如何在linux中获取小于/小于提供日期的数据

我有一个sample.txt文件，文件中的数据如下所示：

LINE 1|2017-02-19|LINE 3|LINE 4| LINE 5 | LINE 6
LINE 1|2017-02-06|LINE 3|LINE 4| LINE 5 | LINE 6
LINE 1|2017-03-06|LINE 3|LINE 4| LINE 5 | LINE 6
LINE 1|2017-02-07|LINE 3|LINE 4| LINE 5 | LINE 6
LINE 1|2017-03-25|LINE 3|LINE 4| LINE 5 | LINE 6
LINE 1|2017-02-06|LINE 3|LINE 4| LINE 5 | LINE 6

现在我想从sample.txt文件中获取数据，其中

2017-02-19
2017-02-06
2017-03-06
2017-02-07    
2017-03-25
2017-02-06

            > 2017-02-19

所以我的输出应该是

LINE 1|2017-03-06|LINE 3|LINE 4| LINE 5 | LINE 6
LINE 1|2017-03-25|LINE 3|LINE 4| LINE 5 | LINE 6

使用GNU和 将FS设置为| 日期-d[date]+[OUTPUT_FORMAT]…将字段2$2格式化为%s之后的秒数，并将其传递到变量d Awk将测试d是否小于2017-02-19，其自历元起的秒数为1487480400，如果为真，将打印整个记录$0 密码 注意：通过更改“打印$0”，您只能打印某些字段；要打印$1；，打印$1、$2；，甚至打印出你想要的任何东西，报价为2美元

查找自给定日期起的秒数

---

例如，您可以使用命令行中的date-d2017-02-19+%s查找给定日期后的秒数。

逐行读取文件， 隔离日期， 比较格式化为数字的日期， 跳过该行，或打印该行

#!/bin/ksh

compareDay=$( date -d '2017-02-19' "+%Y%j" )    # build a number from <year><day of year>
cat sample.txt | while read line; do
    day="${line#*\|}"           # remove from the beginning to first |
    day="${day%%\|*}"           # remove from the end (%%=all) up from |
    # compare, skip if condition not met
    [[ $( date -d $day "+%Y%j" ) -gt $compareDay ]] || continue
    echo "$line"                # output the remaining line meeting the condition
done

可能类似于awk'$0>第1行| 2017-02-20'sample.txt

#!/bin/ksh

compareDay=$( date -d '2017-02-19' "+%Y%j" )    # build a number from <year><day of year>
cat sample.txt | while read line; do
    day="${line#*\|}"           # remove from the beginning to first |
    day="${day%%\|*}"           # remove from the end (%%=all) up from |
    # compare, skip if condition not met
    [[ $( date -d $day "+%Y%j" ) -gt $compareDay ]] || continue
    echo "$line"                # output the remaining line meeting the condition
done

cat sample.txt | while read line; do
    [[ $line > 'LINE 1|2017-02-20' ]] || continue
    echo "$line"
done