1:56 p.m.
I've hit a problem I can't quite figure out which a bash script I'm
writing. I'm trying to copy backup files in the format
2013-August-18--1123.zip to an NFS share. I want to have the script copy
the file with just the date. In bash I've setup vars that get the current
date:
# Date variables
log_year=`date "+%Y"`
log_month=`date "+%B"`
log_day=`date "+%d"`
# Filename format YYYY_MM_DD--HHMM.zip
filename=$log_year"-"$log_month"-"$log_day"--"/*".zip"
The problem is I don't really care about the stuff after the '--'. I.e.
from the CLI I'd just 'ls 2013-August-18--*.zip' to get all the files with
that date in the file name. How can I do that in a bash script?
--
Caesar si viveret, ad remum dareris.
Mark Haney
Software Developer/Consultant
mark.haney(a)gmail.com
8:33 a.m.
On 08/19/2013 08:56 PM, Mark Haney wrote:
I've hit a problem I can't quite figure out which a bash
script I'm writing. I'm trying to copy backup files in the format
2013-August-18--1123.zip to an NFS share. I want to have the script copy the file with
just the date. In bash I've setup vars that get the current date:
# Date variables
log_year=`date "+%Y"`
log_month=`date "+%B"`
log_day=`date "+%d"`
Partially offtopic hint:
avoid multiple date invocations, you can get wrong results at the midnight
of the last day of the month
(and it's much worse when doing hours,minutes,seconds).
Just do:
log_ymd=`date +%Y-%B-%d`
--
Roberto Ragusa mail at robertoragusa.it
6:54 a.m.
That's what I missed. I used the wrong single quotes ' instead of `. Git.
Should've caught that right off. It's been a while since I had a chance to
script anything in bash, since I've worked mostly in Windows the last year
or so and I'm rusty. Thanks.
On Aug 19, 2013 6:18 PM, "Bill Oliver" <vendor(a)billoblog.com> wrote:
On Mon, 19 Aug 2013, Mark Haney wrote:
I've hit a problem I can't quite figure out which a bash script I'm
> writing. I'm trying to copy backup files in the format
> 2013-August-18--1123.zip to an NFS share.
> I want to have the script copy the file with just the date. In bash
> I've setup vars that get the current date:
> # Date variables
> log_year=`date "+%Y"`
> log_month=`date "+%B"`
> log_day=`date "+%d"`
>
> # Filename format YYYY_MM_DD--HHMM.zip
>
filename=$log_year"-"$log_**month"-"$log_day"--"/*".zip"
>
> The problem is I don't really care about the stuff after the '--'. I.e.
> from the CLI I'd just 'ls 2013-August-18--*.zip' to get all the files
with
> that date in the
> file name. How can I do that in a bash script?
>
>
> --
> Caesar si viveret, ad remum dareris.
>
> Mark Haney
> Software Developer/Consultant
> mark.haney(a)gmail.com
>
>
>
You mean something like this?
************************
#!/bin/sh
# Date variables
log_year=`date "+%Y"`
log_month=`date "+%B"`
log_day=`date "+%d"`
# Filename format YYYY_MM_DD--HHMM.zip
filename="$log_year-$log_**month-$log_day--*.zip"
filelist=`ls $filename`
echo $filelist
***************************
billo
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4:33 a.m.
hello mark,
On 08/19/2013 01:56 PM, Mark Haney wrote:
I've hit a problem I can't quite figure out which a bash
script I'm
writing. I'm trying to copy backup files in the format
2013-August-18--1123.zip to an NFS share. I want to have the script copy
the file with just the date. In bash I've setup vars that get the current
date:
# Date variables
log_year=`date "+%Y"`
log_month=`date "+%B"`
log_day=`date "+%d"`
# Filename format YYYY_MM_DD--HHMM.zip
filename=$log_year"-"$log_month"-"$log_day"--"/*".zip"
why are you showing "/*" in your "filename=" statement?
such would indicate date as a directory and _all_ files in a directory that
ends with "zip".
if you meant to _delimit_ the "*", then "\" is used to delimit.
i believe actual 'filename' should be;
$log_year"-"$log_month"-"$log_day"*.zip"
The problem is I don't really care about the stuff after the
'--'.
you better care about it as it is written, as, for what you want, 'filename'
would be as i present.
