BASH 脚本，无需替换即可将变量传递到新脚本中
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BASH script to pass variables without substitution into new script
提问by interoperate
As part of a system build script I have a script that creates various files and configurations.
作为系统构建脚本的一部分，我有一个创建各种文件和配置的脚本。
However one part of the build script creates a new script that contains variables that I don't want resolved when the build script runs. Code snippet example
但是，构建脚本的一部分创建了一个新脚本，其中包含我不希望在构建脚本运行时解析的变量。代码片段示例
cat - > /etc/profile.d/mymotd.sh <<EOF
hostname=`uname -n`
echo -e "Hostname is $hostname"
EOF
I have tried all sorts of combinations of ' and " and ( and [ but I cannot get the script to send the content without substituting the values and placing the substitutes in the new script rather than the original text.
我已经尝试了 ' 和 " 和 ( 和 [ 的各种组合，但是如果不替换值并将替换值放置在新脚本而不是原始文本中，我就无法让脚本发送内容。
Ideas?
想法？
回答by Jonathan Callen
The easiest method, assuming you don't want anything to be substituted in the here doc, is to put the EOF marker in quotes, like this:
最简单的方法，假设您不想在此处的文档中替换任何内容，则将 EOF 标记放在引号中，如下所示：
cat - > /etc/profile.d/mymotd.sh <<'EOF'
hostname=`uname -n`
echo -e "Hostname is $hostname"
EOF
回答by ott--
Easiest is to escape the $
最简单的就是逃避$
echo -e "Hostname is \$hostname"
echo -e "主机名是 \$hostname"
