bash 找 。-exec echo {} \; = 缺少`-exec'的参数
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find . -exec echo {} \; = missing argument to `-exec'
提问by Pascal Polleunus
Why doesn't this work? (echo is not the real command)
为什么这不起作用？（回声不是真正的命令）
$ find . -type d -exec echo {}?\;
find: missing argument to `-exec'
I managed to do that anyway like this:
无论如何，我设法这样做：
$ for f in `find . -type d`; do echo $f; done
回答by bakalov
This work for me.
这对我有用。
find . -type f -exec file '{}' \;
Braces are enclosed in single quote marks to protect them from interpretation as shell script punctuation.
大括号用单引号括起来，以防止它们被解释为 shell 脚本标点符号。
回答by djhaskin987
The following line is from the EXAMPLES section of man find:
以下行来自于的示例部分man find：
find . -type f -exec file '{}' \;
It looks to me like the {}part needs to be in single quotes.
在我看来，该{}部分需要用单引号括起来。