I.e. from the CLI I'd just 'ls 2013-August-18--*.zip' to
get all the
files with that date in the file name. How can I do that in a bash script?
not if you use the "/*".
problem just might be that you have worked with it too long and not
really seeing what you are doing. ;=)
ls 2013-August-18--*.zip
should show _all_ files for 18 august, 2013, regardless of time, in
'current' directory.
if not, something is bad wrong.
when i time stamp a backup file, i use;
yyyy-mmdd-hhmm-filename
or
filename-yyyy-mmdd-hhmm
then, when i collect them for transfer to backup tape media, i use;
find . yyyy-mmdd-* | sort | cpio -o -[arguments] /dev/tape
or
find . *-yyyy-mmdd | sort | cpio -o -[arguments] /dev/tape
which finds _all_ files of *-yyyymmdd or yyyymmdd-* and they are backed up.
[and yes, i use *yyyymmdd* for various reasons]
not that much difference if you are using multi "-".
from 1 of my script files;
find $1 -depth | sort | cpio -o -Bcmv > [path to backup file]$2.cpio
and then backup the *.cpio file.
"$1" and "$2" are passed to script from command line.
i have a rather extensive collection of script files that i use for tape
backup that i have used for years for both argument passing and no argument
passing for regularly preformed backups with out problems.
i can not access the extensive list of "find|sort|cpio" at this time.
see attached for a short list. it should give you an idea of using find,
sort, and cpio.
hth.
--
peace out.
in a world with out fences, who needs gates.
sl6.3 linux
tc.hago.
g
.
2:29 a.m.
Hi Bill,
On Mon, Aug 19, 2013 at 07:56:03PM -0500, Bill Oliver wrote:
On Tue, 20 Aug 2013, Suvayu Ali wrote:
>On Mon, Aug 19, 2013 at 05:18:40PM -0500, Bill Oliver wrote:
>>
>># Filename format YYYY_MM_DD--HHMM.zip
>>filename="$log_year-$log_month-$log_day--*.zip"
>>
>>filelist=`ls $filename`
>
>Please don't do this. A much better solution is:
>
>filelist=($log_year-$log_month-$log_day--*.zip)
>
>See: <http://mywiki.wooledge.org/ParsingLs>
>
[...chomp...chomp...chomp...]
# Filename format YYYY_MM_DD--HHMM.zip
filelist=($log_year-$log_month-$log_day--*.zip)
******************
I get:
$ ./jnk2.sh
2013-August-19--a.zip
For some reason, the second file (013-August-19--ba.zip) is not returned...
My fault, I should have been more verbose.
The var=(..) creates an array variable; whereas the var=$(..) construct
runs a sub-shell and stores the output as a string (the last one is
equivalent to var=`..`).
So with my example you can get your list back by using the array.
$ echo ${filelist[@]}
Putting files in an array also lets you loop over them easily.
for file in ${filelist[@]};
do
some_cmd;
done
(or in oneline: for file in ${filelist[@]}; do some_cmd; done)
As you can see to access the whole array, I use the [@] index. You can
also use [*] there. There is a subtle difference between the two when
in side double quotes, otherwise they are equivalent. You can read all
about it in the bash info page under array (info "(bash) Arrays").
To access individual files in the array you can use: ${filelist[1]}
That said, I do not think storing the files in a variable is necessary
at all. This is how I would do it.
for file in $log_year-$log_month-$log_day--*.zip;
do
some_cmd;
done
Simple and easy!
Hope this helps,
--
Suvayu
Open source is the future. It sets us free.
