JSON.parse 中不允许使用哪些字符？

Question

所以在收到我的回复后，我一直收到解析错误。有非法字符吗？

这是回复

[{"businessID": ChIJ49DlQ5NiwokRQ_noyKqlchQ,"latitude": 40.733038,"longitude":-73.6840691,"address":"1201
Jericho Turnpike, New Hyde Park","businessname":"SUBWAY®Restaurants"},{"businessID": ChIJZfl6R5NiwokRZo7PU4NPoMY
,"latitude": 40.7329359,"longitude":-73.684513,"address":"1113 Jericho Turnpike, New Hyde Park","businessname"
:"Gino's"},{"businessID": ChIJcbpnRJNiwokRrtbOKe7HQo0,"latitude": 40.733049,"longitude":-73.684006,"address"
:"1203 Jericho Turnpike, New Hyde Park","businessname":"Wong's Garden"},]

这是我处理响应的函数。我确定它在警报之前就中断了，因为警报没有被触发

var datad = $(msg).text();
    console.log(datad);
    var resultstring = datad.replace(',]',']');
    var JsonParseData = JSON.parse(resultstring);
        alert(JsonParseData); ///BREAKING BEFORE THIS LINE

Answer 1

几个错误。

需要将字符串放在双引号中(")。将 "businessID": ChIJ49DlQ5NiwokRQ_noyKqlchQ 替换为 "businessID":"ChIJ49DlQ5NiwokRQ_noyKqlchQ"
删除下一行末尾的 , "businessname":"Wong's Garden"},]

Answer 2

JSON的键值需要引号。 JSON 数据中的引号少了，终于多了一个逗号和一个回车 return

这样是对的：

[{"businessID":"ChIJ49DlQ5NiwokRQ_noyKqlchQ","latitude":"40.733038","longitude":"-73.6840691","address":"1201 Jericho Turnpike, New Hyde Park","businessname":"SUBWAY®Restaurants"},{"businessID":"ChIJZfl6R5NiwokRZo7PU4NPoMY","latitude":"40.7329359","longitude":"-73.684513","address":"1113 Jericho Turnpike, New Hyde Park","businessname":"Gino's"},{"businessID":"ChIJcbpnRJNiwokRrtbOKe7HQo0","latitude":"40.733049","longitude":"-73.684006","address":"1203 Jericho Turnpike, New Hyde Park","businessname":"Wong's Garden"}]

Answer 3

您的回复是无效的 json 格式，原因有两个：

"businessID" 的值需要引号。
JSON 的最后一个对象后不应有逗号（您的替换字符串函数修复了此问题）。

我推荐你使用这个 JSON 工具包：

http://jsonviewer.stack.hu/（这命令我的 json 虽然不正确）
https://jsonformatter.org/（我经常用这个，我的最爱）

JSON.parse 中不允许使用哪些字符？

What characters arent allowed in a JSON.parse?

javascript

php

google-geocoding-api

google-places-api