所以我收到了错误
警告:mysqli_error()需要1个参数,0在第21行的C:\ wamp \ www \ xxx \ xxx中给出
在我的页面上,我不明白为什么。
我知道这可能不足以让你们为我解决这个问题,但也许你可以向我倾斜正确的方向?
以下是发生错误的整个PHP块:
<?php
require_once("nbbc/nbbc.php");
$bbcode = new BBCode;
$sql = "SELECT * FROM posts ORBER BY id DESC";
$res = mysqli_query($dbCon, $sql) or die(mysqli_error());
$posts = "";
if(mysqli_num_rows($res) > 0) {
while($row = mysqli_fetch_assoc($res)) {
$id = $row['id'];
$title = $row['title'];
$content = $row['content'];
$date = $row['date'];
$admin = "<div><a href='del_post.php?pid=$id'>Delete</a> <a href='edit_post.php?pid=$id'>Edit</a></div>";
$output = $bbcode->Parse($content);
$posts .= "<div><h2><a href='view_post.php?pid=$id'>$title</a></h2><h3>$date</h3><p>$output</p>$admin</div>";
}
echo $posts;
}else {
echo "There are no posts to be displayed.";
}
?>
答案 0 :(得分:0)
程序mysqli_error(...)
需要 mysqli链接($dbCon
),请参阅此处:
http://php.net/manual/en/mysqli.error.php
答案 1 :(得分:0)
你的语法错了。您需要将$dbCon
传递给mysqli_error()
从
改变$res = mysqli_query($dbCon, $sql) or die(mysqli_error());
要
$res = mysqli_query($dbCon, $sql) or die(mysqli_error($dbCon));
<强>语法强>
程序风格
string mysqli_error ( mysqli $link )
面向对象的风格
string $mysqli->error;
<强>更新强>
您的查询中有拼写错误。将ORBER
更改为ORDER
从
改变$sql = "SELECT * FROM posts ORBER BY id DESC";
要
$sql = "SELECT * FROM posts ORDER BY id DESC";