11:46 a.m.
hello reberto,
On 08/20/2013 08:33 AM, Roberto Ragusa wrote:
On 08/19/2013 08:56 PM, Mark Haney wrote:
> I've hit a problem I can't quite figure out which a bash script I'm
writing. I'm trying to copy backup files in the format 2013-August-18--1123.zip to an
NFS share. I want to have the script copy the file with just the date. In bash I've
setup vars that get the current date:
>
> # Date variables
> log_year=`date "+%Y"`
> log_month=`date "+%B"`
> log_day=`date "+%d"`
Partially offtopic hint:
avoid multiple date invocations, you can get wrong results at the midnight
of the last day of the month
(and it's much worse when doing hours,minutes,seconds).
Just do:
log_ymd=`date +%Y-%B-%d`
or, op can use date form as;
find $1 -depth | sort | cpio -o -Bcv > \
/home/users/hold/0000.0001.cpio2file/$2.cpio 2> \
/home/users/hold/0000.0001.cpio2file/$2.err-list.`date +%Y.%m%d.%H%M.%S`
to sort by month's in numerical and not all mixed up, which is why i
prefer using number for month.
--
peace out.
in a world with out fences, who needs gates.
sl6.3 linux
tc.hago.
g
.
2:24 p.m.
On 08/19/2013 12:15 PM, Mark Haney wrote:
How can I do that in a bash script. I can probably use find, but
this
seems much simpler and/or quicker than finding one file.
Are you looking for files with today's date, or with a specific date?
In the former case, you can use the date command to put today's date
into a variable and dissect it; in the latter, you can put the desired
date, in the proper format, on the command line. And, if you're
comfortable with perl, you might want to use it for the date manipulation.
2:15 p.m.
On Mon, Aug 19, 2013 at 1:56 PM, Mark Haney <mark.haney(a)gmail.com> wrote:
I've hit a problem I can't quite figure out which a bash
script I'm writing.
I'm trying to copy backup files in the format 2013-August-18--1123.zip to an
NFS share. I want to have the script copy the file with just the date. In
bash I've setup vars that get the current date:
# Date variables
log_year=`date "+%Y"`
log_month=`date "+%B"`
log_day=`date "+%d"`
# Filename format YYYY_MM_DD--HHMM.zip
filename=$log_year"-"$log_month"-"$log_day"--"/*".zip"
The problem is I don't really care about the stuff after the '--'. I.e. from
the CLI I'd just 'ls 2013-August-18--*.zip' to get all the files with that
date in the file name. How can I do that in a bash script?
If you just want the list you can do basically the same thing you do
from the CLI.
ls $log_year"-"$log_month"-"$log_day"--*.zip
If you want to act on each file individually then maybe a loop.
Assuming your script is running in the same directory as the files
something like this maybe?
for f in $log_year"-"$log_month"-"$log_day"--*.zip; do
// do whatever you want with $f here
done
John
2:15 p.m.
Let me try again. I want to copy a file with a filename like
2013-August-18--1123.zip to another folder. However, because the last 4
digits of the filename won't always be the same, I want to be able to
search in a bash script for '2013-August-18--*.zip'. I.e. match all files
with '2013-August-18--' in the filename. From a console I can do this by:
ls 2013-August-18--*.zip
How can I do that in a bash script. I can probably use find, but this
seems much simpler and/or quicker than finding one file.
Does that make more sense?
On Mon, Aug 19, 2013 at 3:03 PM, Joe Zeff <joe(a)zeff.us> wrote:
On 08/19/2013 11:56 AM, Mark Haney wrote:
>
> # Filename format YYYY_MM_DD--HHMM.zip
>
filename=$log_year"-"$log_**month"-"$log_day"--"/*".zip"
>
I think that's supposed to be "\*" but ICBW.
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--
Caesar si viveret, ad remum dareris.
Mark Haney
Software Developer/Consultant
mark.haney(a)gmail.com
2:01 p.m.
