Question

我在数据属性中有json。它在http://jsonlint.com/等在线验证器中有效。它仍然存在错误。有什么问题？

HTML

<div class="seo" data-seo-controller='{"values":{"child_count":"0","type":"tables"},"title":{"field":"\u00e5\u00e4\u00f6 llasdas","template":"This page has {{child_count}} {{type}}!","prefix":"","suffix":" - Products","fallback":"\u00e5\u00e4\u00f6 llasdas","full":"\u00e5\u00e4\u00f6 llasdas - Products","full-replaced":"\u00e5\u00e4\u00f6 llasdas - Products"},"description":{"field":"rwerwe wer &quot; &#039; ewrerte fsd :&#039;","fallback":"rwerwe wer &quot; &#039; ewrerte fsd :&#039;","full":"rwerwe wer &quot; &#039; ewrerte fsd :&#039;","full-replaced":"rwerwe wer &quot; &#039; ewrerte fsd :&#039;","template":"","prefix":"","suffix":"","limit":155},"url":{"edit":"http:\/\/localhost\/seo\/panel\/pages\/saved-value\/url","preview":"seo\/saved-value"}}'>

的jQuery

$( document ).ready(function() {
   var json = $('.seo').attr('data-seo-controller');
   controller = jQuery.parseJSON( json );
   console.log(controller);
});

小提琴

https://jsfiddle.net/q89hph0L/

Uncaught SyntaxError：意外的令牌＆＃39;

更新

如果复制将json粘贴到字符串中并解析它，似乎有效。更新了小提琴：https://jsfiddle.net/q89hph0L/4/

那么，它不能将json作为数据属性吗？

Answer 1

"由浏览器呈现为"，请参阅：How to escape double quotes in title attribute。它与您的用例相同。

解决方案是将"替换为\"，使用jQuery.data时，您甚至不必使用JSON.parse()解析字符串。更新了jsfiddle：https://jsfiddle.net/0qryjrya/。

jQuery解析有效json中的json错误

1 个答案: