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Copy(append ) the contents of file1 and file2 to file3

 
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# 1  
Old 12-10-2014
Copy(append ) the contents of file1 and file2 to file3
Platform : Oracle linux 6.5

I have two log files with the following contents

Code:
[root@tblmanh236 test]# ls -l
total 8
-rw-r--r--. 1 root root 75 Dec 10 20:55 myLogfile1.log
-rw-r--r--. 1 root root 51 Dec 10 20:57 myLogfile2.log

[root@tblmanh236 test]#
[root@tblmanh236 test]# cat myLogfile1.log
hello world
jaded zombies acted quaintly but kept driving the oxen forward

[root@tblmanh236 test]#
[root@tblmanh236 test]#
[root@tblmanh236 test]#
[root@tblmanh236 test]#
[root@tblmanh236 test]# cat myLogfile2.log
The quick brown fox jumped over the lazy dog
hello
[root@tblmanh236 test]#

I want the contents of myLogfile1.log and myLogfile2.log to be copied(appended) to a single logfile called named mySingleLogfile.log .
How can I do this ?
# 2  
Old 12-10-2014
Hello kraljic,

You can try following for same.
Code:
cat myLogfile1.log myLogfile2.log > Output_file

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
kraljic
# 3  
Old 12-10-2014
In this specific case (only the required two files are in the directory)
Code:
cat * > mySingleLogfile.log

should do the thing. If there were other files than the two you want to merge, then RavinderSingh13's approach might be the right thing.

I think it could be even abbreviated like this
Code:
cat myLogfile[12].log > mySingleLogfile.log

Up to 9 files cat myLogfile*.log > mySingleLogfile.log should work too, but if you had, say 15 logfiles, e.g. myLogfile1.log myLogfile2.log .. myLogfile15.log, then cat myLogfile*.log > output.log will likely produce wrong output (Wrong order! Compare the output of ls myLogfile*.log and ls myLogfile*.log | sort -V) .... In this case you can try this, provided your sort supports the -V option:
Code:
ls myLogfile*.log | sort -V | xargs -I{} cat {} > mySingleLogfile.log

Hope this helps.
This User Gave Thanks to junior-helper For This Post:
kraljic
# 4  
Old 12-10-2014
Thank You Ravinder, Junior-Helper

cat myLogfile[12].log > mySingleLogfile.log solution suits my requirements.

In my case, the logfile names are assigned to variables first
Code:
[root@tblmanh236 test]# export LOG1=myLogfile1.log
[root@tblmanh236 test]# export LOG2=myLogfile2.log
[root@tblmanh236 test]# export LOG3=myLogfile3.log

And the below works well



Code:



[root@tblmanh236 test]# cat $LOG1 $LOG2 $LOG3 > mySingleLogfile.log
[root@tblmanh236 test]#

But, When I am trying to do the equivalent of myLogfile[12].log > mySingleLogfile.log with variables, I am getting the following error because of the [ ] brackets around 123 , I guess

Code:
[root@tblmanh236 test]# cat $LOG[123] > mySingleLogfile.log
cat: [123]: No such file or directory
[root@tblmanh236 test]#
[root@tblmanh236 test]# cat '$LOG[123]' > mySingleLogfile.log
cat: $LOG[123]: No such file or directory
[root@tblmanh236 test]#
[root@tblmanh236 test]#
[root@tblmanh236 test]# cat "$LOG[123]" > mySingleLogfile.log
cat: [123]: No such file or directory
[root@tblmanh236 test]#

## Escaping the brackets didn't work either
[root@tblmanh236 test]# cat $LOG\[123\] > mySingleLogfile.log
cat: [123]: No such file or directory

How can I get this working ?
# 5  
Old 12-10-2014
Quote:
[root@tblmanh236 test]# cat $LOG[123] > mySingleLogfile.log
cat: [123]: No such file or directory
I am getting the following error because of the [ ] brackets around 123 , I guess
No. Let me explain that: It seems like you exported variables LOG1 LOG2 and LOG3, but here you are "catting" $LOG[123]. Most likely, LOG is not defined at all and hence empty, and only [123] remains and cat cannot find a file with name [123]. The shell treated it like a literal filename [123] because the shell was not able to expand it to something appropriate. [123] alone would match only files with name 1 2 and 3

One potential solution:
Code:
export LOG=myLogfile
cat $LOG[123].log > mySingleLogfile.log

Is it viable?

Edit: This should work too:
Code:
export LOG=myLogfile[123].log
cat $LOG > mySingleLogfile.log

Last edited by junior-helper; 12-10-2014 at 01:29 PM..
This User Gave Thanks to junior-helper For This Post:
kraljic
# 6  
Old 12-10-2014
Thank You. It is viable. I have multiple log file variables like LOG1 , LOG2, ... declared in my scripts, I guess I will have to change it to what you've suggested If there is no other way

Code:
export LOG=myLogfile
cat $LOG[123].log > mySingleLogfile.log

You are not that junior. Are you ? Smilie
# 7  
Old 12-10-2014
Quote:
Originally Posted by kraljic
I have multiple log file variables like LOG1 , LOG2, ... declared in my scripts, I guess I will have to change it to what you've suggested If there is no other way
Not sure, I don't know you script and don't know how you want to handle it exactly, but if you wanted to cat files already stored in LOG1 LOG2 and LOG3, (without doing the stunt with LOG I mentioned before), with bash you could try this:
Code:
for i in {1..3}; do
 x=LOG$i
 cat ${!x} >>mySingleLogfile.log
done

I hope this doesn't complicate the situation further Smilie
Quote:
You are not that junior. Are you ? Smilie
Smilie

Edit: Yet another option/variant:
Code:
for i in $LOG1 $LOG2 $LOG3; do
 cat $i >>mySingleLogfile.log
done

Last edited by junior-helper; 12-10-2014 at 02:18 PM..
This User Gave Thanks to junior-helper For This Post:
kraljic
 
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