regex - 在正则表达式中使用\t 似乎不适用于所有选项卡

Question

文件的某些行似乎与正则表达式中的\t 不匹配。有谁知道为什么？

让我们以您可以从 http://download.geonames.org/export/dump/countryInfo.txt 下载的示例文件为例。 .

$ wget http://download.geonames.org/export/dump/countryInfo.txt
--2011-02-03 16:24:08--  http://download.geonames.org/export/dump/countryInfo.txt
Resolving download.geonames.org... 178.63.52.141
Connecting to download.geonames.org|178.63.52.141|:80... connected.
HTTP request sent, awaiting response... 200 OK
Length: 31204 (30K) [text/plain]
Saving to: `countryInfo.txt'

100%[===================================================================================================================================================================================================>] 31,204      75.0K/s   in 0.4s    

2011-02-03 16:24:10 (75.0 KB/s) - `countryInfo.txt' saved [31204/31204]

$ cat countryInfo.txt | grep -E 'AD.AND'
AD  AND 200 AN  Andorra Andorra la Vella    468 84000   EU  .ad EUR Euro    376 AD###   ^(?:AD)*(\d{3})$    ca  3041565 ES,FR   
sdalouche@samxps:/tmp$ cat countryInfo.txt | grep -E 'AD\tAND'
(no result)

output of vi :set list
AD^IAND^I200^IAN^IAndorra^IAndorra la Vella^I468^I84000^IEU^I.ad^IEUR^IEuro^I376^IAD###^I^(?:AD)*(\d{3})$^Ica^I3041565^IES,FR^I$

Answer 1

尝试使用 -P选项而不是 -E :

cat countryInfo.txt | grep -P 'AD\tAND'

这将使用 Perl 风格的正则表达式，它将捕获 \t .

$ echo -e '-\t-' | grep -E '\t'
(no result)
$ echo -e '-\t-' | grep -P '\t'
-   -