如何在 Bash 中保留带引号的字符串中的换行符？
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How do I preserve newlines in a quoted string in Bash?
提问by bob dobbs
I'm creating a script to automate the creation of apache virtual hosts. Part of my script goes like this:
我正在创建一个脚本来自动创建 apache 虚拟主机。我的部分脚本是这样的：
MYSTRING="<VirtualHost *:80>
ServerName $NEWVHOST
DocumentRoot /var/www/hosts/$NEWVHOST
...
"
echo $MYSTRING
However, the line breaks in the script are being ignored. If I echo the string, is gets spat out as one line.
但是，脚本中的换行符将被忽略。如果我回显字符串，它会作为一行吐出。
How can I ensure that the line breaks are printed?
如何确保打印换行符？
回答by Martin
Add quotes to make it work:
添加引号以使其工作：
echo "$MYSTRING"
Look at it this way:
这样看：
MYSTRING="line-1
line-2
line3"
echo $MYSTRING
this will be executed as:
这将被执行为：
echo line-1 \
line-2 \
line-3
i.e. echowith three parameters, printing each parameter with a space in between them.
即echo有三个参数，打印每个参数之间有一个空格。
If you add quotes around $MYSTRING, the resulting command will be:
如果在 周围添加引号$MYSTRING，则生成的命令将是：
echo "line-1
line-2
line-3"
i.e. echowith a single string parameter which has three lines of text and two line breaks.
即echo具有单个字符串参数，其中包含三行文本和两个换行符。
