bash 将参数传递到 Simple ShellScript 时出现问题（找不到命令）
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Trouble Passing Parameter into Simple ShellScript (command not found)
提问by Spencer
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
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”。该脚本名为 test.sh
if [ = ""]
then
echo "none"
else
echo
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 foundbefore the script continues on and correctly echos test. What am I doing wrong?
如果我在没有参数的情况下运行脚本，一切正常。但是，如果我运行此命令source test.sh -test，则会-bash: [test: command not found在脚本继续运行并正确响应测试之前收到此错误。我究竟做错了什么？
回答by shellter
you need spaces before/after '[',']' chars, i.e.
在 '[',']' 字符之前/之后需要空格，即
if [ "" = "" ] ; then
#---^---------^ here
echo "none"
else
echo ""
fi
And you need to wrap your reference (really all references) to $1 with quotes as edited above.
并且您需要使用上面编辑的引号将您的参考文献（实际上是所有参考文献）包装到 $1 中。
After you fix that, you may also need to give a relative path to your script, i.e.
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
When you get a shell error message has you have here, it almost always helps to turn on shell debugging with set -vxbefore 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.
当您在此处收到 shell 错误消息时set -vx，在导致您遇到问题的行之前或在脚本顶部附近打开 shell 调试几乎总是有帮助的。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。
I hope this helps.
我希望这有帮助。
