检查bash变量是否等于0 [英] Check if bash variable equals 0

问题描述

我有一个bash变量深度，我想测试它是否等于0.如果是的话，我想停止执行脚本.到目前为止，我有:

I have a bash variable depth and I would like to test if it equals 0. In case yes, I want to stop executing of script. So far I have:

zero=0;

if [ $depth -eq $zero ]; then
    echo "false";
    exit;
fi

不幸的是，这导致:

 [: -eq: unary operator expected

(由于翻译可能不准确)

(might be a bit inaccurate due to translation)

请，我如何修改脚本以使其正常工作?

Please, how can I modify my script to get it working?

推荐答案

好像未设置您的depth变量.这意味着在bash将变量的值代入表达式后，表达式[ $depth -eq $zero ]变为[ -eq 0 ].这里的问题是-eq运算符被错误地用作仅具有一个参数(零)的运算符，但是它需要两个参数.这就是为什么您收到一元运算符错误消息的原因.

Looks like your depth variable is unset. This means that the expression [ $depth -eq $zero ] becomes [ -eq 0 ] after bash substitutes the values of the variables into the expression. The problem here is that the -eq operator is incorrectly used as an operator with only one argument (the zero), but it requires two arguments. That is why you get the unary operator error message.

正如 Doktor J 在对此答案的评论中提到的，避免检查中未设置变量的问题的安全方法是将变量包含在""中.请参阅他的评论以获取解释.

As Doktor J mentioned in his comment to this answer, a safe way to avoid problems with unset variables in checks is to enclose the variables in "". See his comment for the explanation.

if [ "$depth" -eq "0" ]; then
   echo "false";
   exit;
fi

与[命令一起使用的未设置变量对bash来说是空的.您可以使用以下测试来验证这一点，因为xyz为空或未设置，因此所有测试均评估为true:

An unset variable used with the [ command appears empty to bash. You can verify this using the below tests which all evaluate to true because xyz is either empty or unset:

• if [ -z ] ; then echo "true"; else echo "false"; fi
• xyz=""; if [ -z "$xyz" ] ; then echo "true"; else echo "false"; fi
• unset xyz; if [ -z "$xyz" ] ; then echo "true"; else echo "false"; fi

• if [ -z ] ; then echo "true"; else echo "false"; fi
• xyz=""; if [ -z "$xyz" ] ; then echo "true"; else echo "false"; fi
• unset xyz; if [ -z "$xyz" ] ; then echo "true"; else echo "false"; fi

这篇关于检查bash变量是否等于0的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！