“意外的运算符"来自[$ var -eq 1] [英] "Unexpected operator" from [ $var -eq 1 ]

问题描述

我在bash脚本中使用了这个条件

  if [$ USE_TEST -eq 1];然后回声你好"科幻 

我在其中传递 USE_TEST 作为环境变量.

如果我传递env变量，那么shell将执行正常，但是如果我没有该变量，那么我会得到这个

  [:-eq:意外的运算符 

我该如何解决

解决方案

这是引号很重要的事情之一；当未定义$ USE_TEST时，您的语句将扩展为:

  if [-eq 1];然后 

常用解决方法:

  if ["$ USE_TEST" -eq 1];然后 

...尽管这也意味着切换到字符串比较，因为 -eq 将以空字符串中断:

 如果[[$ USE_TEST"= 1];然后 

但是您也可以考虑使用 [[(这是内置的bash)或 $ {USE_TEST:-0} (指定默认值).

I am using this conditional in my bash script

if [ $USE_TEST -eq 1 ]; then
echo "Hello"
fi

where i pass USE_TEST as an environment variable.

If i pass env variable then shell executes ok but if i don't have that variable then i get this

[: -eq: unexpected operator

How can i fix that

解决方案

This is one of those matter where quotes are important; when $USE_TEST is not defined, your statement expands to:

if [ -eq 1 ]; then

The common fix:

if [ "$USE_TEST" -eq 1 ]; then

...though that would also imply switching to a string comparison, because -eq will break with an empty string:

if [ "$USE_TEST" = 1 ]; then

But you may also consider using [[ (which is a bash builtin) or ${USE_TEST:-0} (which specifies a default value).

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