如何从文件[shell]中某些特定行的末尾删除\\ n？

Question

In Redhat I have file.csv that has below data:

170033101;20170302;;;"Free text 1"
170033101;20170302;;;"Free text 2"
170033101;20170302;;;"Free 
text 3"
170033101;20170302;;;"Free text 4"

I want to create another corrected file (Correct_file.csv) after removing the wrong \\n from the file to be as below:

170033101;20170302;;;"Free text 1"
170033101;20170302;;;"Free text 2"
170033101;20170302;;;"Free text 3"
170033101;20170302;;;"Free text 4"

My solution:

I made the below shell script to find rows preceding those lines that don't start with 170, and then create sed.txt that has a sed line for each wrong row to replace the \\n with space.

I am unable to perform the sed command or tr command to remove a specific line based on the line number

My Script:

>sed.txt;
for i in `grep -nv '^[1706]' $1|cut -f 1 -d \:`
do
if [ $i -eq 1 ]
then
continue
else
j=`expr $i - 1`
echo $j"s/\n//" >>sed.txt
fi
done
sed -f sed.txt $1 >$2

I call the script and pass 2 parameters 1- old file 2- new corrected file, and the new file is exactly as the old with no correction.

Answer 1

You can use this awk command that works bases on the fact if a line ends with " or not:

awk '!/"$/{p=$0; next} p!=""{$0 = p $0; p=""} 1' file

170033101;20170302;;;"Free text 1"
170033101;20170302;;;"Free text 2"
170033101;20170302;;;"Free text 3"
170033101;20170302;;;"Free text 4"

Answer 2

sed returns the new string, so you don't need to echo it. Simply call it sed .. >> data.txt

The following sed statement will replace the new line at the end of a line with nothing. You need to pass only the lines you want to translate

sed ':a;N;$!ba;s/\n//g' <LINE INPUT>

If you pass it a file, it will read the whole file in a loop, and replace the newline(s) with a space.

Answer 3

You can use this sed :

sed '/^170/{:loop; N;/\n170/{P;D;t}; s/\n//g;b loop}' file

Input:

$ cat file
170033101;20170302;;;"Free text 1"
170033101;20170302;;;"Free text 2"
170033101;20170302;;;"Free 
text 3"
170033101;20170302;;;"Free 
text 
4"

Test:

$ sed '/^170/{:loop; N;/\n170/{P;D;t}; s/\n//g;b loop}' file > correct_file.csv
170033101;20170302;;;"Free text 1"
170033101;20170302;;;"Free text 2"
170033101;20170302;;;"Free text 3"
170033101;20170302;;;"Free text 4"

Answer 4

When i want to work with \\n i prefer simple perl over sed:

$ cat file1
170033101;20170302;;;"Free text 1"
170033101;20170302;;;"Free text 2"
170033101;20170302;;;"Free
text 3"
170033101;20170302;;;"Free text 4" 

$ perl -pe 's/[^"]\n/ /g' file1
170033101;20170302;;;"Free text 1"
170033101;20170302;;;"Free text 2"
170033101;20170302;;;"Free text 3"
170033101;20170302;;;"Free text 4"

This perl oneliner replaces with a single space every new line \\n not followed by quotes "

PS: You can add >newfile at the end of the command to send the "corrected" output to a newfile or you can even edit the current file in place using -i perl switch.

Answer 5

try following awk too once.

awk '{printf("%s%s",$0 !~ /^[0-9]+/?"":(NR>1?RS:""),$0)} END{print ""}'  Input_file

Checking here if any line if not starting from digits then printing new line there by RS(record separator) making sure it shouldn't happen on very first line else printing nothing. in END section of awk printing NULL which will print a new line at last.