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php - mysqli_num_rows 语句的问题

转载 作者:行者123 更新时间:2023-11-29 01:38:55 25 4
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我很难弄清楚为什么我的 $numrows 变量出错。我收到这个错误..

警告:mysqli_num_rows() 期望参数 1 为 mysqli_result,给定为 null

对于这一行...

$numrows = mysqli_num_rows($stmt);

使我的 else 声明“该主题不存在显示”。

我的 $numrows 变量有什么问题?

$con = mysqli_connect("localhost", "root", "", "");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$cid = $_GET['cid'];
$tid = $_GET['tid'];
//$userid = ( isset( $_SESSION['user'] ) ? $_SESSION['user'] : "" );

//Prepared SELECT stmt
$stmt = $con->prepare("SELECT * FROM forum_topics WHERE `category_id`=? AND id=? LIMIT 1");
if ( !$stmt || $con->error ) {
die('Select topics prepare() failed: ' . htmlspecialchars($con->error));
}
if(!$stmt->bind_param('ii', $cid, $tid)) {
die('Select topics bind_param() failed: ' . htmlspecialchars($stmt->error));
}
if(!$stmt->execute()) {
die('Select topics execute() failed: ' . htmlspecialchars($stmt->error));
}
$numrows = mysqli_num_rows($stmt);
if($numrows == 1){
echo "<table width='100%'>";
if ( $_SESSION['user'] ) {
echo "<tr><td colspan='2'><input type='submit' value='Add Reply' onClick=\"window.location =
'forum_post_reply.php?cid=".$cid."$tid=".$tid."'\"> <hr />";
} else {
echo "<tr><td colspan='2'><p>Please log in to add your reply</p><hr /></td></tr>";
}
while($row = mysqli_fetch_assoc($stmt)){
//Prepared SELECT stmt to get forum topics
$stmt2 = $con->prepare("SELECT * FROM forum_posts WHERE `category_id`=? AND topic_id=?");
if ( !$stmt2 || $con->error ) {
die('Select topics prepare() failed: ' . htmlspecialchars($con->error));
}
if(!$stmt2->bind_param('ii', $cid, $tid)) {
die('Select topics bind_param() failed: ' . htmlspecialchars($stmt2->error));
}
if(!$stmt2->execute()) {
die('Select topics execute() failed: ' . htmlspecialchars($stmt2->error));
}
while($row2 = mysqli_fetch_assoc($stmt2)){
echo "<tr><td valign='top' style='border: 1px solid #000000;'>
<div style='min-height: 125px;'>".$row['topic_title']."<br />
by ".$row2['post_creator']." - " .$row2['post_date']. "<hr />" . $row2['post_content'] ."</div></td>
<td width='200' valign='top' align='center' style='border: 1px solid #000000;'>User Info Here!</td></tr>
<tr><td colspan='2'><hr /></td></tr>";
}
}
} else {
echo "<p>This topic does not exist.</p>";
}

最佳答案

您正在混合面向对象过程 风格。应该是——

$numrows = $stmt->num_rows;

DOCS

关于php - mysqli_num_rows 语句的问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31509119/

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