将本地源更改为数据库源

时间:2013-12-30 15:46:12

标签: javascript bootstrap-typeahead typeahead.js

如何将源(第13行)从本地更改为我的数据库?

我希望从数据库中检索建议。

我也希望建议样式与typeahead.js类似。

这是我目前的代码:

<html>

    <link href="//netdna.bootstrapcdn.com/bootstrap/3.0.0/css/bootstrap.min.css" rel="stylesheet" media="screen">
    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
    <script src="//netdna.bootstrapcdn.com/bootstrap/3.0.0/js/bootstrap.min.js"></script>
    <link href="//raw.github.com/jharding/typeahead.js-bootstrap.css/master/typeahead.js-bootstrap.css" rel="stylesheet" media="screen">
    <script src="//twitter.github.com/typeahead.js/releases/latest/typeahead.min.js"></script>

    <script type="text/javascript">
        $(document).ready(function() {
            $('.q.typeahead').typeahead({                               
                name: "q",                
                local: ['italy', 'france', 'england', 'united states', 'brazil', 'spain']                                              
            });
        });
    </script>

    <script type="text/javascript">

        //document.getElementById("suggestion")

            function getSuggestion(q) {

            var ajax;
            if(window.XMLHttpRequest)//for ie7+, FF, Chrome
                ajax = new XMLHttpRequest();//ajax object
            else
                ajax = new ActiveXObject("Microsoft.XMLHTTP");//for ie6 and previous
            ajax.onreadystatechange = function() {
                if(ajax.status === 200 && ajax.readyState === 4) {
                    //if result are not there then don't display them
                    if(ajax.responseText === "")
                        document.getElementById("suggestion").style.visibility = "hidden";
                    else {
                        document.getElementById("suggestion").style.visibility = "visible";
                        document.getElementById("suggestion").innerHTML = ajax.responseText;
                    }
                }
            };
            ajax.open("GET", "suggestion.php?q=" + q, false);
            ajax.send();
        }
    </script>

    <input type="text" class="q typeahead" id="citiesInput" />

</html>

提前致谢。

0 个答案:

没有答案