bash sed 用双引号替换变量
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Sed replace variable in double quotes
提问by Ken J
I've created a bash script that takes a parameter. I want to pass that parameter to sed to replace an existing string with another which is composed of the variable:
我创建了一个带参数的 bash 脚本。我想将该参数传递给 sed 以用另一个由变量组成的字符串替换现有字符串：
variable=
echo $variable
sed -i -e 's/name="master"/name="$variable"/g' test
The problem is that the script is not replacing $variable with the parameter, it's just replacing the string with "$variable":
问题是脚本没有用参数替换 $variable，它只是用 "$variable" 替换字符串：
<host name=""$variable"" xmlns="urn:jboss:domain:3:0:>
How can I replace a string in quotes with the variable?
如何用变量替换引号中的字符串？
回答by Sven Marnach
Variable expansion does not happen within single quotes. Do it in double quotes:
单引号内不会发生变量扩展。用双引号做：
sed -i -e 's/name="master"/name="'"$variable"'"/g' test
回答by Avinash Raj
Change your code like his,
像他一样改变你的代码，
sed -i -e 's/name="master"/name="'"$variable"'"/g' test
回答by farrell47
just do like this:
只是这样做：
var=apple
sed -i "s/pineapple/$"
