AppleScript 无法获得 api 响应的引用形式

Question

我今天早些时候制作了一个 AppleScript 来显示 Geektools 的 YouTube 订阅者数量，但我希望它更容易被人们使用，并试图让它根据文件名工作（例如，获取子计数-PewDiePie.scpt 并输出 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 中取出数字时它给了我错误回应

工作（原始）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 returnNumbersInString

returnNumbersInString(apiResponse)

损坏的可自定义代码

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 nameFilter

set AppleScript's text item delimiters to {"."}
set nameFilter to the text items of channelName
set channelName to item 1 of nameFilter

set 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 returnNumbersInString

returnNumbersInString(apiResponse)

每次我做第二个时它输出错误

Can’t get quoted form of {"{
 \"items\": [
  {
   \"statistics\": {
    \"subscriberCount\": \"76957805\"
   }
  }
 ]
}"}.

它从网站获取信息后立即失败，这没有任何意义，因为 none 获取网站 link 之外的代码已更改，任何人都可以帮我解决这个问题

Answer 1

您已将 do shell script 命令括在此处的大括号中：

set apiResponse to {do shell script curlCommand}

因此，apiResponse 现在是一个包含 JSON 字符串的列表，而不是简单的 JSON 字符串。删除大括号，使该行显示为：

set apiResponse to do shell script curlCommand

AppleScript 无法获得 api 响应的引用形式

AppleScript can't get quoted form of api response

applescript

youtube-api

youtube-data-api