line feeds in csv - Shell Programming and Scripting

gowrish · #1 08-31-2005

hi all,

i have csv file with three comma separated columns
i/p file
First_Name, Address, Last_Name
XXX, "456 New albany \n newyork, Unitedstates \n 45322-33", YYY\n
ZZZ, "654 rifle park \n toronto, canada \n 43L-w3b", RRR\n

is there any way i can remove \n (newline) from the second column
between the two double quotes"..."

vgersh99 · #2 08-31-2005

will this do?

Code:

sed 's/\\n//g' csvFile

gowrish · #3 08-31-2005

thanks vgersh99
I tired that ! but it didn't work . all it does is removes all the \n from the csvfile.
( i.e ) all the data will be in one single line .

Rakesh Ranjan · #4 08-31-2005

Just change the pattern that vgersh99 said try something like
sed 's/\\n/\"^["]*\"'
\" for checking starting quotes ^["] for checking anything other than " (inside [] no metachr except '-' means anything so no need for \") and lastly \\n for newline. g as suggested by vgersh stands for globaly and so should not be used here.
I haven't run this (away from any Linux box) so in case of problem just fiddle with above pattern & if u still have problem plz. do tell me.

vgersh99 · #5 09-01-2005

Quote:

Originally Posted by gowrish

thanks vgersh99
I tired that ! but it didn't work . all it does is removes all the \n from the csvfile.

that's right.

Quote:

Originally Posted by gowrish

( i.e ) all the data will be in one single line .

not really. given your sample file, the output would be: [on separate lines]

Code:

First_Name, Address, Last_Name
XXX, "456 New albany  newyork, Unitedstates  45322-33", YYY
ZZZ, "654 rifle park  toronto, canada  43L-w3b", RRR

gowrish · #6 09-01-2005

thanks man !

I tried ur suggestion, it didn't workout---i got this error msg!
sed: command garbled: s/\\n/\"^["]*\"
click on this link, there was a similar problem in this site , they had the solution too but not clear enough for me to solve my problem..
Remove Carriage returns between strings in a field
I didn't understand how they are calling the file, how the loop exactly workings
Please lookat the solution,please suggest me on this .
thanks
Gowrish

vgersh99 · #7 09-01-2005

Quote:

Originally Posted by Rakesh Ranjan

Just change the pattern that vgersh99 said try something like
sed 's/\\n/\"^["]*\"'
\" for checking starting quotes ^["] for checking anything other than " (inside [] no metachr except '-' means anything so no need for \") and lastly \\n for newline. g as suggested by vgersh stands for globaly and so should not be used here.
I haven't run this (away from any Linux box) so in case of problem just fiddle with above pattern & if u still have problem plz. do tell me.

I think you're confusing the syntax of 's':

Code:

s/regular expression/replacement/flags
                   Substitute   the   replacement  string   for
                  instances  of  the  regular  expression in the
                  pattern  space.   Any  character  other   than
                  backslash  or newline can be used instead of a
                  slash to delimit the RE and  the  replacement.
                  Within the RE and the replacement, the RE del-
                  imiter itself can be used as a literal charac-
                  ter if it is preceded by a backslash.
                  An ampersand (&) appearing in the  replacement
                  will  be  replaced  by the string matching the
                  RE.  The special meaning of & in this  context
                  can   be   suppressed   by   preceding  it  by
                  backslash.  The characters \n, where  n  is  a
                  digit, will be replaced by the text matched by
                  the  corresponding  backreference  expression.
                  For each backslash (\) encountered in scanning
                  replacement from beginning to end, the follow-
                  ing  character  loses  its special meaning (if
                  any).  It is unspecified what special  meaning
                  is  given  to any character other than &, \ or
                  digits.

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