将参数传递到简单 ShellScript 时出现问题（未找到命令）

发布于 2024-12-12 17:48:15 字数 324 浏览 0 评论 0原文

我正在尝试编写一个简单的 shell 脚本，如果有，则打印出第一个参数，如果没有，则打印“none”。该脚本称为 test.sh

    if [$1 = ""]
    then
        echo "none"
    else
        echo $1
    fi

如果我运行不带参数的脚本，一切正常。但是，如果我运行此命令 source test.sh -test，则会在脚本继续运行并正确回显测试之前收到此错误 -bash: [test: command not found。我做错了什么？

原文

I am trying to write a simple shell-script that prints out the first parameter if there is one and prints "none" if it doesn't. The script is called test.sh

    if [$1 = ""]
    then
        echo "none"
    else
        echo $1
    fi

If I run the script without a parameter everything works. However if I run this command source test.sh -test, I get this error -bash: [test: command not found before the script continues on and correctly echos test. What am I doing wrong?

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要走就滚别墨迹 2024-12-19 17:48:15

你需要在'[',']'字符之前/之后有空格，

if [ "$1" = "" ] ; then
#---^---------^ here
   echo "none"
else
    echo "$1"
fi

即你需要将你的引用（实际上是所有引用）包装到$1，并使用上面编辑的引号。

修复该问题后，您可能还需要提供脚本的相对路径，即

source ./test.sh -test
#------^^--- there

当您收到 shell 错误消息时，使用 set -vx 来打开 shell 调试几乎总是有帮助的> 在引起麻烦的行之前，或者非常靠近脚本顶部的地方。然后您可以看到正在执行的每一行/代码块，以及 shell 正在使用的变量的值。

我希望这有帮助。

you need spaces before/after '[',']' chars, i.e.

if [ "$1" = "" ] ; then
#---^---------^ here
   echo "none"
else
    echo "$1"
fi

And you need to wrap your reference (really all references) to $1 with quotes as edited above.

After you fix that, you may also need to give a relative path to your script, i.e.

source ./test.sh -test
#------^^--- there

When you get a shell error message has you have here, it almost always helps to turn on shell debugging with set -vx before the lines that are causing your trouble, OR very near the top your script. Then you can see each line/block of code that is being executed, AND the value of the variables that the shell is using.

I hope this helps.

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