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使用 PHP/MySQL 創建 JSON 數據的正確方法

[英]Proper way to create JSON data with PHP/MySQL

根據以下答案更新:

根據下面的答案,我現在有以下 PHP 腳本:

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);
}

它返回以下內容:

{
    "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"
            }
        }
    }
}

它應該根據從數據庫返回的行數返回許多項目,為什么我只得到 1 個項目?

試試這個:

<?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;

?>

是的,它應該相當簡單,類似於

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

然后在你的while循環中你可以擁有,

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

最后對數組進行編碼,

json_encode($the_data);

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