Using grep, treating two lines as one? - linux

Question

Suppose I have a configuration file that can be in one of two format below (short example, but basically the first format is a line that is too long that you have to use a line continuation character, while the second format is just simply a long line without the line continuation)
data1=x data2=y data3=z \
datakey
second format
data=1 data2=y data3=z datakey
I want to match the exact line data1=x data2=y data3=x datakey for both situation. Is there simple way of doing that?

Answer 1

read interprets \ as the line continuation character:
while read line ; do
if [[ $line == 'data=1 data2=y data3=z datakey' ]] ; then
echo "$line"
fi
done

Answer 2

I would use sed to create an output without the ending \:
sed -e ':begin;/\\$/{N;bbegin};s/\\\n//g' your_file
Then you could grep it:
sed -e ':begin;/\\$/{N;bbegin};s/\\\n//g' your_file | grep your_pattern
You can even do this all in sed:
sed -n -e ':begin;/\\$/{N;bbegin};s/\\\n//g;/your_pattern/p' your_file
UPDATE:
To explain above:
:begin sets a tag to which I can branch (goto) with the b command.
/\\$/{N;bbegin} if the current line ends with a \ (/\\$/), append the next line to the buffer (N) and goto begin (bbegin).
Then, when the lines does not end with a \, remove all the \ and the line break (thes/\\n//g`).
Then -n option tells sed no to print the line at the end of the script.
/your_pattern/p prints the line if it matches your_pattern.
UPDATE2:
We could even do better and show the original lines of your file:
sed -n -e ':begin;/\\$/{N;bbegin};h;s/\\\n//g;/your_pattern/{g;p}' your_file
What this does is before removing the \ and the line break, it saves the data in the hold space (h) and if the line matches, it prints the data that was saved (g copies the hold space to the pattern space that is printed).

Answer 3

Maybe grep is not a best tool for problems like that.
You could join all lines ends with \ and then grep that output as usual:
Suppose you have a file:
$> cat text
1
2
fasdfasdf
data1=x data2=y data3=z \
datakey
fasfd
sdf
So you can join all lines ends with \:
$> awk '{line = $0}; /.*\\/ {split($0,tmp,"\\"); line = tmp[1]; getline; line = line $0}; { print line }' text
1
2
fasdfasdf
data1=x data2=y data3=z datakey
fasfd
sdf

Answer 4

awk -v i="data1=x data2=y data3=z datakey" '{x=x" "$0}END{y=match(x," "i);if(y) print "yes its a match"}' temp
tested below:
> cat temp
data1=x data2=y data3=z
datakey
> awk -v i="data1=x data2=y data3=z datakey" '{x=x" "$0}END{y=match(x," "i);if(y) print "yes its a match"}' temp
yes its a match
>

Answer 5

I like the sed example above, and didn't notice the awk ones already
submitted. Here is awk (nawk) version that keeps the formatting and deals
with multiple continuation lines and end-of-file.
nawk -v re="search-string" \
'{ls=ls""$0;lp=lp""$0}
/\\$/{ls=substr(ls,1,length(ls)-1);lp=lp"\n";next}
ls~re{print lp}
{lp=ls=""}
END{if (ls ~ re)print substr(lp,1,length(lp)-1)}' input-file
Explanation:
re=search-string sets the regular expression to look for.
ls=ls""$0 Concatenate new line to ls (line-search), also store
original line to lp (line-print) as per sed example above.
/\\$/ checks to see if input has continuation character and
if it does substr() removes the extra character from ls, lp has a
newline appended to keep original formatting. Lastly next
causes awk to read the next line and start from the first rule.
ls~i searches line-search for re and if it matches prints lp,
could add a switch to print ls if preferred.
Lastly (not penultimately) reset ls & lp when no continuation line.
END is special rule, used here to detect that ls still has a
value assigned, meaning the continuation line is at the EOF. Searched for the regex, when it
matches it cuts off the additional newline added above.
Now, no one would write '\' at the end of a line would they?