JSON.parse中不允许使用什么字符？

Question

因此，收到我的回应后，我不断收到解析错误。有没有非法字符？

这里是响应

[{"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"},] 

这里是我的功能是处理响应。我知道肯定它的提前警报，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。

1. 需要在双引号中输入字符串（"）。更换"businessID": ChIJ49DlQ5NiwokRQ_noyKqlchQ与"businessID":"ChIJ49DlQ5NiwokRQ_noyKqlchQ"

2. 在下面一行的末尾"businessname":"Wong's Garden"},]

Answer 2

JSON的键值需要有引号删除,。你必须在JSON数据更少的报价，最后你有一个更逗号和一个更回车

这样是正确的：

[{"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

你的反应是有两个原因的无效JSON格式：

• 的 “businessID” 的值需要加上引号。
• JSON的最后一个对象（您的替换字符串函数修复此问题）后不应该有逗号。

，我建议你使用这个JSON工具包：

• http://jsonviewer.stack.hu/（该订单的json虽然是不正确的）
• https://jsonformatter.org/（我用这个有很多，我最喜欢的）