Bash脚本不评估文件名中的变量 [英] Bash script doesn't evaluate variable in filename

问题描述

我有一个bash脚本，可以创建我的文件夹的备份.它应使用版本号命名tar gz文件，但不能:

I have a bash script which creates a backup of my folder. It should name the tar gz file using a version number but it doesn't:

#!/bin/bash

ver='1.2.3'
tar czf $ver_myfolder.tar.gz .

预期输出:

1.2.3_myfolder.tar.gz

1.2.3_myfolder.tar.gz

实际输出:

_myfolder.tar.gz

_myfolder.tar.gz

如果我像这样附加变量:

If I append the variable like this:

#!/bin/bash

ver='1.2.3'
tar czf myfolder-$ver.tar.gz .

有效

推荐答案

您应在此处使用${var}，因为您将在其后附加下划线，该下划线被视为变量名称的有效字符.由于这个事实，shell不知道您是在扩展变量名$ver还是$ver_myfolder:

You should use ${var} here since you are appending underscore after it which is considered a valid character for variable names. Due to that fact shell doesn't know whether you're expanding variable name $ver or $ver_myfolder:

ver='1.2.3'
tar czf "${ver}_myfolder.tar.gz" .

由于未设置$ver_myfolder，您将获得一个空值.

Since $ver_myfolder is unset, you get an empty value.

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