Question

我在我的应用程序中使用typeahead，我已将其包含在

中

<link rel="stylesheet" href="https://twitter.github.io/typeahead.js/css/examples.css"/> 
<script src="https://twitter.github.io/typeahead.js/releases/latest/typeahead.bundle.js"></script>

但我仍然得到typeahead不是函数错误。我甚至包括了jquery 1.10.2版本文件。以下是我的代码

<script>
var docarray=[];
var condarray=[];
var specarray=[];
   $(document).ready(function() {
    var doctorjsonurl = "http://xxxxxxxxxxx/doctorfeed/doctortypeahead.jsp";//just for security
    $.ajax({
        type :"GET",
        dataType: "json",
        url : "/childrensplugin/proxy.jsp?reqUrl="+encodeURIComponent(doctorjsonurl),
        success : function(data){
         for(i=0;i<data.doctornames.length;i++){
              docarray.push(data.doctornames[i]);
            }
            var substringMatcher = function(strs) {
                return function findMatches(q, cb) {
                    var matches, substringRegex;                       
                    matches = [];                        
                    substrRegex = new RegExp(q, 'i');                        
                    $.each(strs, function(i, str) {
                        if (substrRegex.test(str)) {
                            matches.push(str);
                        }
                    });
                    cb(matches);
                };
            };
            $('#the-doc .typeaheaddoc').typeahead({
                hint: true,
                highlight: true,
                minLength: 3
            },
            {
                name: '',
                source: substringMatcher(docarray)
            });
            },
        error: function (xhr, ajaxOptions, thrownError) {
         //   alert(xhr.status);
         //   alert(thrownError);
        }
    });
var condjsonurl = "http://xxxxxxxx/doctorfeed/condtreatsforcondtreatsearchtypeahead.jsp";
    $.ajax({
        type :"GET",
        dataType: "json",
        url : "/childrensplugin/proxy.jsp?reqUrl="+encodeURIComponent(condjsonurl),
        success : function(data){

         for(i=0;i<data.allconditions.length;i++){
              specarray.push(data.allconditions[i]);
            }

            for(i=0;i<data.alltreatments.length;i++){
              specarray.push(data.alltreatments[i]);
            }   

            var substringMatcher = function(strs) {
                return function findMatches(q, cb) {
                    var matches, substringRegex;                       
                    matches = [];                       
                    substrRegex = new RegExp(q, 'i');                        
                    $.each(strs, function(i, str) {
                        if (substrRegex.test(str)) {
                            matches.push(str);
                        }
                    });
                    cb(matches);
                };
            };
            $('#the-cond .typeaheadcond').typeahead({
                hint: true,
                highlight: true,
                minLength: 1
            },
            {
                name: '',
                source: substringMatcher(specarray)
            });
            },
        error: function (xhr, ajaxOptions, thrownError) {
         //   alert(xhr.status);
         //   alert(thrownError);
        }
    });

});
</script>

有谁能告诉我这里我缺少的东西..

Answer 1

将标记放在头部
优先将代码置于

未捕获的TypeError：$（...）。typeahead不是函数错误

1 个答案: