使用jQuery，PHP和MySQL为单选按钮加载JSON数据

小编典典

使用jQuery，PHP和MySQL为单选按钮加载JSON数据

ajax

我正在尝试填充第三组单选按钮，作为以下脚本的补充：http : //www.electrictoolbox.com/json-data-jquery-php-
radio-buttons/

由于某种原因，我似乎无法用相应的数据填充第三组。它只是保持空白:(

调用该 populateFruittype()函数仅会返回[ ]，而populateFruitVariety()正确返回json数据。

getdata.php（数据库连接/获取数据）

<?php

$dsn = "mysql:host=localhost;dbname=mydb";
$username = "username";
$password = "password";
$pdo = new PDO($dsn, $username, $password);

$rows = array();

if(isset($_GET['fruitName'])) {
    $stmt = $pdo->prepare("SELECT DISTINCT variety FROM fruit WHERE name = ? ORDER BY variety");
    $stmt->execute(array($_GET['fruitName']));
    $rows[] = $stmt->fetchAll(PDO::FETCH_ASSOC);
}

if(isset($_GET['fruitVariety'] )) {
    $stmt = $pdo->prepare("SELECT DISTINCT fruittype FROM fruit WHERE variety = ? ORDER BY fruittype");
    $stmt->execute(array($_GET['fruitVariety']));
    $rows[] = $stmt->fetchAll(PDO::FETCH_ASSOC);
}

echo json_encode($rows);

?>

的HTML

<!DOCTYPE html>
<html>
<head>
    <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"/>
    <title>Toevoegen</title>
    <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.1/jquery.min.js"></script>

    <script language="javascript" type="text/javascript">

        function populateFruitVariety() {

            var fruitName = $('input[name=fruitName]:checked').val();

            $.getJSON('getdata.php', {fruitName: fruitName}, function(fruit) {

                var html = '';
                $.each(fruit, function(index, array) {
                    html = html + '<label><input type="radio" name="fruitVariety" value="' + array['variety'] + '" />' + array['variety'] + '</label> ';
                });
                $('#varieties').html(html);

            });

        }

        function populateFruittype() {

            var fruitVariety = $('input[name=fruitVariety]:checked').val();

            $.getJSON('getdata.php', {fruitVariety: fruitVariety}, function(fruit) {

                var html = '';
                $.each(fruit, function(index, array) {
                    html = html + '<label><input type="radio" name="fruitType" value="' + array['fruittype'] + '" />' + array['fruittype'] + '</label> ';
                });
                $('#fruittype').html(html);

            });

        }

        $(function() {
            $('input[name=fruitName]').change(function() {
                populateFruitVariety();
            });
        });

        $(function() {
            $('input[name=fruitVariety]').change(function() {
                populateFruittype();
            });
        });


    </script>

</head>
<body>


<form>

    <div>
        <strong>Fruit:</strong>
        <label><input type="radio" name="fruitName" value="Apple"/>Apple</label>
        <label><input type="radio" name="fruitName" value="Banana"/>Banana</label>
        <label><input type="radio" name="fruitName" value="Orange"/>Orange</label>
        <label><input type="radio" name="fruitName" value="Pear"/>Pear</label>
    </div>
    <div>
        <strong>Variety:</strong>
        <span id="varieties"></span>
    </div>
    <div>
        <strong>Type:</strong>
        <span id="fruittype"></span>
    </div>
</form>
</body>
</html>

可以在以下位置找到数据库查询和内容：http: //www.electrictoolbox.com/mysql-example-
table/

只需添加：

`fruittype` varchar(50) NOT NULL

并用一些自定义值填充

阅读 219

2020-07-26

共1个答案

小编典典

问题解决了。

.change(function() 必须是 .live('click', function()

2020-07-26