find file and print only contents with a hit by grep

Shell Programming and Scripting
1

Hi,

can someone help me.

I have some files and search a content in this files. If i have a hit I will print a output:

filename:content

But are more hits in one file: The output is always filename:content

E.G. Seach about "three"

file1 {one, two, three, four, three}
file2 {three, five, six, seven}
file3 {seven, eight}

Output:
file1:three
file1:three
file2:three

I search a unix statement which makes this output. I searching without success for a week now.
THANKS.

2

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3 
 grep "xxx" *datei* /dev/null
4

Guess you just need the file name with content include some key words, use -l option in grep:

grep -l tree file*
5 
$> grep -o three file*
file1:three
file1:three
file2:three
6

If your grep does not have -o etc. and this is regarding to your former post Searching Problem, you can also just slightly modify the line I already gave in the there to get only the matches in the files, not the complete files as output:

$> cat file1
eins
zwei
OCT
drei
vier
$> cat file2
16-SEP-2009 09:11:47 10.65.4.24
16-SEP-2009 09:11:47 10.3.4.11
30-SEP-2009 10:11:47 10.3.4.11
1-OCT-2009 10:11:47 10.65.4.24
6-OCT-2009 10:11:47 10.3.4.11
6-OCT-2009 12:31:01 10.3.4.11
16-OCT-2009 11:11:47 10.65.4.24
17-OCT-2009 11:11:47 10.65.4.24
18-OCT-2009 11:11:47 10.3.4.11
$> find . -type f -exec grep -l OCT {} \; | while read LINE; do awk -v l=$LINE '/OCT/ {print l":"$0}' $LINE; done
./file2:1-OCT-2009 10:11:47 10.65.4.24
./file2:6-OCT-2009 10:11:47 10.3.4.11
./file2:6-OCT-2009 12:31:01 10.3.4.11
./file2:16-OCT-2009 11:11:47 10.65.4.24
./file2:17-OCT-2009 11:11:47 10.65.4.24
./file2:18-OCT-2009 11:11:47 10.3.4.11
./file1:OCT