Question

我正在练习使用命令read：

$ ls * | while read line; do printf "test %s\n"  $line; done
test data1.txt
test data5.txt
test data6.txt
test data7.txt

它可以正常工作，但是

$ ls * | while read line; do file $line; done
data1.txt: cannot open `data1.txt' (No such file or directory)
data5.txt: cannot open `data5.txt' (No such file or directory)
data6.txt: cannot open `data6.txt' (No such file or directory)
data7.txt: cannot open `data7.txt' (No such file or directory)
data8.txt: cannot open `data8.txt' (No such file or directory)

文件名合法

$ ls
data1.txt  data5.txt  data6.txt  data7.txt  data8.txt

我的代码有什么问题。

Answer 1

我怀疑您的文件未命名为"data1.txt"，实际上更像是"data1.txt "。请注意最后的微妙空间。这很可能是脚本中的一个很小的错字造成的。也可能是无法打印的字符的使用。您可以使用ls --quoting-style=shell-escape进行检查。

示例：

$ touch "data1.txt "
$ touch "data2.txt"
$ echo "a" > "data3.txt "
$ ls data* | while read line; do file $line; done
data1.txt: cannot open (No such file or directory)
data2.txt: empty
data3.txt: cannot open (No such file or directory)

         ^
Notice the missing spaces

此行有几个问题：

Never parse ls
Always quote your variables
read removes all leading and trailing whitespace characters。 这实际上是这里的主要问题

因此，如果您想解决read问题，请执行以下操作：

$ ls data* | while IFS= read -r line; do file "$line"; done
data1.txt : empty
data2.txt: empty
data3.txt : ASCII text

         ^
Notice the spaces, this is due to read

但是正确的方法是：

$ for f in data*; do file "$f"; done
data1.txt : empty
data2.txt: empty
data3.txt : ASCII text

命令`file`无法打开* .txt

1 个答案: