bash Linux 脚本中 if 表达式中的 -z 值是什么意思？
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What means the -z value in an if expression on a Linux script?
提问by AndreaNobili
In this script I found this ifexpression:
在这个脚本中，我发现了这个if表达式：
if [ -z ]; then
echo "Usage: createpkg.sh <rev package>"
exit
else
CURRENT_VERSION=
fi
My problem is that I can't find what exactly means this -zvalue.
我的问题是我找不到这个-z值的确切含义。
From the content of the echo I can deduct that (maybe) $1variable represents the sotware version. and that (maybe) -zis a void value. So if I execute the script without passing to it the version of the software that I would packing it print me the correct procedure to execute the script.
从回声的内容中，我可以$1推断出（也许）变量代表软件版本。这（也许）-z是一个空值。因此，如果我执行脚本而不将其传递给它，我将打包它的软件版本会打印出执行脚本的正确程序。
But I am not sure about the real meaning of the -zvalue.
但我不确定该-z值的真正含义。
回答by fedorqui 'SO stop harming'
From man test:
来自man test：
-z STRING
the length of STRING is zero
So the condition:
所以条件：
if [ -z ]; then
means "if the variable $1 is empty". Where $1is probably the first parameter of the script: if you execute it like ./script <parameter1> <parameter2>, then $1=parameter1, $2=parameter2and so forth.
表示“如果变量 $1 为空”。哪里$1可能是脚本的第一个参数：如果你像 那样执行它./script <parameter1> <parameter2>，那么$1=parameter1，$2=parameter2等等。
回答by devnull
help testtells:
help test告诉：
String operators:
-z STRING True if string is empty.
In your example, the script would print Usage: createpkg.sh <rev package>and exit if an argument was not supplied.
在您的示例中，Usage: createpkg.sh <rev package>如果未提供参数，脚本将打印并退出。
