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
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