当我在另一个中调用我的php脚本时,选择js不起作用

时间:2014-05-07 07:46:46

标签: javascript php

我有这个代码在独立运行时工作正常(我可以在下拉列表中搜索)

<?php

include_once ("db_connection.php");

$conn = testdb_connect ();

$sth = $conn->prepare('SELECT testCase FROM tooldata ');

$sth->execute();

    while($row = $sth->fetch(PDO::FETCH_ASSOC))
        { 
        foreach($row as $key) 
            {
                $var=explode('_', $key);
                $feature[] =  $var[0];  
            } 
        }
$feature = array_intersect_key($feature, array_unique(array_map('strtolower', $feature)));
    foreach($feature as $newarr) 
        {
            $newFeature[]=$newarr;
        }
?>

<html>
<head>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
    <link href="../select2-3.4.8/select2.css" rel="stylesheet"/>
    <script src="../select2-3.4.8/select2.js"></script>
    <script>
        $(document).ready(function() { $("#feature").select2(); });
    </script>
</head>

<body>
<h2>Select feature</h2>
<form method="get" action="feature.php">
<select name="feature" id="feature">
        <?php
        ?>
        <option value > Select Feature</option>
        <?php
        foreach($newFeature as $feat)
        {
            ?>
        <option value="<?php echo $feat;?>"><?php echo $feat;?></option>
            <?php
       }
         ?>
</select>
<input type="submit" name="submit" value="Feature">
</body>
</html>

当我在其他php脚本中包含这个脚本时说main.php,当我运行main.php时,我在下拉列表中的搜索不起作用,我在网络选项卡中得到了这个 - 未捕获TypeError:undefined is不是第7行和第37行的功能,分别是 include(“feature_list.php”); }

可能的原因是什么? 请指导

0 个答案:

没有答案