为什么使用变量时我的sed命令失败? [英] Why is my sed command failing when using variables?

问题描述

我使用bash尝试为日期插入变量，并在日志文件中搜索该日期，然后将输出发送到文件.如果我这样对日期进行硬编码，它将起作用:

using bash, I'm trying to insert a variable for the date and search a log file for that date, then send the output to a file. If I hardcode the date like this it works:

sed -n '/Nov 22, 2010/,$p' $file >$log_file

但是如果我这样做，它将失败:

but if I do it like this it fails:

date="Nov 22, 2010"
sed -n '/$date/,$p' $file >$log_file

我得到的错误是: sed:1:"//Nov 22，2010/，":预期的上下文地址感谢您的帮助.

The error I get is: sed: 1: "/Nov 22, 2010/,": expected context address Thanks for the help.

推荐答案

在shell脚本中，单引号和双引号之间是有区别的.单引号内的变量不扩展( (不同于双引号内的变量)，因此您需要:

In shell scripts, there is a difference between single and double quotes. Variables inside single quotes are not expanded (unlike those in double quotes), so you would need:

date="Nov 22, 2010"
sed -n "/$date/,\$p" $file >$log_file

反斜杠转义了美元符号，该符号应由外壳按字面意义使用，并且具有sed的含义.

The backslash escapes the dollar sign that should be taken literally by the shell and has a meaning to sed.

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