验证 grep 中的特定列答案

【问题标题】：Validating specific column in grep验证 grep 中的特定列
【发布时间】：2020-12-23 18:37:14
【问题描述】：

好吧，这快把我逼疯了。我有一个包含以下内容的文本文件：

"1","2","3","4","text","2020-01-01","2020-12-13","4"
"1","2","3","4","text","2020-12-07","2020-12-03","22"
"1","2","3","4","text","2020-12-12","2020-04-11","21"
"1","2","3","4","text","2020-05-21","2020-03-23","453"

等等

我想过滤 第二个日期 在 12 月的行，我尝试了以下操作：

grep '.*(\d{4}-\d{2}-\d{2}).*(2020-12-).*' > output.txt
grep '.*\d{4}-\d{2}-\d{2}.*2020-12-.*' > output.txt
grep -P '.*\d{4}-\d{2}-\d{2}.*2020-12-.*' > output.txt

但似乎没有任何效果。有什么方法可以使用 grep、egrep、sed 或 awk 完成此任务？

【问题讨论】：

见：Filtering lines in grep

标签： regex sed grep

【解决方案1】：

您需要使用grep 的-P 选项来启用与perl 兼容的正则表达式，请尝试以下操作。使用您展示的示例进行编写和测试。

grep -P '("\d+",){4}"[a-zA-Z]+","2020-12-\d{2}"' Input_file

说明：以上为补充说明，以下仅作说明之用。

grep             ##Starting grep command from here.
-P               ##Mentioning -P option for enabling PCRE regex with grep.
'("\d+",){4}     ##Looking for " digits " comma this combination 4 times here.
"[a-zA-Z]+",     ##Then looking for " alphabets ", with this one.
"2020-12-\d{2}"  ##Then looking for " 2020-12-07 date " which OP needs.
' Input_file     ##Mentioning Input_file name here.

【讨论】：

或者，如果您只想在文件的第 6 个字段中查找日期 2020-12-07，那么我们也可以简单地为此执行 awk -F'","' '$6=="2020-12-12"' Input_file。
请注意，OP 想要 12 月的任何日期。不过，我同意 awk 可能更好，因为这里有不同的列。
@KenY-N，当然，感谢您的告知，我现在已经更改了正则表达式，以捕捉 12 月的任何日期。
我试了一下，做了个小改动，如：grep -P '.*"\d{4}-\d{2}-\d{2}","2020-12-\ d{2}"' in.csv > december_2020.csv 忽略第一个日期列...它有效！我猜 -P 有魅力，谢谢！

【解决方案2】：

我建议使用替代解决方案awk，因为输入数据使用通用分隔符按行和列结构化：

awk -F, '$7 ~ /-12-/' file

"1","2","3","4","text","2020-01-01","2020-12-13","4"
"1","2","3","4","text","2020-12-07","2020-12-03","22"

【讨论】：

是的，这个也试过了，它也有效。谢谢。

【解决方案3】：

简写为grep -P 或egrep：

$ cat test.txt
"1","2","3","4","text","2020-01-01","2020-12-13","4"
"1","2","3","4","text","2020-12-07","2020-12-03","22"
"1","2","3","4","text","2020-12-12","2020-04-11","21"
"1","2","3","4","text","2020-05-21","2020-03-23","453"
$
$ grep -P '^"([^"]*","){6}2020-12-' test.txt
"1","2","3","4","text","2020-01-01","2020-12-13","4"
"1","2","3","4","text","2020-12-07","2020-12-03","22"
$
$ egrep '^"([^"]*","){6}2020-12-' test.txt
"1","2","3","4","text","2020-01-01","2020-12-13","4"
"1","2","3","4","text","2020-12-07","2020-12-03","22"

解释：

^" - 期待 " 开始
([^"]*","){6} - 扫描除" 之外的所有字符，然后是","；重复 6 次
2020-12- - 期待 202012-

【讨论】：

【解决方案4】：

问题出在：

egrep '.*\d{4}-\d{2}-\d{2}.2020-12-.' > output.txt
                          ^ HERE

. 只匹配一个字符，但你想跳过","，所以改为：

egrep '.*\d{4}-\d{2}-\d{2}.+2020-12-.' > output.txt
                          ^^ HERE

. 变为 .+。

【讨论】：

这绝不是这里唯一的问题。 grep 通常根本不支持 \d，而您正在任何列中寻找匹配项。