Bash 基本名称语法
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Bash basename syntax
提问by Unknown
I'm trying to write a script that takes the basename of an argument, then checks if there is an extension in that argument. If there is, it prints the extension.
我正在尝试编写一个脚本，该脚本采用参数的基本名称，然后检查该参数中是否有扩展名。如果有，它会打印扩展名。
Here is my code:
这是我的代码：
file=basename
ext=${file%*}
echo ${file#"$stub"}
echo $basename
I'm echoing the final $basename $1 to check what the output of basename is.
我正在回显最终的 $basename $1 以检查 basename 的输出是什么。
Some tests reveal:
一些测试显示：
testfile.sh one.two
./testfile: line 2: one.two: command not found
one.two
testfile.sh ../tester
./testfile: line 2: ../tester: No such file or directory
../tester
So neither $basename $1 are working. I know it's a syntax error so could someone explain what I'm doing wrong?
所以 $basename $1 都不起作用。我知道这是一个语法错误，所以有人可以解释我做错了什么吗？
EDIT:
编辑：
I've solved my problem now with:
我现在已经解决了我的问题：
file=$(basename "" )
stub=${file%.*}
echo ${file#"$stub"}
Which reduces my argument to a basename, thank you all.
这将我的论点简化为基本名称，谢谢大家。
回答by chepner
First, your syntax is wrong:
首先，你的语法是错误的：
file=$( basename "" )
Second, this is the correct expression to get the file name's (last) extension:
其次，这是获取文件名（最后一个）扩展名的正确表达式：
ext=${file##*.}
回答by Janito Vaqueiro Ferreira Filho
If you want to assign to a variable the output of a command, you must execute it, either with back quotes or using a special parenthesis quoting:
如果要将命令的输出分配给变量，则必须使用反引号或使用特殊括号引用来执行它：
file=`basename ""`
file=$(basename "")
To remove a filename's extension you should do
要删除文件名的扩展名，您应该这样做
ext=${file#*.}
this will get the value of $file, then remove everything up to the period (that's why it is necessary). If your filename contains periods, then you should use ext=${file##*.}, where the ##tells bash to delete the longest string that matches instead of the shortest, so it will delete up to the last period, instead of the first one.
这将获得 的值$file，然后删除该期间之前的所有内容（这就是必要的原因）。如果您的文件名包含句点，那么您应该使用ext=${file##*.},##告诉 bash 删除匹配的最长字符串而不是最短的字符串，因此它将删除到最后一个句点，而不是第一个。
Your echo $basename $1is wierd. It is telling bash to print the value of a variable called $basenameand the value of the variable $1, which is the first argument to the script (or function).
你的echo $basename $1很奇怪。它告诉 bash 打印被调用的变量的值和变量$basename的值$1，这是脚本（或函数）的第一个参数。
If you want to print the command you're trying to execute, do this:
如果要打印要执行的命令，请执行以下操作：
echo basename
If you're trying to print the output of the command, you can do one of:
如果您尝试打印命令的输出，您可以执行以下操作之一：
echo $(basename "")
echo "$file"
basename ""
Hope this helps =)
希望这会有所帮助 =)
