嗨,我想在同一页面上创建页面意味着我想将数据库中的图像显示到 PHP 页面,并且我想每页显示 20 张图像,因此当用户单击 2,3,4,5 转到下一页时,这里是我正在尝试我的代码,但它给了我一些错误。
PHP 代码
<?php
$con=mysqli_connect("localhost","root","123","user");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; };
$start_from = ($page-1) * 20;
$sql = "SELECT * FROM save_data ORDER BY ID DESC LIMIT $start_from, 20";
$rs_result = mysql_query ($sql,$con);
while ($row = mysql_fetch_assoc($rs_result)) {
$post_id = $row['ID'];
$title = $row['Title'];
$image = $row['Name'];
?>
<center>
<a href="pictures.php?title=<?php echo $title; ?>">
<h3><?php echo $title; ?></h3></a>
<a href="pictures.php?title=<?php echo $title; ?>">
<img src='uploads/<?php echo $image; ?>' width='140' height='140'></a>
</center>
<?php
};
?>
<?php
$sql = "SELECT COUNT(ID) FROM save_data";
$rs_result = mysql_query($sql,$con);
$row = mysql_fetch_row($rs_result);
$total_records = $row[0];
$total_pages = ceil($total_records / 20);
for ($i=1; $i<=$total_pages; $i++) {
echo "<a href='pages.php?page=".$i."'>".$i."</a> ";
};
?>
我遇到的错误
Warning: mysql_query() expects parameter 2 to be resource, object given in C:\xampp\htdocs\mysql_login\all-images.php on line 15
Warning: mysql_fetch_assoc() expects parameter 1 to be resource, null given in C:\xampp\htdocs\mysql_login\all-images.php on line 17
Warning: mysql_query() expects parameter 2 to be resource, object given in C:\xampp\htdocs\mysql_login\all-images.php on line 41
Warning: mysql_fetch_row() expects parameter 1 to be resource, null given in C:\xampp\htdocs\mysql_login\all-images.php on line 42
我发现一些代码错误:
你正在混合 mysql 和 mysqli, 坚持下去
$rs_result = mysqli_query ($con, $sql);
while($row = mysqli_fetch_row($rs_result)){
echo $row["table field name"];
}