Around 07:56pm on Monday, August 19, 2013 (UK time), Mark Haney scrawled:
I've hit a problem I can't quite figure out which a bash
script I'm
writing. I'm trying to copy backup files in the format
2013-August-18--1123.zip to an NFS share. I want to have the script copy
the file with just the date. In bash I've setup vars that get the current
date:
# Date variables
log_year=`date "+%Y"`
log_month=`date "+%B"`
log_day=`date "+%d"`
# Filename format YYYY_MM_DD--HHMM.zip
filename=$log_year"-"$log_month"-"$log_day"--"/*".zip"
The problem is I don't really care about the stuff after the '--'. I.e.
from the CLI I'd just 'ls 2013-August-18--*.zip' to get all the files with
that date in the file name. How can I do that in a bash script?
Not sure I totally understand what the problem is, but I think if I were
doing what I think you may be trying to do, I'd consider using the find
command.
Steve
--
Website: www.stevesearle.com
19:58:18 up 38 days, 23:30, 1 user, load average: 0.00, 0.01, 0.00
7:56 p.m.
On Tue, 20 Aug 2013, Suvayu Ali wrote:
On Mon, Aug 19, 2013 at 05:18:40PM -0500, Bill Oliver wrote:
>
> # Filename format YYYY_MM_DD--HHMM.zip
> filename="$log_year-$log_month-$log_day--*.zip"
>
> filelist=`ls $filename`
Please don't do this. A much better solution is:
filelist=($log_year-$log_month-$log_day--*.zip)
See: <http://mywiki.wooledge.org/ParsingLs>
Thanks for the comment -- I hadn't thought of that. However, it doesn't
work and it doesn't make sense to me. I thought that putting things in
parentheses simply put it in a shell.
In any case, here's my test case. I created two files:
2013-August-19--a.zip
and
2013-August-19--ba.zip
Then, using this script:
******************
#!/bin/sh
# Date variables
log_year=`date "+%Y"`
log_month=`date "+%B"`
log_day=`date "+%d"`
# Filename format YYYY_MM_DD--HHMM.zip
filename="$log_year-$log_month-$log_day--*.zip"
filelist=`ls $filename`
echo $filelist
******************
I get the following output:
$ !./j
./jnk.sh
2013-August-19--a.zip 2013-August-19--ba.zip
Using this script:
******************
#!/bin/sh
# Date variables
log_year=`date "+%Y"`
log_month=`date "+%B"`
log_day=`date "+%d"`
# Filename format YYYY_MM_DD--HHMM.zip
filelist=($log_year-$log_month-$log_day--*.zip)
******************
I get:
$ ./jnk2.sh
2013-August-19--a.zip
For some reason, the second file (013-August-19--ba.zip) is not returned...
did you mean
filelist=$(ls log_year-$log_month-$log_day--*.zip)
? ... but that puts the evil "ls" back in. The complaint against
"ls"
is that it breaks on wacky file names. That isn't an issue here.
billo
5:48 p.m.
On Mon, Aug 19, 2013 at 05:18:40PM -0500, Bill Oliver wrote:
# Filename format YYYY_MM_DD--HHMM.zip
filename="$log_year-$log_month-$log_day--*.zip"
filelist=`ls $filename`
Please don't do this. A much better solution is:
filelist=($log_year-$log_month-$log_day--*.zip)
See: <http://mywiki.wooledge.org/ParsingLs>
--
Suvayu
Open source is the future. It sets us free.
5:18 p.m.
On Mon, 19 Aug 2013, Mark Haney wrote:
I've hit a problem I can't quite figure out which a bash
script I'm writing. I'm trying to copy backup files in the format
2013-August-18--1123.zip to an NFS share.
