how to grep a pattern which is split in 2 lines.

[linuxguy2011]

Hi,

I need to grep a pattern which can be present in one line or could be split in 2 lines...

Example

getData(xxx, yyy);

getData(xxx,
yyy);

Normal grep wont work in this case. Can anyone please help on this?

There are 100's of files in which i need to search for this pattern so time is also a constrain.

Thanks in advance

linuxguy2011

[barriehie]

So you're really looking for 3 patterns:

pattern1: getData(xxx, yyy);
pattern2: getData(xxx,
pattern3: yyy);

Perl would perhaps be better suited to this task. If pattern1 then print else if pattern2 get next line and concatenate with previous line and print?

[dufftime]

I Googled for "multi line grep":

www . unix . com/unix-dummies-questions-answers/56703-multiline-grep.html

You'll want to replace the regular expression within the #'s with your search string, e.g. replace this:

while ($whole_file =~ m#\<a\>(.*?)\<\/a\>#sg) {

Using your examples, with something like this:

while ($whole_file =~ m#(getData\(.*?\);)#sg) {

You can modify this script to take a search string as a parameter instead of the hard-coded "getData". Use $ARGV[0] in place of getData.

This also looks promising: www . commandlinefu . com/commands/view/5087/multi-line-grep

Sorry about the links - can't post URL's until I have 15 posts!

Go go Perl!

-dufftime

[venka2k_cbe]

You can awk command to grep the two columns.

If you want to grep only first column of a file for certain string. You use following cmd,

awk -F: '{print $1} <filename> |grep <string>

~Venky

[linuxguy2011]

Hi Guys,

Thanks for your replies...I solved the problem using a different way

I used this approch since there were so many files and that too in different directories so i needed a quick way to search

Take the below as example

pattern1: getData(xxx, yyy);
pattern2: getData(xxx,
pattern3: yyy);

i searched for xxx (using grep -n) and appended the output to a file say xxxout

then i searched for yyy (using grep -n) and appeneded the output to a different file say yyyout

once the search went through all the files i can match the lines in xxxout with yyyout line by line.

If the line in xxxout matches with a line in yyyout that means pattern1: getData(xxx, yyy); is found.

if the line does not match then get the line number and file name using some sed operation (sorry i dont have the code at hand now...using my pda) increment the line number by 1 and search for "$filename:$lineNo" in yyyout

if the pattern is found that means the

pattern2: getData(xxx,
pattern3: yyy)

is found.

Reject everything else.

This saves a lot of time when a large number of files are to be searched in lots of directories

I hope i made it clear.

linuxguy2011

[Manko10]

What about

grep -P 'getData\(xxx,\s*yyy\);'

With PCREs enabled, grep also matches newlines as whitespace characters.