Sponsored Content
Operating Systems Solaris Regular Expression in Date ls command Post 302873403 by capnpepper on Tuesday 12th of November 2013 04:50:09 AM
Old 11-12-2013
Try this

Code:
ls -lrt | grep -E amey_in[0-9]{8}.csv

 

10 More Discussions You Might Find Interesting

1. UNIX for Dummies Questions & Answers

using regular expression for directories in find command

Hi, I want to find the files available in a directory /var/user/*/*/data/. I tried using the command "find /var/user/ -path '*/*/data/ -name '*' -type f" it says find: 0652-017 -path is not a valid option and then i tried using "find /var/user/ -name '*/*/data/*' -type f" but its not... (3 Replies)
Discussion started by: vinothbabu12
3 Replies

2. Shell Programming and Scripting

perl regular expression format date

Hi Everyone, Mon 18 Jan 2010 09:52:10 AM MYT the output is 20100118 09:52:10 how to do it in perl regular expression =~ Thanks (3 Replies)
Discussion started by: jimmy_y
3 Replies

3. Shell Programming and Scripting

validate date pattern using Regular Expression

Hi, i am java guy and new to unix. I want to validate date pattern using Regex expression here is the sample program i have written. #!/bin/sh checkDate="2010-04-09" regex="\\d{4}-\\d{2}-\\d{2}\$" echo $regex if ] then echo "OK" else echo "not OK" fi But the ouput is... (2 Replies)
Discussion started by: vvenu88
2 Replies

4. Shell Programming and Scripting

Integer expression expected: with regular expression

CA_RELEASE has a value of 6. I need to check if that this is a numeric value. if not error. source $CA_VERSION_DATA if * ] then echo "CA_RELESE $CA_RELEASE is invalid" exit -1 fi + source /etc/ncgl/ca_version_data ++ CA_PRODUCT_ID=samxts ++ CA_RELEASE=6 ++ CA_WEEK_NO=7 ++... (3 Replies)
Discussion started by: ketkee1985
3 Replies

5. Shell Programming and Scripting

Required help in perl regular expression substitution for this date format

Hi, I have written a small perl script to handle particular date format using perl, but it is not substituting the whole string. Can some one please check on what is the issue with the code. $_ = "Date: November 25, 2010 09:02:01 PM";... (1 Reply)
Discussion started by: sarbjit
1 Replies

6. Programming

Perl: How to read from a file, do regular expression and then replace the found regular expression

Hi all, How am I read a file, find the match regular expression and overwrite to the same files. open DESTINATION_FILE, "<tmptravl.dat" or die "tmptravl.dat"; open NEW_DESTINATION_FILE, ">new_tmptravl.dat" or die "new_tmptravl.dat"; while (<DESTINATION_FILE>) { # print... (1 Reply)
Discussion started by: jessy83
1 Replies

7. Shell Programming and Scripting

Regular Expression in Find command [KSH]

Hello, I am trying to use regex wtih find command in KSH. For some reason it is not working as expected. Input: comm_000_abc_0102.c comm_000_abc.c 456_000_abc_1212.cpp 456_000_abc_.cpp Expected Output: comm_000_abc_0102.c kkm_000_abc_8888.cpp (Basically I want to find all... (6 Replies)
Discussion started by: vinay4889
6 Replies

8. UNIX for Advanced & Expert Users

sed: -e expression #1, char 0: no previous regular expression

Hello All, I'm trying to extract the lines between two consecutive elements of an array from a file. My array looks like: problem_arr=(PRS111 PRS213 PRS234) j=0 while } ] do k=`expr $j + 1` sed -n "/${problem_arr}/,/${problem_arr}/p" problemid.txt ---some operation goes... (11 Replies)
Discussion started by: InduInduIndu
11 Replies

9. Shell Programming and Scripting

Regular expression in grep command

Here is the content of a file: abcdefgh 1234 When I do: grep a?c <file> I expect the output to show "abcdefgh". Its not happening. Any ideas? "a?c" should mean either ac or c. This should mean the first line is a match. Yet its not happening. I have tried with -e option in grep, with... (1 Reply)
Discussion started by: Rameshck
1 Replies

10. Shell Programming and Scripting

Grep command to search a regular expression in a line an only print the string after the match

Hello, one step in a shell script i am writing, involves Grep command to search a regular expression in a line an only print the string after the match an example line is below /logs/GRAS/LGT/applogs/lgt-2016-08-24/2016-08-24.8.log.zip:2016-08-24 19:12:48,602 ERROR... (9 Replies)
Discussion started by: Ramneekgupta91
9 Replies
REGEX(3)						     Library Functions Manual							  REGEX(3)

NAME
re_comp, re_exec - regular expression handler SYNOPSIS
char *re_comp(s) char *s; re_exec(s) char *s; DESCRIPTION
Re_comp compiles a string into an internal form suitable for pattern matching. Re_exec checks the argument string against the last string passed to re_comp. Re_comp returns 0 if the string s was compiled successfully; otherwise a string containing an error message is returned. If re_comp is passed 0 or a null string, it returns without changing the currently compiled regular expression. Re_exec returns 1 if the string s matches the last compiled regular expression, 0 if the string s failed to match the last compiled regular expression, and -1 if the compiled regular expression was invalid (indicating an internal error). The strings passed to both re_comp and re_exec may have trailing or embedded newline characters; they are terminated by nulls. The regular expressions recognized are described in the manual entry for ed(1), given the above difference. SEE ALSO
ed(1), ex(1), egrep(1), fgrep(1), grep(1) DIAGNOSTICS
Re_exec returns -1 for an internal error. Re_comp returns one of the following strings if an error occurs: No previous regular expression, Regular expression too long, unmatched (, missing ], too many () pairs, unmatched ). 3rd Berkeley Distribution May 15, 1985 REGEX(3)
All times are GMT -4. The time now is 10:29 AM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy