bash 将多行输出放在变量中
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place a multi-line output inside a variable
提问by ufk
I'm writing a script in bash and I want it to execute a command and to handle each line separately. for example:
我正在用 bash 编写一个脚本，我希望它执行一个命令并分别处理每一行。例如：
LINES=$(df)
echo $LINES
it will return all the output converting new lines with spaces.
它将返回所有输出转换为空格的新行。
example:
例子：
if the output was supposed to be:
如果输出应该是：
1
2
3
then I would get
然后我会得到
1 2 3
how can I place the output of a command into a variable allowing new lines to still be new lines so when I print the variable i will get proper output?
如何将命令的输出放入允许新行仍然是新行的变量中，以便在打印变量时获得正确的输出？
回答by kubanczyk
Generally in bash $vis asking for trouble in most cases. Almost always what you really mean is "$v"in double quotes.
通常在 bash$v中，大多数情况下都是自找麻烦。几乎总是你真正的意思是"$v"双引号。
LINES="`df`" # double-quote + backtick
echo "$LINES"
OLDPATH="$PATH"
回答by Jan Hudec
No, it will not. The $(something)only strips trailing newlines.
不，不会。该$(something)只带尾随换行。
The expansion in argument to echo splits on whitespace and than echo concatenates separate arguments with space. To preserve the whitespace, you need to quote again:
echo 的参数扩展在空格上拆分，然后 echo 将单独的参数与空格连接起来。要保留空格，您需要再次引用：
echo "$LINES"
Note, that the assignment does notneed to be quoted; result of expansion is not word-split in assignment to variable and in argument to case. But it can be quoted and it's easier to just learn to just always put the quotes in.
请注意，这样的决定也并不需要被引用; 扩展的结果在赋值给变量和在参数中没有分词case。但是它可以被引用，而且学习总是把引号放在里面会更容易。
