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not work:(
awk '!/^""$/ {if ( $0 ~ /^"[^,].+/) { printf NR==1?$0:"\n"$0} else { printf $0}} END { printf "\n"} ' file.lis
awk: syntax error near line 1
awk: illegal statement near line 1

maybe...

sure:

"2721348","2013-10-06","2013-10-06 15:54:37","Tree","1","0","1","1","paper","sumer","","ppd","no","no","anonymous","195","oo","Seria SP, kete","000000000126 ","Red...

thanks its work by 1 line file.

Can you help what can i do, if i have a file with 1 million lines which include 15 bad characters(apostrophe and enter).

Hi experts,

I have a csv file where the delimiter is comma and the text between apostrophe.

Like this:

"aaa","bbb","ccc","ddd","eee","fff","ggg"


But now i have a lot of enter(end of...

Thank You!

Thats not my file. My original file is read only. Your work around is very good! Thank You!

od -c output:
0000000 " " , " " , " " , " " , " " , "
0000020 " , " " , " " , " " \n " " , " "
0000040 , " " , " " , ...

input-output:
alma@app2:/home/snayper$ cat aaa.txt
"","","","","","","","",""
"","","","","","","","",""alma@app2:/home/snayper$ sed 's/"[^"]*"//g' ./aaa.txt
,,,,,,,,
alma@app2:/home/snayper$

Its not work if you make a text file by vi editor. Try to make a file in windows whit this text:
"","","","","","","","",""

After you try again but whit 2 rows, like this:...

Hi Everyone!

I have a problem my script. I'm useing sed to find the separate character in my text
sed 's/"[^"]*"//g' ./myfile.txt

My problem:
When the last row is doesn't has enter in the end...

i pasted into to command line:

nawk -F, S=4000 '{ for ( i = 1; i <= NF; i++ ) { if ( length ( $i ) > S ) { $i = substr ( $i, 1, S ) OFS substr ( $i, S + 1 ) } } print }' OFS=, textfile.lis

and...

Thanks Your answer!

i got this message:
awk: syntax error near line 1
awk: bailing out near line 1

I'm use solaris(maybe important).

Hi everybody!

I have a large comma separated text file. This file incluede some column which largest than 4000 character(this is not the last column). I tried with sed but i can't do it.

I...

thanks a lot!

i use this:

sed 's/"[^"]*"//g' file | perl -nle 'print length'

Hi all,

I have this input:
"203324780",,"89321213261247090146","VfdsD150","0D","fd3221","V0343","aaa","Direkt","fsa","2015.02.27","39833,54454,21214",,,"fd","NORMAL","D","10fd","1243...

@vidyadhar85

Thanks Master!

now i get this:

looks like better :)

0002012Mar282012Apr192012Aug162012Aug292012Oct012012Oct080Dec120Feb260Mar110Apr030Apr250Apr260Apr300May100May240May310May310May310Jun020Jun030Jun04

but...

@vidyadhar85

not work

i get this:
%d%s%02d
%d%s%02d 2012 Mar 28
%d%s%02d 2012 Apr 19
%d%s%02d 2012 Aug 16
%d%s%02d 2012 Aug 29
%d%s%02d 2012 Oct 1
%d%s%02d 2012 Oct 8
%d%s%02d 09:28 Dec...

Hi,

I wold like to replace this 2012 Mar 02 to this 20120302.

i have already this code:

ls -lrt | awk '{ print $8$6$7 }'| sed -e s/[0-9]*:[0-9]*/'2013'/g
-e s/'Jan'/01/g -e s/'Feb'/02/g...