the lines that interest you, which are way easier to define
grep -i '^[a-z]*$' file.txt
(match lines containing only letters and empty lines, and -i option makes grep case-insensitive)
to remove empty lines as well:
grep -i '^[a-z]+$' file.txt
caution when using Windows text files, as there's a carriage return at the end of the line, so nothing would match depending on grep versions (tested on windows here and it works)
but just in case:
grep -iP '^[a-z]*\r?$'
(note the P option to enable perl expressions or \r is not recognized)
You can use this sed:
sed '/^[A-Za-z0-9]\+$/!d' file
(OR)
sed '/[^A-Za-z0-9]/d' file
$ awk '!/[^[:alpha:]]/' file.txt
a
b
c

Bash Script - Sed Command

I created a script to add 100 space characters at the end of each line:
#!/bin/ksh
sed -i -e 's/\n/ - 100 spaces - /' $1
but it doesn't work and I think it is because of \n. Any thoughts?
Sed processes the content of the lines without the newline. Your code never sees a newline, so it cannot replace it.
Match the end of the string:
sed -i -e 's/$/ - 100 spaces - /' $1
Although Karoly has already pointed out the error in your script, you could also save yourself typing 100 spaces by using a condition and break to make a sort of loop
sed ':1;/ \{100\}$/!{s/$/ /;b1}' file
Will print 100 space at the end of the line
If there are already spaces at the end and you want to add 100
sed 's/$/1/;:1;/1 \{100\}$/!{s/$/ /;b1};s/1\( *\)$/\1/' file
Just a suggestion to avoid typing 100 spaces (although I'm sure that by this time, you already have!) - use perl:
perl -pe 's/$/" " x 100/e' file
As the other answers have stated, this matches the end of the line and replaces with 100 spaces, using the e modifier to allow you to write " " x 100 and let perl do the work for you.
As with sed, you can modify the file in-place using the -i switch - as with sed, I'd suggest using something like -i.bak to create a backup file file.bak.
Could be the \n as that's undefined by POSIX and so implementation dependent across seds but you also say you want add spaces to the end of a line while your script is trying to replace the end of line with spaces.
In any case this is not a job for sed, just use [s]printf("%100s","") in awk to create a string of 100 blanks, e.g.:
$ echo "foo bar" | awk '{printf "%s%10s%s\n", $1, "", $2}'
foo bar

need to replace 2 things from a csv file; one is Y with 'Y and the the newline charater by ',' value

I need to replace 2 things from a csv file using Unix shell scripting ; one is Y with 'Y and the newline character by ',' value..
my csv:(values will be one below another vertically and all values start with Y)
YC1234
YC5678
expected output is NEEDED IN A LINE horizontally like :
'YC1234','YC5678'
kindly help as i am new to shell scripting...
i tried sed by its difficult in removing newline
cat master_upd.csv | sed -e 's//'\'','\''/g' | sed 's/\n/ /'
tr '\n' ',' < master.csv
The first command 'cat' is probably unnecessary.
Here is a thread on replacing newlines with something else using sed: How can I replace a newline (\n) using sed?
sed -e 's/^.*$/'\''&'\''/' master_upd.csv | sed ':a;N;$!ba;s/\n/,/g'
The first sed puts in the single quotes, the second one replaces newlines with commas. Notice that I removed the cat and just told sed to open your file directly.
awk -v q="'" 'NR>1{printf ","}{printf "%s",q$0q}END{print ""}' file

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