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; ?>
