Question

当我将任何多行文本设置为fish中的变量时，它会删除新行字符并用空格替换它们，如何阻止它执行此操作？最小的完整示例：

~ ) set lines (cat .lorem); set start 2; set end 4;
~ ) cat .lorem 
once upon a midnight dreary while i pondered weak and weary
over many a quaint and curious volume of forgotten lore
while i nodded nearly napping suddenly there came a tapping
as of some one gently rapping rapping at my chamber door
tis some visiter i muttered tapping at my chamber door
~ ) echo $lines | sed -ne $start\,{$end}p\;{$end}q # Should print lines 2..4
~ ) cat .lorem | sed -ne $start\,{$end}p\;{$end}q
over many a quaint and curious volume of forgotten lore
while i nodded nearly napping suddenly there came a tapping
as of some one gently rapping rapping at my chamber door
~ ) echo $lines
once upon a midnight dreary while i pondered weak and weary over many a quaint and curious volume of forgotten lore while i nodded nearly napping     suddenly there came a tapping as of some one gently rapping rapping at my chamber door tis some visiter i muttered tapping at my chamber door

Answer 1

fish在换行符上拆分命令替换。这意味着$lines是一个列表。您可以阅读更多about lists here。

将列表传递给命令时，列表中的每个条目都成为单独的参数。 echo空格分隔其论点。这解释了你所看到的行为。

请注意，其他shell在此处执行相同的操作。例如，在bash中：

lines=$(cat .lorem)
echo $lines

如果要阻止拆分，可以暂时将IFS设置为空：

begin
   set -l IFS
   set lines (cat .lorem)
end
echo $lines

现在$lines将包含换行符。

正如faho所说，read也可以使用，而且更短一些：

read -z lines < ~/.lorem
echo $lines

但请考虑在新行上拆分是否真的符合您的要求。正如faho暗示的那样，您的sed脚本可以替换为数组切片：

set lines (cat .lorem)
echo $lines[2..4] # prints lines 2 through 4

Answer 2

这不仅仅是删除换行符，而是拆分换行符。

您的变量$ lines现在是一个列表，每行都是该列表中的一个元素。

见

set lines (cat .lorem)
for line in $lines
    echo $line
end
echo $lines[2]
printf "%s\n" $lines[2..4]

多行变量删除换行符 - 鱼

2 个答案: