使用awk命令添加新列會將整個內容添加到最后一個？

Question

我試圖在每行中使用awk將當前日期字段追加到管道分隔的內容中。 而不是想要的，整個16列都被追加到第17（新）位置。 我試圖改變事情，但沒有幫助。 我認為有一些基本錯誤。 我可以在這里幫忙嗎？

 awk ' BEGIN{FS="|";}
{
    $17=$(date +"%d-%m-%Y");

    printf("%s|%s|%s|%s|%s|%s|%s|%s|%s|%s|%s|%s|%s|%s|%s|%s|%s\n", $1, $2, $3, $4, $5, $6, $7, $8, $9, $10,
     $11, $12, $13, $14, $15, $16, $17);

}' /Users/temp/dispn/content3.txt | less > /Users/temp/dispn/content4.txt

輸入（16個字段）：

EX122YED| Buy online |example.com/EX122YED |example.com/EX122YED.jpg|White|new|00.00|in stock|XYZ|Accessories|Accessories|Male|30-40|y|EX122YED|0.0

我的輸出

EX122YED| Buy online |example.com/EX122YED |example.com/EX122YED.jpg|White|new|00.00|in stock|XYZ|Accessories|Accessories|Male|30-40|y|EX122YED|0.0|EX122YED| Buy online |example.com/EX122YED |example.com/EX122YED.jpg|White|new|00.00|in stock|XYZ|Accessories|Accessories|Male|30-40|y|EX122YED|0.0

預期（17個字段）：

EX122YED| Buy online |example.com/EX122YED |example.com/EX122YED.jpg|White|new|00.00|in stock|XYZ|Accessories|Accessories|Male|30-40|y|EX122YED|0.0|06-22-2014

Answer 1

我猜這里的問題是$(date +"%d-%m-%Y"); 腳本中的bash $0被解釋為$0因為bash不會擴展它。 為了防止這種情況，您可以使用雙引號定義awk腳本，如下所示：

awk "{print \$0\"|\"$(date +\"%d-%m-%Y\")}"

這需要使用各種轉義的$和"字符來使其起作用。

如果您不需要使用awk但也可以使用sed ，則可以使用：

sed "s/$/|$(date +"%d-%m-%Y")/"

Answer 2

將日期分配給變量，然后整行打印，

awk -F'|' -v var=$(date +"%d-%m-%Y") '{print $0,var}' OFS="|"

例：

$ echo 'EX122YED| Buy online |example.com/EX122YED |example.com/EX122YED.jpg|White|new|00.00|in stock|XYZ|Accessories|Accessories|Male|30-40|y|EX122YED|0.0' | awk -F'|' -v var=$(date +"%d-%m-%Y") '{print $0,var}' OFS="|"
EX122YED| Buy online |example.com/EX122YED |example.com/EX122YED.jpg|White|new|00.00|in stock|XYZ|Accessories|Accessories|Male|30-40|y|EX122YED|0.0|22-06-2014

如果要使用month-date-year格式，則將date命令更改為date +"%m-%d-%Y"