使用PHP / MySQL创建JSON数据的正确方法

发布时间：2020-12-13 13:29:38 所属栏目：PHP教程来源：网络整理

导读：根据以下答案更新：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()); $db = mysql_select_db('db_name',$connec

根据以下答案更新：

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","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？

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;

?>

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!