[英]How can I resolve the “Resource ID #8” error message in PostgreSQL/PHP?
[英]How can I resolve the “Resource ID #8” error message in MYSQL/PHP?
function show_post($user_id)
{
global $database;
$post = array();
$sql = sprintf("SELECT body,stamp FROM post WHERE user_id = '%s' ORDER BY stamp DESC ",$user_id);
$result = $database->query($sql);
while( $data = $database -> fetch_array($result))
{
$post[] = array(
'user_id' => $user_id,
'body' => $data->'body',
'stamp' => $data->'stamp'
);
}
return $post;
}
$posts = $link->show_post($_SESSION['user_id']);
if(count($posts))
{
?>
<table border='1' cellspacing='0' cellpadding='5' width='500'>
<?php
foreach($posts as $key => $list)
{
echo "<tr valign='top'>\n";
echo "<td>".$list['user_id'] ."</td>\n";
echo "<td>".$list['body'] ."<br/>\n";
echo "<small>".$list['stamp'] ."</small></td>\n";
echo "</tr>\n";
}
?>
</table>
<?php
}else
{
echo "nothing entered";
}
这将返回:资源ID#8
您能协助我找到问题吗?
添加此内容,看看它回显了什么
function show_post($user_id)
{
global $database;
$post = array();
$sql = sprintf("SELECT body,stamp FROM post WHERE user_id = '%s' ORDER BY stamp DESC ",$user_id);
$result = $database->query($sql);
while( $data = $database -> fetch_array($result))
{
echo '<pre>'; print_r($data); echo '</pre><br />';
$post[] = array(
'user_id' => $user_id,
'body' => $data->'body',
'stamp' => $data->'stamp'
);
}
return $post;
}
除非那个print_r($ data)打印出奇怪的东西,否则我看不出这段代码有什么问题。 可能与您使用的数据库类有关。
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