How can I print a specific field from a specific line in a delimited type file

Question

I have a sorted, delimited type file and I want to extract a specific field in specific line.

This is my input file: 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

I want to extract 27092011084688 (value from 4th line , 3rd column ).

I used awk 'NR==4' but it gave me whole 4th line.

Answer 1

Fairly straightforward:

awk -F',' 'NR == 4 { print $3 }' somefile.csv

Using , as a field separator, take record number 4 and print field 3 in somefile.csv .

Answer 2

$ sed -n "4p" somefile.csv | cut -d, -f3

Edit

What's this?

• -n turns of normal output
• 4p prints the 4th row
• -d, makes cut use , as delimiter
• -f3 makes cut print the 3rd field

Answer 3

使用awk一种方法：

awk -F, 'NR==4 { print $3 }' file.txt

Answer 4

使用以下内容：

awk -F ',' 'NR==4 {print $3}'

Answer 5

perl替代csv文件中第4行的print element 3：

perl -F, -lane 'print $F[2] if $. == 4' somefile.csv