"test

本文介绍了"test -v $ TESTENV"即使定义了TESTENV，也始终为假的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在Fedora 27下运行Bash 4.4.19，我有一个简单的脚本:

I'm running Bash 4.4.19, under Fedora 27, and I have a simple script:

#!/bin/bash TESTENV="Hello" echo "$TESTENV" if [ -v $TESTENV ]; then echo "Yup" fi echo "Done"

运行此命令时，它会打印:

When I run this, it prints:

Hello ./myscript: line 3: [: Hello: binary operator expected Done

因此，我在第3行添加了多余的括号，现在看起来像这样:

So, I add the extra bracket to line 3, which now looks like:

if [[ -v $TESTENV ]]; then

但是会产生:

Hello Done

有什么用?我希望以上两者之一都能看到是".我已经尝试将"$ TESTENV" 用引号引起来，并且还尝试了 -z 运算符-但其行为是相同的.

What gives? I expected to see "Yup" for one/both of the above. I've tried wrapping "$TESTENV" in quotes, and I've also tried the -z operator instead - but the behavior is the same.

推荐答案

对于 $ ，您使用的是变量的 value ，而不是变量名本身.您可能是说:

With $, you are using the value of the variable, not the variable name itself. You probably meant:

if [[ -v TESTENV ]]; then echo "Yup" fi

如果您定义了名为 Hello 的变量(例如 Hello = hi )，则 if 条件将会成功，并且您可能没有注意到根本没有问题:)

If you defined variable named Hello (such as Hello=hi) the if condition would have succeeded and you might not have noticed the problem at all :)