防止 bash 在不引用所有内容的情况下进行解释
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Prevent bash from interpreting without quoting everything
提问by lzap
I need to output some text as bash script, but in a script. I use cat for this, but it has one drawback. It interprets variables and stuffduring it is being written. I do want to prevent this.
我需要输出一些文本作为 bash 脚本，但在脚本中。我为此使用 cat，但它有一个缺点。它在写入过程中解释变量和内容。我确实想防止这种情况。
How to do that without quoting all varibles(my script is failrly long)? Example
如何在不引用所有变量的情况下做到这一点（我的脚本很长）？例子
cat >/tmp/script << EOF
$HOSTNAME
# lots of other stuff I do NOT want to escape like $VARIABLE
# ...
EOF
cat /tmp/script
myhostname.mylan
I want:
我想要：
cat /tmp/script
$HOSTNAME
Edit: Please note my script (here only $HOSTNAME) is very long, I dont want to change it all. Also single quoting does not workwith <<
编辑：请注意我的脚本（这里只有$HOSTNAME）很长，我不想改变它。同时单引号不起作用与<<
cat >/tmp/script '<< EOF
$HOSTNAME
EOF'
File not found: EOF'
What's the trick? Thanks.
有什么诀窍？谢谢。
回答by torek
If you want everything quoted:
如果你想引用所有内容：
cat << 'EOF'
stuff here with $signs is OK
as are `backquotes`
EOF
See the section on "here documents" in the manual.
请参阅手册中有关“此处文档”的部分。
回答by Eran Ben-Natan
Escape the $:
转义 $：
cat >/tmp/script << EOF
$HOSTNAME
EOF
回答by user unknown
Use sed:
使用 sed：
sed -n '20,30p' "#!/bin/bash
cat >/dev/null << EOF
3
4 $HOSTNAME
5 ls
6 $(ls -l)
7
8 echo 'foo
9 bar'
10
11 echo "Foo
12 $((4+4)) Bar"
EOF
sed -n '3,12p' "echo '
$HOSTNAME
...
' >> /tmp/script
"
echo "fine?"
"
to print line 20 to 30, SSCE:
打印第 20 到 30 行，SSCE：
##代码##working with head/tail should work too.
使用头/尾也应该工作。
You will have to adjust the numbers, if you work on it and insert or delete lines.
如果您处理并插入或删除行，您将不得不调整数字。
回答by WilQu
Try with echo :
尝试使用 echo ：
##代码##