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linux - 如何使用 grep 命令打印列中搜索关键字产生的行,

转载 作者:太空宇宙 更新时间:2023-11-04 04:41:14 25 4
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这是 filename.csv 文件的示例

"Sort Order","Common Name","Formal Name","Type","Sub Type"

"1","Afghanistan","Islamic State of Afghanistan","Independent State"
"2","Albania","Republic of Albania","Independent State"
"3","Algeria","People's Democratic Republic of Algeria","Independent State"
"4","Andorra","Principality of Andorra","Independent State"
"5","Angola","Republic of Angola","Independent State"

那么用通用名称搜索 angola 并像这样打印它的 grep 命令是什么:

"5","Angola","Republic of Angola","Independent State"

我知道我们可以使用:

grep '"Algeria"' filename.csv

但是,如果我更具体一点呢?假设 Algolia 可能存在于其他列中,但是我们只需要在“通用名称”列中打印该名称即可。

我试过了

grep '"Common Name | Algeria"' filename.csv

好像不行。

最佳答案

您可以尝试使用下面的 grep 命令来打印第二列中包含字符串 Angola 的行。

grep -E '^"[^"]*","Angola"' file

这可以通过 awk 轻松完成,

awk -F, '$2=="\"Angola\""' file

关于linux - 如何使用 grep 命令打印列中搜索关键字产生的行,,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26923059/

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