1

我正在尝试使用 JQuery 创建一个自动完成字段，该字段从 mysql 数据库接收其自动完成。

索引.php： 

        <!-- Place favicon.ico and apple-touch-icon.png in the root directory -->

        <link rel="stylesheet" href="css/normalize.css">
        <link rel="stylesheet" href="css/main.css">
        <link rel="stylesheet" href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.9.0/themes/base/jquery-ui.css">
        <link rel="stylesheet" href="css/bootstrap.min.css">
        <script src="js/vendor/modernizr-2.6.2.min.js"></script>
        <script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
        <script src="//ajax.googleapis.com/ajax/libs/jqueryui/1.9.1/jquery-ui.min.js"></script>
        <script>window.jQuery || document.write('<script src="js/vendor/jquery-1.8.2.min.js"><\/script>')</script>
        <script src="js/plugins.js"></script>
        <script src="js/main.js"></script>
        <script type="text/javascript">
        $(function() {

            $("#college").autocomplete({
                source: "search.php",
                minLength: 2
            });
        });
        </script>
    </head>
    <body>

        <form action="<?php echo $_SERVER['PHP_SELF'];?>"  method="post">
            <fieldset>
            <legend>jQuery UI Autocomplete Example - PHP Backend</legend>

            <label for="state">State (abbreviation in separate field): </label>

            <input type="text" id="college"  name="college" />

            <input type="submit" name="submitBtn" value="Submit" />

            </fieldset>
        </form>

        <?php
if (isset($_POST['submit'])) {
echo "<p>";
    while (list($key,$value) = each($_POST)){
    echo "<strong>" . $key . "</strong> = ".$value."<br />";
    }
echo "</p>";
}
?>




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        <![endif]-->

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            s.parentNode.insertBefore(g,s)}(document,'script'));
        </script>
    </body>
</html>

搜索.php： 

/* Connection vars here for example only. Consider a more secure method. */
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$dbname = 'college';

try {
  $conn = new PDO("mysql:host=$dbhost;dbname=$dbname", $dbuser, $dbpass);
}
catch(PDOException $e) {
    echo $e->getMessage();
}

$return_arr = array();

if ($conn)
{
    $ac_term = "%".$_GET['term']."%";
    $query = "SELECT * FROM college_list where name like :term";
    $result = $conn->prepare($query);
    $result->bindValue(":term",$ac_term);
    $result->execute();

    /* Retrieve and store in array the results of the query.*/
    while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
        $row_array['name'] = $row['name'];

        array_push($return_arr,$row_array);
    }


}
/* Free connection resources. */
//$conn = null; 
/* Toss back results as json encoded array. */
echo json_encode($return_arr);

?>

它目前正在从数据库中正确检索，因为它获得了正确数量的自动完成。但是，它们都是空白的。它似乎实际上并没有将“名称”反馈到该领域，我似乎无法弄清楚为什么。有任何想法吗？

4

1 回答 1

1

在没有看到您正在使用的自动完成代码的情况下，我会说您需要一个一维数组而不是您现在生成的二维数组。

您可以尝试更改：

while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
    $row_array['name'] = $row['name'];

    array_push($return_arr,$row_array);
}

到：

while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
    array_push($return_arr, $row['name']);
}

编辑：检查api 文档，您需要以不同的方式构建数组，例如：

while ($row = $result->fetch(PDO::FETCH_ASSOC)) {
    array_push($return_arr, array('label' => $row['name'], 'value' => $row['name']));
}
于 2012-11-17T19:09:33.847 回答