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使用sed更改配置的Bash脚本

[英]Bash script to change config using sed

I'm trying to change the value of malloc to say 1234m via a bash script but not seeing any changes. 我试图通过bash脚本将malloc的值更改为1234m,但没有看到任何更改。 I presume it is an issue with my regex, can anybody see what I've done inncorrectly? 我认为这是我的正则表达式的一个问题,任何人都可以看到我做的正确吗?

String

DAEMON_OPTS="-a :6081 \
             -T localhost:6082 \
             -f /etc/varnish/default.vcl \
             -S /etc/varnish/secret \
             -s malloc,256m"

Code

# get the memory allocation
echo "Enter the memory allocation"
read malloc

# update the default config
sed -ie 's/malloc,.*[0-9m]$/malloc,'$malloc'/gI' /etc/default/varnish

You are missing a " before the $ in the sed pattern. 你错过了"在sed模式中的$之前”。

By the way, your pattern works, but in a different way than you probably intended: the character class [0-9m] matches just one character, m in this case. 顺便说一下,你的模式有效,但方式与你想要的方式不同:字符类[0-9m]只匹配一个字符,在本例中为m The number is being matched by .* . 该号码与.*相匹配。 Better pattern might be malloc,[0-9]\\+m"$ . 更好的模式可能是malloc,[0-9]\\+m"$

Try this sed command: 试试这个sed命令:

On Mac: 在Mac上:

sed -E 's/malloc,[0-9]+m/malloc,'$malloc'/' /etc/default/varnish

or on Linux: 或者在Linux上:

sed -r 's/malloc,[0-9]+m/malloc,'$malloc'/' /etc/default/varnish

以下sed行为我工作,将其括在"而不是'

 sed -ie "s/malloc,[0-9]\+m$/malloc,$malloc/gI" /etc/default/varnish

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