命令`file`无法打开*.txt

Question

我正在练习使用命令read:

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

它工作正常，但

$ 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)

文件名是合法的

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

我的代码有什么问题。

Answer 1

我怀疑您的文件不叫"data1.txt"，实际上更像"data1.txt "。请注意结尾处的细微空间。这很容易是由于脚本中的一个小错误造成的，也可能是使用了不可打印的字符。您可以使用ls --quoting-style=shell-escape对此进行检查。

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

这条单行有几个问题：

1. 从不解析ls
2. 始终引用变量
3. read删除所有前导和尾部空白字符<这实际上是这里的主要问题>

所以，如果你想解决read问题，你可以：

$ 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

但正确的方法是：

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