从bash中的文件路径获取基名 [英] Get basename from filepath in bash

问题描述

我正在尝试使用$ 1变量运行bash命令，该变量包含文件名及其路径.

I am trying to run a bash command using $1 variable which contains a filename and its path.

我只想将文件名记录到output.txt文件中，但是我很难从变量中删除路径.

I want to log only the filename to an output.txt file but I am having difficulty stripping out the path from the variable.

我尝试了以下操作:

file=$(basename $1)
echo ${file} >> output.txt

file=$(basename $1)
echo file >> output.txt

echo "basename $1" >> output.txt

它们都不起作用，返回数据"或文件"

None of them work, either return 'Data' or 'file'

推荐答案

通常来说，这里有 2个陷阱:

• 忽略双引号参数和变量引用- $ 1 或 $ var -和命令替换- $(...)-，使它们易受分词和通配符的影响，这可能会破坏整个命令，或者更糟糕的是，

• neglecting to double-quote parameter and variable references -$1 or $var - and command substitutions - $(...) - making them subject to word splitting and globbing, which could break the overall command, or, worse, introduce security vulnerabilities.

次要的是，确保命令 operand 不会误认为 option ，如果它恰好启动了带有-字符，使用- 表示其后的所有内容均应解释为 operand .

less importantly so, making sure that a command operand isn't mistaken for an option, if it happens to start with a - character, by using -- to signal that everything following it is to be interpreted as an operand.

分配到变量(双引号可选，但推荐):

Assigning to a variable (double-quoting optional, but recommended):

# OK - but note that the $1 inside the $(...) IS double-quoted,
# because command substitutions are "their own world".
file=$(basename -- "$1")

# Better, just to get into the habit of double-quoting:
file="$(basename -- "$1")"

使用变量-始终双引号(除非您明确希望分词和遍历):

Using the variable - always double-quote (unless you explicitly want word-splitting and globbing):

echo "${file}" >> output.txt

---

关于为什么您的尝试不起作用:

• echo文件>>output.txt 无效，因为要引用变量 file ，您必须 $ -前缀(即 $ file ，并且由于上述原因，实际上几乎总是"$ file" ).

• echo file >> output.txt didn't work, because in order to reference variable file you must $-prefix it (i.e., $file and, for the reasons stated above, virtually always as "$file").

echo"basename $ 1">>output.txt 不起作用，因为在双引号字符串内扩展了参数 $ 1 时，您需要命令替换-这里是 $(基本名称"$ 1")-将命令嵌入在双引号字符串中.

echo "basename $1" >> output.txt didn't work, because while parameter $1 is expanded inside the double-quoted string, you need a command substitution - here, $(basename "$1") - to embed commands in a double-quoted string.

file = $(basename $ 1)，由于缺少 $ 1 的双引号，因此 only 与 $ 1 值，不包含任何外壳元字符(包括空格)，以及不以-开头的值； echo $ {file}>>再次由于缺少双引号而导致output.txt 再次无意中更改了 $ file 的值.

file=$(basename $1), due to lack of double-quoting of $1, would only work with $1 values that contain no shell metacharacters (including spaces), and values that don't start with -; echo ${file} >> output.txt - again due to lack of double-quoting - could then inadvertently alter the value of $file again.

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