Question

我正在使用jQuery从php文件中获取一些JSON数据。我的数据是ut8编码并解码它（因为我已经明白，也许我错了）所以它不会显示\ uXXX是使用parseJSON方法。

我的JSON数据是：

({"4":{"id":"88","label":"\u00eb\u00e9\u00fa\u00e4 \u00e0'","parent_id":"1","level":"1","children":[{"id":"92","label":"\u00e4\u00e9\u00f1\u00e8\u00e5\u00f8\u00e9\u00e4","parent_id":"88","level":"2","children":[{"id":"96","label":"\u00ee\u00e9\u00fa\u00e5\u00ec\u00e5\u00e2\u00e9\u00e4 \u00f0\u00e5\u00f8\u00e3\u00e9\u00fa","parent_id":"92","level":"3","children":[]}]},{"id":"90","label":"\u00ee\u00fa\u00ee\u00e8\u00e9\u00f7\u00e4","parent_id":"88","level":"2","children":[{"id":"97","label":"\u00ee\u00f9\u00e5\u00e5\u00e0\u00e5\u00fa","parent_id":"90","level":"3","children":[{"id":"98","label":"\u00f0\u00e5\u00f1\u00e7\u00e0\u00e5\u00fa","parent_id":"97","level":"4","children":[]}]}]}]},"5":{"id":"89","label":"\u00eb\u00e9\u00fa\u00e4 \u00e1'","parent_id":"1","level":"1","children":[]},"6":{"id":"91","label":"\u00eb\u00e9\u00fa\u00e4 \u00e2'","parent_id":"1","level":"1","children":[{"id":"93","label":"\u00e4\u00e9\u00f1\u00e8\u00e5\u00f8\u00e9\u00e4","parent_id":"91","level":"2","children":[{"id":"94","label":"\u00e4\u00e9\u00f1\u00e8\u00e5\u00f8\u00e9\u00e4 \u00e9\u00e4\u00e5\u00e3\u00e9\u00fa","parent_id":"93","level":"3","children":[]},{"id":"95","label":"\u00e4\u00e9\u00f1\u00e8\u00e5\u00f8\u00e9\u00e4 \u00e9\u00e5\u00e5\u00f0\u00e9\u00fa","parent_id":"93","level":"3","children":[]}]}]}});

我的JS是：

    $(document).ready(function(){
$.getJSON(
    'some url.../getJSON.php?callback=?',
    function(data) {
        data = jQuery.paresJSON(data);

        $('#tree1').tree({
            data: data
        });

    }
);

});

这没有输出任何内容，并向我显示上述错误。

任何想法我做错了什么？

先谢谢你们，埃里克

Answer 1

您的代码使用paresJSON代替parseJSON。

修正错字，这应该修复它。

未捕获的TypeError：对象函数（a，b）{return new e.fn.init（a，b，h）}没有方法'paresJSON'

1 个答案: