Sponsored Content
Top Forums UNIX for Dummies Questions & Answers bash pattern matching echo *[! '/' ] doesn't work Post 302080089 by james hanley on Saturday 15th of July 2006 08:43:58 PM
Old 07-15-2006
bash pattern matching echo *[! '/' ] doesn't work

without using ls, just using echo so purely pattern matching

I can say echo */ <-- lists directories

but how would I match files? surely something like *!/ or *[!/] but neither work ?

it seems like there isn't much that I can put in [ ] but surely i should be able to put any ascii character. but how?


thanks
 

10 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

Perl cmds doesn't work in Bash shell . Plz help

Hi everyone, I have a Linux OS in my PC (older version 9). Its default shell is bash. Whenever I try to run some Perl program it throws error ! eg, if I run this simple PERL program , #!/usr/bin/perl printf "\lHello \n"; $var=3 ; printf $var; @list=(1,2,3); printf "@list";... (6 Replies)
Discussion started by: adc22
6 Replies

2. Shell Programming and Scripting

echo $PWD doesn't work

I have entry in the my .profile like below, but still i see $PWD is not defied in my system export PS1=$LOGNAME@`hostname`':'$PWD'>' echo $PWD also gives me nothing, my env list also give no entry for PWD.Can someone help me setting PWD variable. I use /bin/sh (9 Replies)
Discussion started by: yesmani
9 Replies

3. Shell Programming and Scripting

shell script, echo doesn't work

#!/bin/sh something(){ echo "Inside something" echo $1 $2 } val=$(something "Hello " "world") Output expected: Inside somethingHello world But it's not echoing. (4 Replies)
Discussion started by: cola
4 Replies

4. Ubuntu

set completion-ignore-case on doesn't work in bash

completion-ignore-case option doesn't work in my version: /home/user $ echo $BASH_VERSION 3.2.48(1)-release /home/user $ ls -l * -rw-r--r-- 1 user user 0 2009-10-18 00:09 somefile -rw-r--r-- 1 user user 0 2009-10-18 00:09 Somefile /home/user $ set completion-ignore-case on But when I... (2 Replies)
Discussion started by: Sapfeer
2 Replies

5. HP-UX

pgrep doesn't perform full command line pattern matching

Hi! I need to get PID of some particular process and I wonder if I can use pgrep tool for this purpose. The problem is that pgrep doesn't perform pattern matching on the whole command line, even if I use -f key. Parsing output of ps command is not quite convenient... Also deamon, which PID I need... (2 Replies)
Discussion started by: Sapfeer
2 Replies

6. Shell Programming and Scripting

echo doesn't work right

Hi,when I run my first shell script,I got something that doesn't work right. I wrote this code in the script. echo -e "Hello,World\a\n"But the screen print like this: -e Hello,World The "-e" wasn't supposed to be printed out. Can anyone help me out?:wall: Many thanks!:) (25 Replies)
Discussion started by: Demon
25 Replies

7. Shell Programming and Scripting

pipe to grep doesn't work in bash script

Hi, I'm trying to write a script that checks gvfs to see if a mount exists so I can run it from network-manager's status hooks. I thought I'd pipe the output of gvfs-mount -l to grep for the particular mounts I care about. When I do this in a bash script: cmnd="gvfs-mount -l | grep -i... (4 Replies)
Discussion started by: kcstrom
4 Replies

8. Shell Programming and Scripting

Why regex pattern doesn't work in find?

$ find /opt/data_* -maxdepth 3 -type d -name "main*" 2> /dev/null /opt/data_025/maindblogs /opt/data_026/maindblogs /opt/data_027/maindblogs /opt/data_028/maindblogs $ find /opt/data_* -maxdepth 3 -type d -name "rvlogs*" 2> /dev/null /opt/data_002/prod/rvlogs2_archive... (4 Replies)
Discussion started by: urello
4 Replies

9. UNIX for Beginners Questions & Answers

Bash diff date doesn't work

Hi everyone, I've an issue trying to soustracte two dates, e.g: d1=$(date -d "Nov 18, 2017" +%s) d2=$(date +%s) # Today we are 2017-11-16 echo "$(( (d1 - d2) / 86400 ))" Output: 1 I don't understand why it doesn't work. for me, it should give "18 - 16 = 2". Much appreciated... (1 Reply)
Discussion started by: Arnaudh78
1 Replies

10. Shell Programming and Scripting

[Bash] passing variables to executable doesn't work

Bash version 4.4.20 / Ubuntu 16.0.4 Hello, I tried to write a script that gathers some data and passes them to an executable. The executed application answers with an error. The echo output in the script returns correct values. If I copy/paste the last echo command, it get's executed... (2 Replies)
Discussion started by: sushi2k7
2 Replies
DATEFMT_GET_PATTERN(3)							 1						    DATEFMT_GET_PATTERN(3)

IntlDateFormatter::getPattern - Get the pattern used for the IntlDateFormatter

	Object oriented style

SYNOPSIS
public string IntlDateFormatter::getPattern (void ) DESCRIPTION
Procedural style string datefmt_get_pattern (IntlDateFormatter $fmt) Get pattern used by the formatter. PARAMETERS
o $fmt - The formatter resource. RETURN VALUES
The pattern string being used to format/parse. EXAMPLES
Example #1 datefmt_get_pattern(3) example <?php $fmt = datefmt_create( 'en_US', IntlDateFormatter::FULL, IntlDateFormatter::FULL, 'America/Los_Angeles', IntlDateFormatter::GREGORIAN, 'MM/dd/yyyy' ); echo 'pattern of the formatter is : ' . datefmt_get_pattern($fmt); echo 'First Formatted output with pattern is ' . datefmt_format($fmt, 0); datefmt_set_pattern($fmt,'yyyymmdd hh:mm:ss z'); echo 'Now pattern of the formatter is : ' . datefmt_get_pattern($fmt); echo 'Second Formatted output with pattern is ' . datefmt_format($fmt, 0); ?> Example #2 OO example <?php $fmt = new IntlDateFormatter( 'en_US', IntlDateFormatter::FULL, IntlDateFormatter::FULL, 'America/Los_Angeles', IntlDateFormatter::GREGORIAN, 'MM/dd/yyyy' ); echo 'pattern of the formatter is : ' . $fmt->getPattern(); echo 'First Formatted output is ' . $fmt->format(0); $fmt->setPattern('yyyymmdd hh:mm:ss z'); echo 'Now pattern of the formatter is : ' . $fmt->getPattern(); echo 'Second Formatted output is ' . $fmt->format(0); ?> The above example will output: pattern of the formatter is : MM/dd/yyyy First Formatted output is 12/31/1969 Now pattern of the formatter is : yyyymmdd hh:mm:ss z Second Formatted output is 19690031 04:00:00 PST SEE ALSO
datefmt_set_pattern(3), datefmt_create(3). PHP Documentation Group DATEFMT_GET_PATTERN(3)
All times are GMT -4. The time now is 10:56 PM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy