I have a text file (error log file) , which has occurences of an error message like
Code:
ORA-01652: unable to extend temp segment by 8 in tablespace xxxxx
There are around 3000 error messages like this in the error log file. But there are only 7 or 8 distinct tablespace names ( shown as xxxx above). The portion highlighted in red is same for all of the error messages. It is just the xxxx part that varies.
How can I get the distint occurences these error messages ?
Sample output of this file
Code:
SQL> create table vendor_dtl_clone1 tablespace TSTDATA
parallel (degree 4)
as select * from vendor_dtl; 2 3
create table vendor_dtl_clone1 tablespace tstdata
*
ERROR at line 1:
ORA-01652: unable to extend temp segment by 8 in tablespace TSTDATA
.
.
some text
.
some text
.
.
.
.
create table SHIP_TRACK_DTL
(
shp_track_dtl_id number(9,0),
shp_guide_id number(9,0),
orgn_zip varchar2(11),
shp_guide_type varchar2(2),
create_date_time date,
proc_stat_code number(2,0),
error_seq_nbr number(9,0)
)
tablespace brsdata;
create table SHIP_TRACK_DTL tablespace brsdata
*
ERROR at line 1:
ORA-01652: unable to extend temp segment by 8 in tablespace BRSDATA
I am a beginner to scripting, please help me in this regard.
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The table:
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The data:
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Hello,
I have a file which looks like:
%_SPE_RDP_NUM_ECH(7)*************%
%_SPE_RDP_NUM_ECH(70)************%
%_SPE_RDP_NUM_ECH(71)************%
%_SPE_RDP_NUM_ECH(72)************%
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From the below ps output , I want the distinct values of the third field (ie. I need the distinct PPIDs)
$ ps -ef
UID PID PPID C STIME TTY TIME CMD
root 1 0 0 Nov 10 - 48:49 /etc/init
root 1769576 1 0 Nov 10 - 0:07... (1 Reply)
Hi,
I have two files of the following format
file1
chr1:345-456
chr2:123-456
chr2:455-678
chr3:456-789
chr3:444-555
file2
chr1:345-456
chr2:123-456
chr3:456-789
output (2 Replies)
Platform :Oracle Linux 6.4
Shell : bash
The below file has 7 lines , some of them are duplicates. There are only 3 distinct lines. But why is the uniq command still showing 7 ?
I just want the distinct lines to be returned.
$ cat test.txt
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Hi All,
I am working on designing the archival process for my system, where I will have to find distinct file names ( when excluded time_stamp extention ) from given directory and for each file type keep the latest and move all other older to different location ( lets say dir Back ). Below are... (2 Replies)
Discussion started by: freakabhi
2 Replies
LEARN ABOUT DEBIAN
bup-margin
bup-margin(1) General Commands Manual bup-margin(1)NAME
bup-margin - figure out your deduplication safety margin
SYNOPSIS
bup margin [options...]
DESCRIPTION
bup margin iterates through all objects in your bup repository, calculating the largest number of prefix bits shared between any two
entries. This number, n, identifies the longest subset of SHA-1 you could use and still encounter a collision between your object ids.
For example, one system that was tested had a collection of 11 million objects (70 GB), and bup margin returned 45. That means a 46-bit
hash would be sufficient to avoid all collisions among that set of objects; each object in that repository could be uniquely identified by
its first 46 bits.
The number of bits needed seems to increase by about 1 or 2 for every doubling of the number of objects. Since SHA-1 hashes have 160 bits,
that leaves 115 bits of margin. Of course, because SHA-1 hashes are essentially random, it's theoretically possible to use many more bits
with far fewer objects.
If you're paranoid about the possibility of SHA-1 collisions, you can monitor your repository by running bup margin occasionally to see if
you're getting dangerously close to 160 bits.
OPTIONS --predict
Guess the offset into each index file where a particular object will appear, and report the maximum deviation of the correct answer
from the guess. This is potentially useful for tuning an interpolation search algorithm.
--ignore-midx
don't use .midx files, use only .idx files. This is only really useful when used with --predict.
EXAMPLE
$ bup margin
Reading indexes: 100.00% (1612581/1612581), done.
40
40 matching prefix bits
1.94 bits per doubling
120 bits (61.86 doublings) remaining
4.19338e+18 times larger is possible
Everyone on earth could have 625878182 data sets
like yours, all in one repository, and we would
expect 1 object collision.
$ bup margin --predict
PackIdxList: using 1 index.
Reading indexes: 100.00% (1612581/1612581), done.
915 of 1612581 (0.057%)
SEE ALSO bup-midx(1), bup-save(1)BUP
Part of the bup(1) suite.
AUTHORS
Avery Pennarun <apenwarr@gmail.com>.
Bup unknown-bup-margin(1)