sed循环中的参数 [英] Arguments inside sed loop

问题描述

我正在尝试了解sed命令和循环.我需要输入部分文本(20行)并将其附加到具有文件名的csv中.这是我的代码

Im trying to understand sed command and loops. I need to take part of the text (20 lines) and append it to the csv with the file name. here is my code

for i in ~/mspl/*.doc
do 
    catdoc "$i" > auto.txt
    sed -e '1,20!d' auto.txt > auto1.txt
    sed -e '1s/^/"$i" ;/' auto1.txt > auto2.txt
    sed -e '20s/$/~/' auto2.txt > auto3.txt
    cat auto3.txt >> lines.csv
done

问题是我的第二个"i"参数没有转换为csv中的文件名.

the problem is that my second "i" argument doesn't convert to the file name in csv.

在线

sed -e '1s/^/"$i" ;/' auto1.txt > auto2.txt

请告诉我我的错误是什么?

Please tell what is my mistake here?

推荐答案

问题是变量不会在单引号'中扩展.使用双引号代替":

The issue is that variables are not expanded within single quotes '. Use double quotes instead ":

# this is assuming you did want to add explicit " around your $i
sed -e "1s/^/\"$i\" ;/" auto1.txt > auto2.txt

如果双引号字符串需要包含显式双引号，则必须转义它们( \").如果不能使用双引号(例如，如果您有非常复杂的命令)已经具有多个级别的引用)，您随时可以退出变量周围的引用:

If your double-quoted string needs to contain explicit double quotes, you have to escape them (\"). If using double quotes is not an option (e.g. if you have a very complex command with multiple levels of quoting already) you can always exit the quoting around your variable:

sed -e '1s/^/"'$i'" ;/' auto1.txt > auto2.txt

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