Script to manipulate logfile text

Shell Programming and Scripting
1

Hi guys, I was wandering if a Shell guru could give me some advice on tackling a problem.

I have used a mixture of grep, cut and awk to get data from a log file in the following format:

14/11/08 10:39:      Checking currenly    :  Enabled
14/11/08 10:39:      Records allocated    :  221
14/11/08 10:39:      Records Freed         :  0
14/11/08 10:39:      Records Aged          :  56

list continues in same format etc.......

What I would now like to do is pipe it into another bit of script which would leave it in a comma delimited format with date timestamp so i can throw it into a sql data base, format like so.

14/11/08 , 10:39 , Enabled , 221 , 0 , 56

I have come a little stuck and am not sure how I would do this so any advice would be much appreciated.

2

Just to add a quick note the formatting has slightly messed after posting to the forum, everything is in straight clear cut colums with set character positions

3

How do you know how far to carry the
Enabled , 221 , 0 , 56 ?

Are you grouping things with the same date/time, or until some other message comes up?

4

Those values are taken from the last column until the end of the file, the date time stamp is the same due to earlier text processing. thats why I just want to end up with

14/11/08 , 10:39 , Enabled , 221 , 0 , 56

5

Try this:

awk '{
s=$1" , "substr($2,1,5)" , "$6
for(i=1;i<4;i++) {
  getline;s=s" , "$6
}
print s}' file

Regards

6

Could you possibly run me through what each line of the code is doing so i can alter it if need be regards ross

7

And thanks works great on the text I provided above

8

Ok one more question how would I modify the above if the format above was as such like below.

14/11/08 10:39:  Records allocated................ 221
14/11/08 10:39:  Records Freed.................... 0
14/11/08 10:39:  Records Aged..................... 20
14/11/08 10:39:  Records allocated................ 221
14/11/08 10:39:  Records Freed.................... 0
14/11/08 10:39:  Records Aged..................... 20
9

Ok, here we go:

awk '{
s=$1" , "substr($2,1,5)" , "$6
for(i=1;i<4;i++) {
  getline;s=s" , "$6
}
print s}' file

s=$1" , "substr($2,1,5)" , "$6

On the 1st, 5th, 9th line etc. assign field 1, the first 5 characters of the 2nd field and the 6th field to string s using " , " as delimiters.

for(i=1;i<4;i++) {
  getline;s=s" , "$6
}

Get the next 3 lines and add the 6th field after the string s.

print s

Print the string s and get the next line.
After printing the first string the next line is the 5th line since we have read 3 strings in the loop after the 1st string.

Regards

10

And what should be the output?

Regards

11

Same format as before

14/11/08 , 10:39 , 221 , 0 , 20 , 221 , 0 , 20

thanks for all your help

12

perl:

open FH,"<a.txt";
while(<FH>){
tr/\n//d;
my @temp=split(/ +/,$_);
my $str=sprintf("%s,%s",$temp[0],$temp[1]);
$hash{$str}=sprintf("%s,%s",$hash{$str},$temp[5]);
}
foreach $key (keys %hash){
print $key,$hash{$key},"\n";
}