Question

我正在尝试在每一行中使用awk将当前日期字段追加到我的管道分隔内容中。而不是预期的，整个16列被附加到第17（新）位置。我试过改变一些事情，但没有帮助。我认为有一些基本的错误。我可以在这里得到帮助吗？

 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

输入（16个字段）：

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

我的输出

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

预期（17个领域）：

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

我想这里的问题是脚本中的$(date +"%d-%m-%Y");被解释为$0因为bash没有扩展它。为了防止这种情况，您可以使用双引号定义awk脚本，如下所示：

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

需要所有类型的转义$和"字符才能使其正常工作。

如果您不需要使用awk，但也可以使用sed，则可以使用：

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

Answer 2

将日期指定给变量并使用整行打印

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

示例：

$ 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

如果您想要month-date-year格式，请将date命令更改为date +"%m-%d-%Y"

使用awk命令添加新列会将整个内容添加到最后一个？

2 个答案: