Question

根据以下答案进行了更新：

根据以下答案，我现在有以下PHP脚本：

header('Content-type:application/json');

function getdata($the_query)
{
    $connection = mysql_connect('server', 'user', 'pass') or die (mysql_error());
    $db = mysql_select_db('db_name', $connection) or die (mysql_error());

    $results = mysql_query($the_query) or die(mysql_error());

    header('Content-type:application/json');

    $the_data['rss']['channels']['title'] = $title;
    $the_data['rss']['channels']['link'] = $link;
    $the_data['rss']['channels']['description'] = $description;

    while($row = mysql_fetch_array($result))
    {
        extract($row);

        $the_data['rss']['channels']['items']['title'] = $item_title;
        $the_data['rss']['channels']['items']['link'] = "$item_link;
        $the_data['rss']['channels']['items']['date'] = $item_date;
        $the_data['rss']['channels']['items']['description'] = $item_description;
    }   

    mysql_close($connection);

    return json_encode($the_data);
}

返回以下内容：

{
    "rss":
    {
        "channels":
        {
            "title":"title goes here",
            "link":"link goes here",
            "description":"description goes here",
            "items":
            {
                "title":"'title goes here",
                "link":"link goes here",
                "date":"date goes here",
                "description":"description goes here"
            }
        }
    }
}

它应该根据从数据库返回的行数返回许多项目，为什么我只获得1项？

Answer 1

试试这个：

<?php
$channel = array(
     'title' => 'title goes here',
     'link' => 'link here',
     'description' => 'description',
     'items' => array()
);
while($row = mysql_fetch_array($results))
{
    extract($row);
    $channel['items'][] = array(
        'title' => $title,
        'link' => $link,
        'guid' => $guid,
        'pubDate' => $date,
        'description' => $description
    );
}   
$channels = array($channel);
$rss = (object) array('rss'=> array('channels'=>$channels));
$json = json_encode($rss);
echo $json;

?>

Answer 2

是的，它应该相当简单，类似于

$the_data['rss']['channels']['title'] = $title;
$the_data['rss']['channels']['link'] = $link;
$the_data['rss']['channels']['description'] = $desc;

然后在你的while循环内，你可以拥有，

$the_data['rss']['channels']['items'][] = $row;

最后对数组进行编码，

json_encode($the_data);

使用PHP / MySQL创建JSON数据的正确方法

2 个答案: