Uncaught TypeError: Object function (a,b){return new e.fn.init(a,b,h)} hat keine Methode 'paresJSON'

bin ich mit Hilfe von jQuery, um einige der JSON-Daten aus php-Datei.
meine Daten ut8 Kodieren und zu Dekodieren ( wie ich das verstanden hab, maybe im wrong) so ist es
nicht zeigen, dass die \uXXX ist die Verwendung parseJSON Methode.

mein JSON-Daten :

({"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":[]}]}]}});

und mein JS ist:

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

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

    }
);

});

diese Ausgänge nichts und zeigt mir den oben genannten Fehler.

irgendeine Idee, was habe ich falsch gemacht?

Vielen Dank im Voraus u Jungs,
Eric

  • der code verwendet paresJSON statt parseJSON
  • das scheint wie eine Antwort
  • als Antwort Hinzugefügt
InformationsquelleAutor eric.itzhak | 2011-11-17
Schreibe einen Kommentar