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Need 10 lines before the first occurance of a pattern

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# 1  
Old 01-27-2010
Question Need 10 lines before the first occurance of a pattern
Here is the text file:
Code:
1This is a text file
2this is a text file
3This is a text file
4this is a text file
5This is a text file
6this is a text file
7This is a text file
8this is a text file
9This is a text file
10this is a text file
11This is a text file
12this is a text file
13This is a text file
14this is a text file
check
15This is a text file
16this is a text file
check
17This is a text file
18this is a text file
19This is a text file
20this is a text file

I need to search for first occurrence of "check" and need to show 10 lines above this pattern.

The result should be:
Code:
5This is a text file
6this is a text file
7This is a text file
8this is a text file
9This is a text file
10this is a text file
11This is a text file
12this is a text file
13This is a text file
14this is a text file

Please help!
Last edited by Yogesh Sawant; 01-27-2010 at 04:02 PM.. Reason: added code tags
# 2  
Old 01-27-2010
If your grep version don't support the -B option, this should do the job:

Code:
awk '
/check/ {
  for(i=0;i<10;i++) {
    print a[(NR+i)%10]
  }
  exit
}
{a[NR%10]=$0}
' file

# 3  
Old 01-27-2010
Or a (mostly) sed solution:

Code:
sed -n '/check/q;p' input_file | tail -10

# 4  
Old 01-27-2010
Bug its working
hey its working... thanks so much!
# 5  
Old 02-26-2010
MySQL
see the following grep command.
Code:
grep --max-count 1 -B 10 'check' input  | grep -v 'check'

Output:
Code:
5This is a text file
6this is a text file
7This is a text file
8this is a text file
9This is a text file
10this is a text file
11This is a text file
12this is a text file
13This is a text file
14this is a text file

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