未捕获的syntaxerror意外令牌U JSON

Question

当我在chrome中运行我的页面时,我收到此错误"未捕获的syntaxerror意外令牌U".在Firefox中,我得到了"JSON.parse:意外的角色".我从php文件返回json数据,返回的json字符串有效.我用http://jsonlint.com/检查了它.任何帮助将不胜感激...谢谢.

这是返回的JSON字符串

[
    ["1","Pan Africa Market","\"1521 1st Ave, Seattle, WA\"","47.608941","-122.340145","restaurant"],
    ["2","The Melting Pot","14 Mercer St, Seattle, WA","47.624562","-122.356442","restaurant"],
    ["3","Ipanema Grill","1225 1st Ave, Seattle, WA","47.606366","-122.337656","restaurant"],
    ["4","Sake House","230 1st Ave, Seattle, WA","47.612825","-122.34567","bar"],
    ["5","Crab Pot","1301 Alaskan Way, Seattle, WA","47.605961","-122.34036","restaurant"],
    ["6","Mexican Kitchen","2234 2nd Ave, Seattle,WA","47.613975","-122.345467","bar"],
    ["7","Wingdome","1416 E Olive Way, Seattle, WA","47.617215","-122.326584","bar"],
    ["8","Piroshky Piroshky","1908 Pike pl, Seattle, WA","47.610127","-122.342838","restaurant"]
]

Answer 1

当JSON.parse实际给出的值时,通常会看到该错误undefined.所以,我会检查试图解析它的代码 - 很可能你没有解析这里显示的实际字符串.

Answer 2

我在验证时收到此消息(在MVC项目中).对我来说,添加ValidationMessageFor元素修复了这个问题.

确切地说,jquery.validate.unobtrusive.js中的第43行导致了这个问题:

  replace = $.parseJSON(container.attr("data-valmsg-replace")) !== false;

Answer 3

JSON.parse 的参数可能什么都不返回（即 JSON.parse 的值是undefined）！

当我从 xyz.sol 文件解析 Compiled solidity 代码时，它发生在我身上。

import web3 from './web3';
import xyz from './build/xyz.json';

const i = new web3.eth.Contract(
  JSON.parse(xyz.interface),
  '0x99Fd6eFd4257645a34093E657f69150FEFf7CdF5'
);

export default i;

被拼错为

JSON.parse(xyz.intereface)

什么都没有返回！

Answer 4

发生此错误的最常见情况是使用模板，该模板先生成控件，然后更改方式id和/或name通过“覆盖”默认模板来生成，例如

@Html.TextBoxFor(m => m, new {Name = ViewData["Name"], id = ViewData["UniqueId"]} )

然后忘记更改ValidationMessageFor为

@Html.ValidationMessageFor(m => m, null, new { data_valmsg_for = ViewData["Name"] })    

希望这可以节省您一些时间。