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Remove EOL selectively

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# 1  
Old 11-10-2011
Remove EOL selectively
Hi,

I have a text as below
Code:
test1 test2 test3\
test4 test5 test6 test7
newtest1 newtest2\
newtest3 newtest4 newtest5

And need this to be replaces to
Code:
test1 test2 test3 test4 test5 test6 test7
newtest1 newtest2 newtest3 newtest4 newtest5

So my requirement is to remove the EOL if the last character is \.

I tried my luck and searches but but unsuccessfull.. help here is very much appreciated.

Thanks
Praveen

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# 2  
Old 11-10-2011
Not sure what OS you're using or if the file contents will be static like the example you gave but this may work:

Code:
awk '{sub(/\\/, "")}!(NR%2){print $0" "p}{p=$0}' file.txt

Output:
test4 test5 test6 test7 test1 test2 test3
newtest3 newtest4 newtest5 newtest1 newtest2

Hope this helps point you in the right direction.
# 3  
Old 11-10-2011
Try something like this:
Code:
sed "/\\\\$/ { N; s/\\\\\n/ /}" filename

# 4  
Old 11-17-2011
Sorry , awk is not working as expected ....
Code:
#cat test_sudo
User_Alias      TEST=test1,test2,test3,test4,test5,\
                     test6,test7,ltest8,test9


#awk '{sub(/\\/, "")}!(NR%2){print $0" "p}{p=$0}' test_sudo
                     test6,test7,ltest8,test9 User_Alias        TEST=test1,test2,test3,test4,test5,

#sed "/\\\\$/ { N; s/\\\\\n/ /}" test_sudo
sed: 0602-404 Function /\\$/ { N; s/\\\n/ /} cannot be parsed.

This is an AIX machine
Last edited by Franklin52; 11-17-2011 at 04:04 AM.. Reason: Code tags
# 5  
Old 11-17-2011
Let a simple shell script do the job:
Code:
]$ cat test_sudo
User_Alias      TEST=test1,test2,test3,test4,test5,\
                     test6,test7,ltest8,test9
$ while read line
> do
> echo "$line"
> done
User_Alias      TEST=test1,test2,test3,test4,test5,                     test6,test7,ltest8,test9

The read shell-builtin has the capability to deal with \-NL by itself.
# 6  
Old 11-17-2011
Another approach:
Code:
awk '/\\$/{printf substr($0,1,length-1) FS; next}1' file


Use nawk or /usr/xpg4/bin/awk on Solaris.
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