bash 中的 Printf 示例不会创建换行符
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Printf example in bash does not create a newline
提问by WolfHumble
Working with printfin a bash script, adding no spaces after "\n"does not create a newline, whereas adding a space creates a newline, e.?g.:
与工作printf在bash脚本，加入无空格后"\n"不创建换行符，而添加的空间中形成一个换行符，e.?g：
No space after
"\n"NewLine=`printf "\n"` echo -e "Firstline${NewLine}Lastline"Result:
FirstlineLastlineSpace after
"\n "NewLine=`printf "\n "` echo -e "Firstline${NewLine}Lastline"Result:
Firstline Lastline
后面没有空格
"\n"NewLine=`printf "\n"` echo -e "Firstline${NewLine}Lastline"结果：
FirstlineLastline空间后
"\n "NewLine=`printf "\n "` echo -e "Firstline${NewLine}Lastline"结果：
Firstline Lastline
Question: Why doesn't 1. create the following result:
问题：为什么不 1. 创建以下结果：
Firstline
Lastline
I know that this specific issue could have been worked around using other techniques, but I want to focus on why 1. does not work.
我知道这个特定问题可以使用其他技术解决，但我想集中讨论为什么 1. 不起作用。
Edited: When using echo instead of printf, I get the expected result, but why does printf work differently?
编辑：当使用 echo 而不是 printf 时，我得到了预期的结果，但为什么 printf 的工作方式不同？
NewLine=`echo "\n"`
echo -e "Firstline${NewLine}Lastline"
Result:
结果：
Firstline
Lastline
回答by álvaro González
The backtick operator removes trailing new lines. See 3.4.5. Command substitutionat http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_03_04.html
反引号运算符删除尾随的新行。见3.4.5。命令替换在http://tldp.org/LDP/Bash-Beginners-Guide/html/sect_03_04.html
Note on edited question
注意编辑的问题
Compare:
相比：
[alvaro@localhost ~]$ printf "\n"
[alvaro@localhost ~]$ echo "\n"
\n
[alvaro@localhost ~]$ echo -e "\n"
[alvaro@localhost ~]$
The echo command doesn't treat \nas a newline unless you tell him to do so:
\n除非您告诉他这样做，否则echo 命令不会将其视为换行符：
NAME
echo - display a line of text
[...]
-e enable interpretation of backslash escapes
POSIX 7 specifies this behaviour here:
POSIX 7在此处指定了此行为：
[...] with the standard output of the command, removing sequences of one or more characters at the end of the substitution
[...] 使用命令的标准输出，在替换结束时删除一个或多个字符的序列
回答by Sergio Abreu
Maybe people will come here with the same problem I had: echoing \n inside a code wrapped in backsticks. A little tip:
也许人们会带着我遇到的同样问题来到这里：在用 backsticks 包裹的代码中回显 \n。一个小技巧：
printf "astring\n"
# and
printf "%s\n" "astring"
# both have the same effect.
# So... I prefer the less typing one
The short answer is:
简短的回答是：
# Escape \n correctly !
# Using just: printf "$myvar\n" causes this effect inside the backsticks:
printf "banana
"
# So... you must try \n that will give you the desired
printf "banana\n"
# Or even \\n if this string is being send to another place
# before echoing,
buffer="${buffer}\\n printf \"$othervar\\n\""
One common problem is that if you do inside the code:
一个常见的问题是，如果你在代码里面做：
echo 'Tomato is nice'
when surrounded with backsticks will produce the error
当被 backsticks 包围时会产生错误
command Tomato not found.
The workaround is to add another echo -e or printf
解决方法是添加另一个 echo -e 或 printf
printed=0
function mecho(){
#First time you need an "echo" in order bash relaxes.
if [[ $printed == 0 ]]; then
printf "echo -e \\n"
printed=1
else
echo -e "\r\n\r\\n"
fi
}
Now you can debug your code doing in prompt just:
现在您可以在提示符下调试您的代码：
(prompt)$ `mySuperFunction "arg1" "etc"`
The output will be nicely
输出会很好
mydebug: a value
otherdebug: whathever appended using myecho
a third string
and debuging internally with
并在内部调试
mecho "a string to be hacktyped"
回答by Robert S
$ printf -v NewLine "\n"
$ echo -e "Firstline${NewLine}Lastline"
Firstline
Lastline
$ echo "Firstline${NewLine}Lastline"
Firstline
Lastline
回答by an0nym0usc0ward
It looks like BASH is removing trailing newlines. e.g.
看起来 BASH 正在删除尾随的换行符。例如
NewLine=`printf " \n\n\n"`
echo -e "Firstline${NewLine}Lastline"
Firstline Lastline
NewLine=`printf " \n\n\n "`
echo -e "Firstline${NewLine}Lastline"
Firstline
Lastline
回答by Paused until further notice.
Your edited echoversion is putting a literal backslash-n into the variable $NewLinewhich then gets interpreted by your echo -e. If you did this instead:
您编辑的echo版本将文字反斜杠-n 放入变量中$NewLine，然后由您的echo -e. 如果你这样做：
NewLine=$(echo -e "\n")
echo -e "Firstline${NewLine}Lastline"
your result would be the same as in case #1. To make thatone work that way, you'd have to escape the backslash and put the whole thing in single quotes:
您的结果将与情况 #1 相同。要使之以这种方式工作，您必须转义反斜杠并将整个内容放在单引号中：
NewLine=$(printf '\n')
echo -e "Firstline${NewLine}Lastline"
or double escape it:
或双重逃脱它：
NewLine=$(printf "\\n")
Of course, you could just use printfdirectly or you can set your NewLine value like this:
当然，您可以直接使用，也可以printf像这样设置 NewLine 值：
printf "Firstline\nLastline\n"
or
或者
NewLine=$'\n'
echo "Firstline${NewLine}Lastline" # no need for -e
回答by yabt
We do not need "echo" or "printf" for creating the NewLine variable:
我们不需要“echo”或“printf”来创建 NewLine 变量：
NewLine="
"
printf "%q\n" "${NewLine}"
echo "Firstline${NewLine}Lastline"
回答by Oleg Razgulyaev
Works ok if you add "\r"
如果您添加“\r”，则可以正常工作
$ nl=`printf "\n\r"` && echo "1${nl}2"
1
2
