Shell Programming and Scripting

I have to grep on a few words in a file and then display the line containing those words and the line above it.

For ex -

File1.txt contains...

abc xyz abc
This is a test
Test successful
abc xyz abc
Just a test
Test successful


I find the words 'Test successful' in the file and then I have to display the line containing these words and the line above it.

So my output should be...

This is a test
Test successful
Just a test
Test successful

I am just not sure how to display the line above the 'Test successful' in my result.

grep -B1 "Test successful" yourfile

var=`grep -n "Test successful" <filename> | cut -f1 -d ":"`

for i in $var
do
head -$i <filename> | tail -2
done

use this awk utility

Code:

awk 'c-->0;$0~s{if(b)for(c=b+1;c>1;c--)print r[(NR-c+1)%b];print;c=a}b{r[NR%b]=$0}' b=1 a=0 s="Test successful" filename

where a=no of lines below the pattern
b=no of lines above the pattern
s=pattern

aster007,

that worked very well...

Can you also let me know how can I get the same result from all files in the directory (using '*') rather than putting <filename>?

Code:

for file in *
do
  var=`grep -n "Test successful" $file | cut -f1 -d ":"`
  for i in $var
  do
    head -$i $file | tail -2
  done
done

That worked awsome!

Thank you -- Peterro and aster007.