使用 cut 排除封閉的分隔符

Question

假設我有一個 CSV 檔案：

"col1","col2","col3"
"col4","col5,subtext","col6

我遇到的問題如下：

cut -d, -f1,2 test.txt
"coll1","col2"
"col4","col5

所需的輸出為：

"col1","col2"
"col4","col5,subtext"

Answer 1

Perl 附帶的 ParseWords 模組非常優雅地涵蓋了這一點。下面的例子。

$ perl -MText::ParseWords -nE '@a=quotewords ",",1,$_;say $a[0],",",$a[1]' <test.txt
"col1","col2"
"col4","col5,subtext"
$

Answer 2

如果您有gawkv4 可用，則有一個很好的解決方案使用 awk 解析 csv 並忽略字段內的逗號

例子：

gawk -vFPAT='[^,]*|"[^"]*"' '{print $1 "," $2}' test.txt

Answer 3

另一種perl解決方案，假設所有欄位都被引用

$ perl -F'/"\K,(?=")/' -lane 'print "$F[0],$F[1]"' test.txt 
"col1","col2"
"col4","col5,subtext"

• -F'/"\K,(?=")/'僅當欄位分隔符號前後為逗號時，欄位分隔符號才會為"逗號"
• print "$F[0],$F[1]"列印前兩個字段，分隔符,

grep也 可以用

$ grep -oE '^"[^"]*","[^"]*"' test.txt 
"col1","col2"
"col4","col5,subtext"

如果需要 N 個字段，請使用裡面的grep -oE '^("[^"]*",){1}"[^"]*"'數字{}N-1

Answer 4

你也可以用 awk 嘗試一下，如下所示；

awk -F'","'  '{printf "%s\",\"%s\"\n", $1, $2 }' test.txt 

例如;

user@host$ awk -F'","'  '{printf "%s\",\"%s\"\n", $1, $2 }' test.txt 
"col1","col2"
"col4","col5,subtext"