这个问题在这里已经有了答案:
mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
(31 个回答)
6年前关闭。
我正在尝试用 php 和 mysql 分隔页面，但显示错误
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in D:\XAMPP\htdocs\all.php on line 17
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in D:\XAMPP\htdocs\all.php on line 58
我的 PHP 代码。
<html>
<head>
<title>All </title>
</head>
<body>
<?php
$dbhost = "localhost";
$dbusername = "root";
$dbpassword = "";
$dbname = "drill";
$mysqli = new mysqli ($dbhost, $dbusername, $dbpassword, $dbname);
$strSQL = "SELECT * FROM data ";
$objQuery = mysqli_query($mysqli,$strSQL);
$Num_Rows = mysqli_num_rows($objQuery);
$Per_Page = 2; // Per Page
$Page = $_GET["Page"];
if(!$_GET["Page"])
{
$Page=1;
}
$Prev_Page = $Page-1;
$Next_Page = $Page+1;
$Page_Start = (($Per_Page*$Page)-$Per_Page);
if($Num_Rows<=$Per_Page)
{
$Num_Pages =1;
}
else if(($Num_Rows % $Per_Page)==0)
{
$Num_Pages =($Num_Rows/$Per_Page) ;
}
else
{
$Num_Pages =($Num_Rows/$Per_Page)+1;
$Num_Pages = (int)$Num_Pages;
}
$strSQL .=" order by no ASC LIMIT $Page_Start , $Per_Page";// no is autoincrement
$objQuery = mysqli_query($mysqli,$strSQL);
?>
<table width="600" border="1">
<tr>
<th width="91"> <div align="center">no</div></th>
<th width="98"> <div align="center">campaign </div></th>
<th width="198"> <div align="center">url</div></th>
</tr>
<?php
while($objResult = mysqli_fetch_array($objQuery,MYSQLI_ASSOC))
{
?>
<tr>
<td><div align="center"><?php echo $objResult["no"];?></div></td>
<td><div align="center"><?php echo $objResult["camp"];?></div></td>
<td><div align="center"><?php echo $objResult["url"];?></div></td>
</tr>
<?php
}
?>
</table>
<br>
Total <?php echo $Num_Rows;?> Record : <?php echo $Num_Pages;?> Page :
<?php
if($Prev_Page)
{
echo " <a href='$_SERVER[SCRIPT_NAME]?Page=$Prev_Page'><< Back</a> ";
}
for($i=1; $i<=$Num_Pages; $i++){
if($i != $Page)
{
echo "[ <a href='$_SERVER[SCRIPT_NAME]?Page=$i'>$i</a> ]";
}
else
{
echo "<b> $i </b>";
}
}
if($Page!=$Num_Pages)
{
echo " <a href ='$_SERVER[SCRIPT_NAME]?Page=$Next_Page'>Next>></a> ";
}
mysqli_close($mysqli);
?>
</body>
</html>
最佳答案
尝试
if($objQuery){
$Num_Rows = mysqli_num_rows($objQuery);
}
关于php - 如何用 PHP 和 MYSQL 分离页面？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/35714905/