I am having trouble sorting one file based on another file. I tried the grep -f function and failed. Basically what I have is two files that look like this:
File 1 (the list)
File 2 ( the file that needs to be sorted)
So file 1 is a list that has names in a particular order and I want to sort file 2 according to that order while also extracting the other columns.
So the end output would look like this.
Final file
Thanks
Phil
---------- Post updated at 03:30 PM ---------- Previous update was at 03:29 PM ----------
How to Sort files based on predefined values.?
Normally Sorting happens for the alphabetic or numberic orders..
Is there any way to sort a fields based on the Field values..?
Field10 has :
one
two
three
five
four
six
ten
seven
eight
nine.
in predefined order { one, two, three,... (2 Replies)
Hi ,,
i have the below file...
D 2342135
B 214236
C argjlksd
V lskjrghaklsr
C slkrgj
B sdg4tsd
E aslkgjlkasg
i want to sort the lines into different files based on the starting letter of the line. so that i have different files for lines starting with a letter.
thanks (1 Reply)
Hi,
I am having trouble sorting one file based on another file. I tried the grep -f function and failed. Basically what I have is two files that look like this:
File 1 (the list)
gh
aba
for
hmm
File 2 ( the file that needs to be sorted)
aba 2 4 6 7
for 2 4 7 4... (4 Replies)
Hi
I have a requirement like below
I need to sort the files based on the timestamp in the file name and run them in sorted order and then archive all the files which are one day old to temp directory
My files looks like this
PGABOLTXML1D_201108121235.xml... (1 Reply)
Right now there is no unix direct commad that can sort the files base on its name having numbers:
We can use the following:
In case your file name are like:
abc-UP018.zip
xyz-UP019.zip
ls *|sort -t'-' -k2 (2 Replies)
Hi,
I have a list of log files as follows:
name_date_0001_ID0.log
name_date_0001_ID2.log
name_date_0001_ID1.log
name_date_0002_ID2.log
name_date_0004_ID0.log
name_date_0005_ID0.log
name_date_0021_ID0.log
name_date_0025_ID0.log
.......................................... (4 Replies)
Experts
I have a list of files in the directory
mysample1
mysample2
mysample3
mysample4
mysample5
mysample6
mysample7
mysample8
mysample9
mysample10
mysample11
mysample12
mysample13
mysample14
mysample15 (4 Replies)
I want to extract dates from the files and i have different types of files with pattern. I have list file with the patterns and want to date extract based on it in a sh script
Files in the directory :
file1_20160101.txt
file2_20160101_abc.txt
filexyz20160101.txt
list file with... (2 Replies)
I have the below input data in a file and need to get the output as mentioned below. Need to sort the data by size(Asc/des)/by subdirectory
Below is the input which is there in a file:
120 /root/path2/part-00000-d3700305-428d-4b13-8161-42051f4ac5ed-c000.json
532 ... (3 Replies)
Discussion started by: ajarramuk
3 Replies
LEARN ABOUT PLAN9
zgrep
ZGREP(1) General Commands Manual ZGREP(1)NAME
zgrep - search possibly compressed files for a regular expression
SYNOPSIS
zgrep [ grep_options ] [ -e ] pattern filename...
DESCRIPTION
Zgrep invokes grep on compressed or gzipped files. These grep options will cause zgrep to terminate with an error code:
(-[drRzZ]|--di*|--exc*|--inc*|--rec*|--nu*). All other options specified are passed directly to grep. If no file is specified, then the
standard input is decompressed if necessary and fed to grep. Otherwise the given files are uncompressed if necessary and fed to grep.
If the GREP environment variable is set, zgrep uses it as the grep program to be invoked.
EXIT CODE
2 - An option that is not supported was specified.
AUTHOR
Charles Levert (charles@comm.polymtl.ca)
SEE ALSO grep(1), gzexe(1), gzip(1), zdiff(1), zforce(1), zmore(1), znew(1)ZGREP(1)