使用带有嵌入空格的参数（例如文件名）调用 bash 脚本中的命令
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Calling commands in bash script with parameters which have embedded spaces (eg filenames)
提问by John Dibling
I am trying to write a bash script which does some processing on music files. Here is the script so far:
我正在尝试编写一个对音乐文件进行一些处理的 bash 脚本。这是到目前为止的脚本：
#!/bin/bash
SAVEIFS=$IFS
IFS=printf"\n[johnd:/tmp/tunes] 2 $ ./test.sh
./test.sh: line 9: syntax error near unexpected token `$(f)'
./test.sh: line 9: ` $arr=($(f))'
[johnd:/tmp/tunes] 2 $
"
find `pwd` -iname "*.mp3" -o -iname "*.flac" | while read f
do
echo "$f"
$arr=($(f))
exiftool "${arr[@]}"
done
IFS=$SAVEIFS
This fails with:
这失败了：
while read -r -d '' f; do
echo "$f" # For debugging?
arr+=("$f")
done < <(find . -iname "*.mp3" -o -iname "*.flac" -print0)
exiftool "${arr[@]}"
I have tried many different incantations, none of which have worked. The bottom line is I'm trying to call a command exiftool, and one of the parameters of that command is a filename which may contain spaces. Above I'm trying to assign the filename $fto an array and pass that array to exiftool, but I'm having trouble with the construction of the array.
我试过很多不同的咒语，都没有奏效。最重要的是我正在尝试调用 command exiftool，该命令的参数之一是可能包含空格的文件名。上面我试图将文件名分配$f给一个数组并将该数组传递给exiftool，但是我在构建数组时遇到了麻烦。
Immediate question is, how do I construct this array? But the deeper question is how, from within a bash script, do I call an external command with parameters which may contain spaces?
直接的问题是，我如何构造这个数组？但更深层次的问题是，如何从 bash 脚本中调用带有可能包含空格的参数的外部命令？
回答by michaelb958--GoFundMonica
You actually did have the call-with-possibly-space-containing-arguments syntax right (program "${args[@]}"). There were several problems, though.
您实际上确实拥有正确的 call-with-possably-space-包含-arguments 语法 ( program "${args[@]}")。但是，有几个问题。
Firstly, $(foo)executes a command. If you want a variable's value, use $fooor ${foo}.
首先，$(foo)执行一个命令。如果您想要一个变量的值，请使用$foo或${foo}。
Secondly, if you want to append something onto an array, the syntax is array+=(value)(or, if that doesn't work, array=("${array[@]}" value)).
其次，如果您想将某些内容附加到数组中，则语法为array+=(value)（或者，如果这不起作用，则为array=("${array[@]}" value)）。
Thirdly, pleaseseparate filenames with \0whenever possible. Newlines are all well and good, but filenames can contain newlines.
第三，请\0尽可能将文件名与. 换行符都很好，但文件名可以包含换行符。
Fourthly, readtakes the switch -d, which can be used with an empty string ''to specify \0as the delimiter. This eliminates the need to mess around with IFS.
第四，read采用 switch -d，它可以与空字符串''一起使用\0作为分隔符。这消除了对IFS.
Fifthly, be careful when piping into whileloops - this causes the loop to be executed in a subshell, preventing variable assignments inside it from taking effect outside. There is a way to get around this, however - instead of piping (command | while ... done), use process substitution (while ... done < <(command)).
第五，当管道进入while循环时要小心——这会导致循环在子外壳中执行，防止其内部的变量赋值在外部生效。但是，有一种方法可以解决此问题 -command | while ... done使用进程替换 ( while ... done < <(command))代替管道( )。
Sixthly, watch your process substitutions - there's no need to use $(pwd)as an argument to a command when .will do. (Or if you really must have full paths, try quoting the pwdcall.)
第六，注意你的进程替换 - 不需要将$(pwd)何时用作命令的参数.。（或者，如果您确实必须拥有完整路径，请尝试引用pwd调用。）
tl;dr
tl;博士
The script, revised:
修改后的脚本：
find . -iname "*.mp3" -o -iname "*.flac" -exec exiftool {} +
# Much shorter!
Another way
其它的办法
Leveraging find's full capabilities:
利用find的全部功能：
while read -r -d '' f; do
echo "$f" # For debugging?
arr+=("$f")
done < <(find . -iname "*.mp3" -o -iname "*.flac" -print0)
# Warning: somewhat misleading syntax highlighting ahead
newfilename="$(exiftool "${arr[@]}")"
newfilename="$(manipulate "$newfilename")"
cp -- "$some_old_filename" "$newfilename"
Edit 1
编辑 1
So you need to save the output of exiftool, manipulate it, then copy stuff? Try this:
所以你需要保存 的输出exiftool，操作它，然后复制东西？尝试这个：
shopt -s globstar nullglob
a=( **/*.{mp3,flac} )
exiftool "${a[@]}"
You probably will need to change that last bit - I've never used exiftool, so I don't know precisely what you're after (or how to do it), but that should be a start.
您可能需要更改最后一点 - 我从未使用过exiftool，所以我不确切知道您在追求什么（或如何去做），但这应该是一个开始。
回答by glenn Hymanman
You can do this just with bash:
你可以只用 bash 做到这一点：
##代码##This probably works too: exiftool **/*.{mp3,flac}
这可能也有效： exiftool **/*.{mp3,flac}
