Uncatch TypeError： Object function (a，b){return new e.fn.init(a，b，h)} 没有方法 'paresJSON'

Question

我使用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(正如我所理解的，也许我错了)所以它不会显示uXXX是使用parseJSON方法。

my JSON data is:

({"4":{"id":"88","label":"u00ebu00e9u00fau00e4 u00e0'","parent_id":"1","level":"1","children":[{"id":"92","label":"u00e4u00e9u00f1u00e8u00e5u00f8u00e9u00e4","parent_id":"88","level":"2","children":[{"id":"96","label":"u00eeu00e9u00fau00e5u00ecu00e5u00e2u00e9u00e4 u00f0u00e5u00f8u00e3u00e9u00fa","parent_id":"92","level":"3","children":[]}]},{"id":"90","label":"u00eeu00fau00eeu00e8u00e9u00f7u00e4","parent_id":"88","level":"2","children":[{"id":"97","label":"u00eeu00f9u00e5u00e5u00e0u00e5u00fa","parent_id":"90","level":"3","children":[{"id":"98","label":"u00f0u00e5u00f1u00e7u00e0u00e5u00fa","parent_id":"97","level":"4","children":[]}]}]}]},"5":{"id":"89","label":"u00ebu00e9u00fau00e4 u00e1'","parent_id":"1","level":"1","children":[]},"6":{"id":"91","label":"u00ebu00e9u00fau00e4 u00e2'","parent_id":"1","level":"1","children":[{"id":"93","label":"u00e4u00e9u00f1u00e8u00e5u00f8u00e9u00e4","parent_id":"91","level":"2","children":[{"id":"94","label":"u00e4u00e9u00f1u00e8u00e5u00f8u00e9u00e4 u00e9u00e4u00e5u00e3u00e9u00fa","parent_id":"93","level":"3","children":[]},{"id":"95","label":"u00e4u00e9u00f1u00e8u00e5u00f8u00e9u00e4 u00e9u00e5u00e5u00f0u00e9u00fa","parent_id":"93","level":"3","children":[]}]}]}});

和我的JS是:

    $(document).ready(function(){
$.getJSON(
    'some url.../getJSON.php?callback=?',
    function(data) {
        data = jQuery.paresJSON(data);
        $('#tree1').tree({
            data: data
        });
    }
);
});

此命令不输出任何内容，并显示上述错误。

知道我做错了什么吗?

提前感谢你们，Eric

Answer 1

您的代码使用paresJSON而不是parseJSON。

改正这个错别字，这样就能改正了