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compare text in log file

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# 1  
Old 10-13-2011
compare text in log file
Hi all,

I am posting the thread similar to previous posts but here my scenario is i need to know what the files size from start date. Here end time file size will be 0. Also need to know how much time does it took to complete in seconds.

log file:

Name1 START 11:36:45 13102011 File: 2015443 size: 187654 File: 0 size: 0

Name1 END 11:38:47 13102011 File: 0 size: 0 File: 0 size: 0

Name2 START 11:39:52 13102011 File: 2008395 size: 28764 File: 0 size: 0

Name2 END 11:40:21 13102011 File: 0 size: 0 File: 0 size: 0

Name1 START 11:41:59 13102011 File: 1056987 size: 467321 File: 0 size: 0

Name1 END 11:42:36 13102011 File: 0 size: 0 File: 0 size: 0

the result should be

Name1 END ENDTIME time_diff_seconds date File: 2015443 size: 187654 File: 0 size: 0

Name2 END ENDTIME time_diff_Seconds date File: 2008395 size: 28764 File: 0 size: 0

Name3 END ENDTIME time-diff_seconds date File: 1056987 size: 467321 File: 0 size: 0

Can any one please me out.

Regards
Olivia
# 2  
Old 10-13-2011
I saw four Name1 , typo?

---------- Post updated at 10:32 AM ---------- Previous update was at 10:22 AM ----------

Code:
awk 'function time(r,s,t,x,y,z) 
      {str1=mktime("2011 1 1" FS r FS s FS t)
       str2=mktime("2011 1 1" FS x FS y FS z)
       diff=str1-str2 
       H=int(diff/3600)
       M=int((diff-H*3600)/60)
       S=diff-H*3600-M*60
       return sprintf ("%02d:%02d:%02d",H,M,S) 
      }
      /START/ {a[$1]=$3;file[$1]=$6;size[$1]=$8}
      /END/&&a[$1] { split(a[$1],b,":") 
                          split($3,c,":") 
                          c[1]=c[1]>=b[1]?c[1]:c[1]+24
                          printf "%-6s END %s %s %s File: %s size: %s File: 0 size: 0 \n", $1, $3,time(c[1],c[2],c[3],b[1],b[2],b[3]),$4,file[$1],size[$1]
                          delete a[$1]
                        }' sample.TXT

Name1  END 11:38:47 00:02:02 13102011 File: 2015443 size: 187654 File: 0 size: 0
Name2  END 11:40:21 00:00:29 13102011 File: 2008395 size: 28764 File: 0 size: 0
Name3  END 11:42:36 00:00:37 13102011 File: 1056987 size: 467321 File: 0 size: 0

# 3  
Old 10-13-2011
Thank you so much, no there was a typo in result actually it should be name1 (The same jobs would be running repeatedly as per predefined jobs other scripts) also can you suggest me how to rewrite so that i can get only in seconds rather HH:MM:SS i.e 120seconds if it is 00:02:00.

Quote:
Originally Posted by rdcwayx
I saw four Name1 , typo?

---------- Post updated at 10:32 AM ---------- Previous update was at 10:22 AM ----------

Code:
awk 'function time(r,s,t,x,y,z) 
      {str1=mktime("2011 1 1" FS r FS s FS t)
       str2=mktime("2011 1 1" FS x FS y FS z)
       diff=str1-str2 
       H=int(diff/3600)
       M=int((diff-H*3600)/60)
       S=diff-H*3600-M*60
       return sprintf ("%02d:%02d:%02d",H,M,S) 
      }
      /START/ {a[$1]=$3;file[$1]=$6;size[$1]=$8}
      /END/&&a[$1] { split(a[$1],b,":") 
                          split($3,c,":") 
                          c[1]=c[1]>=b[1]?c[1]:c[1]+24
                          printf "%-6s END %s %s %s File: %s size: %s File: 0 size: 0 \n", $1, $3,time(c[1],c[2],c[3],b[1],b[2],b[3]),$4,file[$1],size[$1]
                          delete a[$1]
                        }' sample.TXT

Name1  END 11:38:47 00:02:02 13102011 File: 2015443 size: 187654 File: 0 size: 0
Name2  END 11:40:21 00:00:29 13102011 File: 2008395 size: 28764 File: 0 size: 0
Name3  END 11:42:36 00:00:37 13102011 File: 1056987 size: 467321 File: 0 size: 0

# 4  
Old 10-13-2011
It will be more simple.
Code:
awk 'function time(r,s,t,x,y,z) 
      {str1=mktime("2011 1 1" FS r FS s FS t)
       str2=mktime("2011 1 1" FS x FS y FS z)
       return str1-str2   
      }
      /START/ {a[$1]=$3;file[$1]=$6;size[$1]=$8}
      /END/&&a[$1] { split(a[$1],b,":") 
                          split($3,c,":") 
                          c[1]=c[1]>=b[1]?c[1]:c[1]+24
                          printf "%-6s END %s %s second(s) %s File: %s size: %s File: 0 size: 0 \n", $1, $3,time(c[1],c[2],c[3],b[1],b[2],b[3]),$4,file[$1],size[$1]
                          delete a[$1]
                        }' sample.TXT

# 5  
Old 10-13-2011
SunOS 5.10 Generic_142900-15 sun4v sparc SUNW,T5240

how can i check the storage type being used in unix solaris sparc system?
please help me its urgnet..

thank you
# 6  
Old 10-14-2011
Thank you so much. So grateful for your help.

Quote:
Originally Posted by rdcwayx
It will be more simple.
Code:
awk 'function time(r,s,t,x,y,z) 
      {str1=mktime("2011 1 1" FS r FS s FS t)
       str2=mktime("2011 1 1" FS x FS y FS z)
       return str1-str2   
      }
      /START/ {a[$1]=$3;file[$1]=$6;size[$1]=$8}
      /END/&&a[$1] { split(a[$1],b,":") 
                          split($3,c,":") 
                          c[1]=c[1]>=b[1]?c[1]:c[1]+24
                          printf "%-6s END %s %s second(s) %s File: %s size: %s File: 0 size: 0 \n", $1, $3,time(c[1],c[2],c[3],b[1],b[2],b[3]),$4,file[$1],size[$1]
                          delete a[$1]
                        }' sample.TXT

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