我想写一些预定义的文本与以下文件:
text="this is line one\n
this is line two\n
this is line three"
echo -e $text > filename
我期待的是这样的:
this is line one
this is line two
this is line three
但我明白了:
this is line one
this is line two
this is line three
我确信在每个\n之后没有空间，但额外的空间是如何出来的?
当前回答
Echo在传递给它的参数之间添加空格。$text受变量扩展和单词分割的影响，所以你的echo命令相当于:
echo -e "this" "is" "line" "one\n" "this" "is" "line" "two\n" ...
你可以看到在“this”之前加了一个空格。你可以删除换行符，并引用$text来保留换行符:
text="this is line one
this is line two
this is line three"
echo "$text" > filename
或者你可以使用printf，它比echo更健壮和可移植:
printf "%s\n" "this is line one" "this is line two" "this is line three" > filename
在支持大括号展开的bash中，你甚至可以这样做:
printf "%s\n" "this is line "{one,two,three} > filename
其他回答
如果你把它放在下面，它就会工作:
AA='first line
\nsecond line
\nthird line'
echo $AA
output:
first line
second line
third line
Echo在传递给它的参数之间添加空格。$text受变量扩展和单词分割的影响，所以你的echo命令相当于:
echo -e "this" "is" "line" "one\n" "this" "is" "line" "two\n" ...
你可以看到在“this”之前加了一个空格。你可以删除换行符，并引用$text来保留换行符:
text="this is line one
this is line two
this is line three"
echo "$text" > filename
或者你可以使用printf，它比echo更健壮和可移植:
printf "%s\n" "this is line one" "this is line two" "this is line three" > filename
在支持大括号展开的bash中，你甚至可以这样做:
printf "%s\n" "this is line "{one,two,three} > filename
如果你试图将字符串转换为变量，另一种简单的方法是这样的:
USAGE=$(cat <<-END
This is line one.
This is line two.
This is line three.
END
)
如果你用制表符缩进字符串(例如，'\t')，缩进将被删除。如果使用空格缩进，则缩进将保留。
注意:最后一个右括号在另一行，这很重要。END文本必须单独出现在一行中。
这里只提一个简单的单行连接，因为它有时很有用。
# for bash
v=" guga "$'\n'" puga "
# Just for an example.
v2="bar "$'\n'" foo "$'\n'"$v"
# Let's simplify the previous version of $v2.
n=$'\n'
v3="bar ${n} foo ${n}$v"
echo "$v3"
你会得到这样的结果
bar foo guga puga
所有前导和结尾空格将被保留
echo "$v3" > filename
有很多方法可以做到这一点。对我来说，将缩进的字符串输送到sed中效果很好。
printf_strip_indent() {
printf "%s" "$1" | sed "s/^\s*//g"
}
printf_strip_indent "this is line one
this is line two
this is line three" > "file.txt"
这个答案是基于Mateusz Piotrowski的回答，但做了一些改进。
