awk for 循环文件中的每一行

Question

我需要弄清楚如何遍历文本文件，并为每一行获取 name 、 project # 和 email 字段，并将它们替换为要发送的电子邮件模板。

所以这是名为 send.txt 的文本文件：

Project 1,Jack,Chen,06,12,test@cs.fiu.edu
Project 2,Emily,Weiss,06,12,eweiss@cs.fiu.edu
Project 3,Mary,Gonzalas,06,12,Mgonz@cs.fiu.edu

这是名为 Reminder.email 的电子邮件模板：

Dear __FULLNAME__: 

This is a kindly reminder that our __Project__ meeting will be held on today. 

Best Regards, 
CIS5027

因此，对于文本文件中的每一行，我需要在此电子邮件模板中替换FULLNAME : 和Project的字段。使用我可以为第一行做的相应值，但是我不能为每一行都做。

这是我的脚本

#

!/bin/sh
#Start your code from here

date= date "+%m/%d/%y"
print $date

#The following line is to scan a file called Event-member.data and return any lines with todays date and save them to a file called sendtoday.txt

grep -n $(date +"%m,%d") Event-member.data > sendtoday.txt

#The following line is to remove the first to characters of the file created above sendtoday.txt and output that to a file called send.txt.

cat sendtoday.txt | sed 's/^..//' > send.txt


#This is where im having trouble. When storing the values for the variables below of name, email, project #. The value of NR==1 thus it never goes through the rest of the lines. I've tried different solutions but none seem to work.

p=$(awk -F ',' 'NR==1{print $1}' send.txt)
n=$(awk -F ',' 'NR==1{print $2}' send.txt)
l=$(awk -F ',' 'NR==1{print $3}' send.txt)
e=$(awk -F ',' 'NR==1{print $6}' send.txt)

echo $p  $n  $l  $e

#This part is to replace the values in the email template using sed and save the modified template as sendnow.txt.  

sed -e "s/__FULLNAME__:/\ $n $l :/g;s/__Project__/\ $p /g" Reminder.email > sendnow.txt

cat sendnow.txt


#Yet to be written ... send out modified email templates.
exit 0

########

这是它产生的输出：

06/12/14
Project 1 Jack Chen test@cs.fiu.edu
Dear  Jack Chen : 

This is a kindly reminder that our  Project 1  meeting will be held on today. 

Best Regards, 
CIS5027

如您所见，它确实正确替换了字段，但仅适用于 Jack Chen。send.txt 文件中有 3 行，因此必须有上述模板的 3 个修改版本。

Answer 1

没有理由为此使用awk。您可以直接在 shell 中使用read. 一般格式是read foo bar将第一个字段另存为，将$foo每行的其余部分另存为$bar. 因此，在您的情况下，您会执行以下操作：

while IFS="," read p n l foo bar e; do 
    sed -e "s/__FULLNAME__:/\ $n $l :/g;s/__Project__/\ $p /g" Reminder.email; 
done < file

这IFS是输入字段分隔符，当设置为,读取逗号分隔的字段时。这使您可以获取每个字段并将其存储在变量中。请注意，我使用了两个额外的变量foo和bar. 这是因为每个字段都需要自己的变量名称，而您有 6 个字段。如果您只给出 4 个变量名称，则第 4 个 ( $e) 将包含字段 4 到最后一个。

现在，您的脚本中还有其他各种语法错误。首先shebang行是错误的，你需要#! /bin/sh，#!和之间不能有空行/bin/sh。此外，为了将命令的输出分配给变量，您需要使用var=`command`or，最好是var=$(command)格式。否则，将命令本身作为字符串而不是其输出分配给变量。最后，print不是你想的那样。您正在寻找printf或echo。因此，编写脚本的更好方法是：

#!/bin/sh

date=$(date "+%m/%d/%y")
echo $date

## The following line is to scan a file called Event-member.data 
## and return any lines with todays date and save them to a file 
## called sendtoday.txt
grep -n $(date +"%m,%d") Event-member.data > sendtoday.txt

## The following line is to remove the first to characters of the file
## created above sendtoday.txt and output that to a file called
## send.txt. 
## I rewrote this to avoid the useless use of cat.
sed 's/^..//' sendtoday.txt > send.txt


## This is where you use read
while IFS="," read p n l foo bar e; do 
    sed -e "s/__FULLNAME__:/\ $n $l :/g;s/__Project__/\ $p /g" Reminder.email > sendnow.txt
    cat sendnow.txt
    ## This is where you need to add the code that sends the emails. Something
    ## like this:
    sendmail $e < sendnow.txt

done < send.txt

exit 0

########