如何从分隔类型文件中的特定行打印特定字段

fai*_*zal 6 csv bash shell awk file

我有一个已排序,分隔的类型文件,我想提取特定行中的特定字段.

这是我的输入文件: somefile.csv

efevfe,132143,27092011080210,howdy,hoodie
adfasdfs,14321,27092011081847,howdy,hoodie
gerg,7659876,27092011084604,howdy,hoodie
asdjkfhlsdf,7690876,27092011084688,howdy,hoodie
alfhlskjhdf,6548,27092011092413,howdy,hoodie
gerg,769,27092011092415,howdy,hoodie
badfa,124314,27092011092416,howdy,hoodie
gfevgreg,1213421,27092011155906,howdy,hoodie
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我想提取27092011084688(第4行,第3列的值).

我用过,awk 'NR==4'但它给了我整整4行.

Joh*_*web 5

非常坦率的:

awk -F',' 'NR == 4 { print $3 }' somefile.csv
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使用,字段分隔符,获取记录编号4并在中打印字段3 somefile.csv