I want to have the script copy the file with just the date. In bash I've setup vars
that get the current date:
# Date variables
log_year=`date "+%Y"`
log_month=`date "+%B"`
log_day=`date "+%d"`
# Filename format YYYY_MM_DD--HHMM.zip
filename=$log_year"-"$log_month"-"$log_day"--"/*".zip"
The problem is I don't really care about the stuff after the '--'. I.e. from
the CLI I'd just 'ls 2013-August-18--*.zip' to get all the files with that
date in the
file name. How can I do that in a bash script?
--
Caesar si viveret, ad remum dareris.
Mark Haney
Software Developer/Consultant
mark.haney(a)gmail.com
You mean something like this?
************************
#!/bin/sh
# Date variables
log_year=`date "+%Y"`
log_month=`date "+%B"`
log_day=`date "+%d"`
# Filename format YYYY_MM_DD--HHMM.zip
filename="$log_year-$log_month-$log_day--*.zip"
filelist=`ls $filename`
echo $filelist
***************************
billo
5:02 p.m.
On 19Aug2013 15:15, Mark Haney <mark.haney(a)gmail.com> wrote:
| Let me try again. I want to copy a file with a filename like
| 2013-August-18--1123.zip to another folder. However, because the last 4
| digits of the filename won't always be the same, I want to be able to
| search in a bash script for '2013-August-18--*.zip'. I.e. match all files
| with '2013-August-18--' in the filename. From a console I can do this by:
|
| ls 2013-August-18--*.zip
|
| How can I do that in a bash script. I can probably use find, but this
| seems much simpler and/or quicker than finding one file.
Your console is almost certainly running a bash already. If it works there,
it should work in a script.
Look (untested):
datepart=`date +%Y-%B-%d`
for match in "$datepart--"*.zip
do
[ -f "$match" ] || continue # if no match we get a bare "*"
... do stuff with the file "$match" ...
done
What about the above doesn't work for you, or isn't what you want?
Cheers,
--
Cameron Simpson <cs(a)zip.com.au>
3:08 p.m.
On 08/19/2013 12:51 PM, Mark Haney wrote:
I can (and am) using the date command in a variable. But, the
filename
is not just backup-08-18-2013.zip, which isn't a problem to do. The
problem is the filename is variable /outside/ the date. For example:
backup-08-18-2013--1215.zip. I can VAR the date parts, but that '1215'
part I need to wildcard. Does that make more sense?
I wasn't sure how you were getting the date. I was, however, clear that
there were parts of the filename that both varied and weren't related to
the date. And, by the way you reply, putting the date in the correct
format on the command line isn't an option. (If you're worried about
getting the format wrong, or having somebody else use the script and
fscking it up, I don't blame you; I'd want to do the parsing inside the
script too.)
2:51 p.m.
I can (and am) using the date command in a variable. But, the filename is
not just backup-08-18-2013.zip, which isn't a problem to do. The problem
is the filename is variable /outside/ the date. For example:
backup-08-18-2013--1215.zip. I can VAR the date parts, but that '1215'
part I need to wildcard. Does that make more sense?
It's really a non-issue now, I've just used find, since it expands
wildcards without a lot of effort involved.
Thanks, guys. (NO sarcasm intended.)
On Mon, Aug 19, 2013 at 3:24 PM, Joe Zeff <joe(a)zeff.us> wrote:
On 08/19/2013 12:15 PM, Mark Haney wrote:
> How can I do that in a bash script. I can probably use find, but this
> seems much simpler and/or quicker than finding one file.
>
Are you looking for files with today's date, or with a specific date? In
the former case, you can use the date command to put today's date into a
variable and dissect it; in the latter, you can put the desired date, in
the proper format, on the command line. And, if you're comfortable with
perl, you might want to use it for the date manipulation.
--
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--
Caesar si viveret, ad remum dareris.
Mark Haney
Software Developer/Consultant
mark.haney(a)gmail.com
3924
days inactive
3925
days old
17 comments
9 participants
participants (9)
-
Cameron Simpson -
g -
inode0 -
Joe Zeff -
Mark Haney -
Roberto Ragusa -
Steve Searle -
Suvayu Ali -
William Oliver