使用变量替换 shell 脚本中的字符串 [英] Replace a string in shell script using a variable

问题描述

我正在使用以下代码替换字符串在 shell 脚本中.

I am using the below code for replacing a string inside a shell script.

echo $LINE | sed -e 's/12345678/"$replace"/g'

但它被替换为 $replace 而不是那个变量的值.

but it's getting replaced with $replace instead of the value of that variable.

谁能告诉我出了什么问题?

Could anybody tell what went wrong?

推荐答案

如果你想解释 $replace，你不应该使用单引号，因为它们会阻止变量替换.

If you want to interpret $replace, you should not use single quotes since they prevent variable substitution.

试试:

echo $LINE | sed -e "s/12345678/${replace}/g"

成绩单:

pax> export replace=987654321
pax> echo X123456789X | sed "s/123456789/${replace}/"
X987654321X
pax> _

请注意确保 ${replace} 没有任何对 sed 有意义的字符(例如 /)因为除非逃脱，否则会引起混乱.但是，如果如您所说，您将一个数字替换为另一个数字，那应该没有问题.

Just be careful to ensure that ${replace} doesn't have any characters of significance to sed (like / for instance) since it will cause confusion unless escaped. But if, as you say, you're replacing one number with another, that shouldn't be a problem.

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