宝剑锋从磨砺出，梅花香自苦寒来。这篇文章主要讲述AppleScript无法获得api响应的引用形式相关的知识，希望能为你提供帮助。
我今天早些时候制作了一个AppleScript，它显示了Geektools的YouTube订阅者数量，但我希望人们可以更轻松地使用它并尝试使其不受文件名称的影响（例如，采用子计数 -PewDiePie.scpt和输出PewDiePie的子计数），我已经从文件名中输入了名称，但是当我尝试从api的响应中取出数字时，它给了我错误
工作（原件）的代码
set apiResponse to (do shell script "curl -s 'https://www.googleapis.com/youtube/v3/channels?part=statistics&
forUsername=PewDiePie&
fields=items%2Fstatistics%2FsubscriberCount&
key=AIzaSyAEQGj2ZcDrTU0ZqzteD8eDVJwB9cpmvEo'")on returnNumbersInString(inputString)
set s to quoted form of inputString
do shell script "sed s/[a-zA-Z\']//g <
<
<
" &
s
set dx to the result
set numlist to {}
repeat with i from 1 to count of words in dx
set this_item to word i of dx
try
set this_item to this_item as number
set the end of numlist to this_item
end try
end repeat
end returnNumbersInStringreturnNumbersInString(apiResponse)
【AppleScript无法获得api响应的引用形式】破坏的可定制代码
set channelName to path to me as text
set oldDelims to AppleScript's text item delimiters
set AppleScript's text item delimiters to {"subcount-"}
set nameFilter to text items of channelName
set channelName to item 2 of nameFilterset AppleScript's text item delimiters to {"."}
set nameFilter to the text items of channelName
set channelName to item 1 of nameFilterset curlLink to "https://www.googleapis.com/youtube/v3/channels?part=statistics&
forUsername=" &
channelName &
"&
fields=items%2Fstatistics%2FsubscriberCount&
key=AIzaSyAEQGj2ZcDrTU0ZqzteD8eDVJwB9cpmvEo"
set curlCommand to "curl -s " &
(quoted form of curlLink)set apiResponse to {do shell script curlCommand}
on returnNumbersInString(inputString)
set s to quoted form of inputString
do shell script "sed s/[a-zA-Z\']//g <
<
<
" &
s
set dx to the result
set numlist to {}
repeat with i from 1 to count of words in dx
set this_item to word i of dx
try
set this_item to this_item as number
set the end of numlist to this_item
end try
end repeat
end returnNumbersInStringreturnNumbersInString(apiResponse)
每次我做第二次，它输出错误
Can’t get quoted form of {"{
"items": [
{
"statistics": {
"subscriberCount": "76957805"
}
}
]
}"}.
它从网站上获取信息后立即失败，这没有任何意义，因为除了如何获得网站链接之外的任何代码都没有改变，任何人都可以帮我解决这个问题
答案你在这里用括号括起你的
do shell script命令：set apiResponse to {do shell script curlCommand}
因此，
apiResponse现在是一个包含JSON字符串的列表，而不仅仅是一个JSON字符串。删除大括号，使行显示：set apiResponse to do shell script curlCommand
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