尋ねる:問題の主張のために、今日は26/03/15だとします。過去 5 日間のデータをどのように「累積」できますか?
しかし、この日付は単なる例です。。理想的には、最終データを蓄積したいと思います。X数日、累積されたら、蓄積された情報をメールでお送りします。私はこのプロセスを一日にしますジ数週間、crontabを介して助けを得る。
コードは次のとおりです。
awk_variables=`echo "$line" | awk -F, '( $13 == "*OUT*" )||( $13 == "*IN*" ){print $1,$5,$10,$12,$13}' my_file.csv > /directory/some_directory/non_sorted_filtered.csv`
cat /directory/some_directory/non_sorted_filtered.csv | while read line
do
awk_variables="$line"
awk_variables_array=($awk_variables)
awk_time=${awk_variables_array[0]}
awk_container_ID=${awk_variables_array[1]}
awk_scan_count=${awk_variables_array[2]}
awk_part_number=${awk_variables_array[3]}
awk_direction==${awk_variables_array[4]}
awk_part_number_edited="${awk_part_number:3}"
awk_direction_edited="${awk_direction:1}"
awk_date=@''$awk_time
date=`date -d $awk_date +"%d/%m/%Y"`
echo -e "Date:$date Scan No:$awk_scan_count Part No:$awk_part_number_edited Direction:$awk_direction_edited"
done
出力は次のとおりです。
Date:19/03/2015 Scan No:12 Part No:XXXX1234567 Direction:*OUT*
Date:19/03/2015 Scan No:13 Part No:XXXX1234567 Direction:*OUT*
Date:19/03/2015 Scan No:14 Part No:XXXX1234567 Direction:*OUT*
Date:19/03/2015 Scan No:17 Part No:XXXX1234567 Direction:*OUT*
Date:19/03/2015 Scan No:18 Part No:WTTO275 Direction:*OUT*
Date:19/03/2015 Scan No:19 Part No:WTTO275 Direction:*OUT*
Date:19/03/2015 Scan No:20 Part No:WTTO275 Direction:*OUT*
Date:20/03/2015 Scan No:22 Part No:XXXX1234567 Direction:*OUT*
Date:20/03/2015 Scan No:23 Part No:XXXX1234567 Direction:*OUT*
Date:20/03/2015 Scan No:24 Part No:XXXX1234567 Direction:*OUT*
Date:20/03/2015 Scan No:25 Part No:XXXX1234567 Direction:*OUT*
Date:20/03/2015 Scan No:26 Part No:XXXX1234567 Direction:*OUT*
Date:24/03/2015 Scan No:37 Part No:WTTO523 Direction:*OUT*
Date:25/03/2015 Scan No:43 Part No:WTTO548 Direction:*OUT*
Date:25/03/2015 Scan No:44 Part No:SP-TMX6BP Direction:*OUT*
Date:25/03/2015 Scan No:45 Part No:WTTO548 Direction:*OUT*
Date:25/03/2015 Scan No:49 Part No:02102 Direction:*OUT*
Date:25/03/2015 Scan No:50 Part No:02103 Direction:*OUT*
Date:25/03/2015 Scan No:51 Part No:02118 Direction:*OUT*
Date:25/03/2015 Scan No:52 Part No:02132 Direction:*OUT*
Date:25/03/2015 Scan No:53 Part No:02133 Direction:*OUT*
Date:25/03/2015 Scan No:54 Part No:02134 Direction:*OUT*
Date:25/03/2015 Scan No:55 Part No:02135 Direction:*OUT*
Date:25/03/2015 Scan No:56 Part No:19178 Direction:*OUT*
Date:25/03/2015 Scan No:57 Part No:19179 Direction:*OUT*
だから、次のデータを送信したいと思います。
Date:20/03/2015 Scan No:22 Part No:XXXX1234567 Direction:*OUT*
Date:20/03/2015 Scan No:23 Part No:XXXX1234567 Direction:*OUT*
Date:20/03/2015 Scan No:24 Part No:XXXX1234567 Direction:*OUT*
Date:20/03/2015 Scan No:25 Part No:XXXX1234567 Direction:*OUT*
Date:20/03/2015 Scan No:26 Part No:XXXX1234567 Direction:*OUT*
Date:24/03/2015 Scan No:37 Part No:WTTO523 Direction:*OUT*
Date:25/03/2015 Scan No:43 Part No:WTTO548 Direction:*OUT*
Date:25/03/2015 Scan No:44 Part No:SP-TMX6BP Direction:*OUT*
Date:25/03/2015 Scan No:45 Part No:WTTO548 Direction:*OUT*
Date:25/03/2015 Scan No:49 Part No:02102 Direction:*OUT*
Date:25/03/2015 Scan No:50 Part No:02103 Direction:*OUT*
Date:25/03/2015 Scan No:51 Part No:02118 Direction:*OUT*
Date:25/03/2015 Scan No:52 Part No:02132 Direction:*OUT*
Date:25/03/2015 Scan No:53 Part No:02133 Direction:*OUT*
Date:25/03/2015 Scan No:54 Part No:02134 Direction:*OUT*
Date:25/03/2015 Scan No:55 Part No:02135 Direction:*OUT*
Date:25/03/2015 Scan No:56 Part No:19178 Direction:*OUT*
Date:25/03/2015 Scan No:57 Part No:19179 Direction:*OUT*
気づく:1日に検査する回数はさまざまです。時には100(たとえば)、時々ゼロです。
ベストアンサー1
これは解決策ではなくヒントに近いです。私が正しく覚えている場合は、ISO 8601
日付形式とstring compare
。次のようなものが役に立ちます。
iso_moment=$(date -u +"%Y-%m-%dT%H:%M:%SZ" -d "-5 days")
iso_date=$(date -d $awk_date +"%Y-%m-%dT%H:%M:%SZ")
if [[ "${iso_date}" > "${iso_moment}" ]]
then
echo -e "Date:$date Scan No:$awk_scan_count Part No:$awk_part_number_edited Direction:$awk_direction_edited"
fi