是什么导致Apple脚本的自定义ZSH函数中引起此错误
我正在尝试制作一个ZSH函数，我可以通过终端使用终端来退出给定的应用程序，就像该应用程序使用＆lt; cmd，q＆gt一样退出给定应用程序。快捷方式（因此不仅仅是pkill）。
我使用了如何从在这里，我确认我的zsh函数/参数使用语法似乎是正确的href =“ https://stackoverflow.com/questions/55127741/how-do-you-pass-arguments-to-custom-zsh-functions#:%7e：text=how%20I%20Would%20WOUDD%20WRITE； ％3A“>此问题和
quitapp() {
osascript -e 'quit app ${1:?"The application must be specified."}'
}
但是，我通过尝试退出Spotify进行了测试，但是当我尝试这样做时，
quitapp Spotify
我遇到了此错误：（
9:10: syntax error: Expected expression, property or key form, etc. but found unknown token. (-2741)
我尝试使用小写和大写，并尝试将Spotify包裹在双引号或不带双引号的情况下，并且总是相同的错误）。
我在做什么错？
I am trying to make a zsh function that I can pass an argument to in order to quit a given app using the terminal in the same way that the app quits with the <CMD,Q> shortcut (so not just pkill).
I used the example of how to quit apps in macOS terminal from here and I confirmed that my Zsh function/argument use syntax seems correct from looking at this question and this documentation
quitapp() {
osascript -e 'quit app ${1:?"The application must be specified."}'
}
However, I tested it by trying to quit Spotify but when I tried to do so like this:
quitapp Spotify
I got this error:
9:10: syntax error: Expected expression, property or key form, etc. but found unknown token. (-2741)
(I tried using both lowercase and uppercase, and tried to enclose Spotify in double quotes or without double quotes, and always the same error).
What am I doing wrong?
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在苦苦挣扎的同时，我偶然发现了 this ，我认识到我的错误是认为字符串会被评估为外壳代码通过之前。
但是我可以解决我的问题（归功于我链接到的SO答案）：
While struggling some more I stumbled upon this and I recognize that my mistake was thinking that string would be evaluated as shell code before being passed.
But I can just solve my question with (credit to the SO answer I linked to):