1

我对 jQuery 很陌生。我正在尝试使用一个名为 Searchable Dropdown 的插件,我从这里获得:http: //jsearchdropdown.sourceforge.net/ 。但我不知道如何使它工作..我做错了什么?

<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js" ></script>
<script src="scripts/jquery.searchabledropdown-1.0.8.min.js" type="text/javascript"></script>
<script>

    $(document).ready(function() {
        $("select").searchable();
    });
</script>

</head>

<?php
echo "<form method='post' action='' id='employeesselection'>
  <select    name='select_employee' id='select_employee'>";
       while($row=mysql_fetch_array($employees)){
            $selected = ($row['Id'] == $_POST['select_employee'])?'selected="selected"':'';
    echo '<option '.$selected.' value="'.$row['Id'].'">'.$row['Etunimi'].' - '. $row['Sukunimi'].'</option>';

}
   echo "</select></form>";

选择工作正常,但在选择中搜索 jQuery 应该启用,不起作用。

4

2 回答 2

0

也试着给你的代码一些语义:

<?php 

// DATABASE CONNECTION
// DATABASE QUERY

 ?>
<html lang="en">
<head>

<!-- CSS -->

</head>

<body>
    <form action="" method="post" id="employeesselection">
        <select name="select_employee" id="select_employee">
            <?php while ( $row = mysql_fetch_array( $employees ) ) {
                $selected = ( $row['id'] == $_POST['select_employee'] ) ? ' selected ' : '';
            ?>      
                <option <?php echo $selected ?> value="<?php echo $row['id'] ?>">
                    <?php echo $row['etunimi'] ?> - <?php echo $row['sukunimi'] ?>
                </option>
            <?php } ?>
        </select>
    </form>

<!-- All Javascript down here -->
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js" ></script>
<script src="scripts/jquery.searchabledropdown-1.0.8.min.js" type="text/javascript"></script>

<script>
    $(document).ready(function() {
            $("select").searchable();
    });
</script>
</body>
</html>

也许错误不在您的javascript中,而是在php中...

祝调试顺利!

于 2013-04-17T12:34:56.977 回答
0

我相信它与 Jquery 1.9.1 有关。

代替

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js" ></script>

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
于 2013-04-30T13:50:34.673 回答