你好，我是 MySQL 和 PHP 新手，正在接受培训，请告诉我哪里错了
我的问题出现在 ad01 和 ad03 中，它们是 echo Advertise Here，但 ad02 正确 echo
<?php require_once('Connections/localhost.php'); ?>
<?php
mysql_select_db($database_localhost, $localhost);
$query_advtDisplay = "SELECT * FROM advt";
$advtDisplay = mysql_query($query_advtDisplay, $localhost) or die(mysql_error());
$row_advtDisplay = mysql_fetch_assoc($advtDisplay);
$totalRows_advtDisplay = mysql_num_rows($advtDisplay);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Download Links</title>
<link href="css/style.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div class="wrapper">
<div class="ad01">
<?php if ($row_advtDisplay['advt-no']=='ad01')
{
echo $row_advtDisplay['advt-content'];
}
else{ echo "Advertise Here";}
?>
</div>
<div class="middlebox">
<div class="ad02">
<?php if ($row_advtDisplay['advt-no']=='ad02')
{
echo $row_advtDisplay['advt-content'];
}
else{ echo "Advertise Here";}
?>
</div>
<div class="linkbox">
<p>Download Links</p>
<ul>
<li><a href="#">Link 01</a></li>
<li><a href="#">Link 02</a></li>
<li><a href="#">Link 03</a></li>
<li><a href="#">Link 04</a></li>
<li><a href="#">Link 05</a></li>
</ul>
<ul>
<li><a href="#">Link 06</a></li>
<li><a href="#">Link 07</a></li>
<li><a href="#">Link 08</a></li>
<li><a href="#">Link 09</a></li>
<li><a href="#">Link 10</a></li>
</ul>
<ul>
<li><a href="#" target="_blank">Link 11</a></li>
<li><a href="#" target="_blank">Link 12</a></li>
<li><a href="#" target="_blank">Link 13</a></li>
<li><a href="#" target="_blank">Link 14</a></li>
<li><a href="#" target="_blank">Link 15</a></li>
</ul>
</div>
<div class="passwordbox">
<p>RAR Password</p>
</div>
</div>
<div class="ad03"><?php if ($row_advtDisplay['advt-no']=='ad03')
{
echo $row_advtDisplay['advt-content'];
}
else{ echo "Advertise Here";}
?></div>
<div class="clear"></div>
</div>
</body>
</html>
<?php
mysql_free_result($advtDisplay);
?>
只有 ad02 正确显示，ad01 和 ad03 不显示仅表编码问题没有问题抱歉英语不好
最佳答案
mysql_fetch_assoc只会返回一行...因此您需要循环查看所有数据。
$just_one_row = mysql_fetch_assoc($advtDisplay);
// how to loop through all the rows
while ($row = mysql_fetch_assoc($advtDisplay)) {
echo $row["advt-no"];
echo $row["advt-content"];
}
// maybe do something like...
$mydata = array();
while ($row = mysql_fetch_assoc($advtDisplay)) {
$mydata[ $row['advt-no'] ] = $row['advt-content'];
}
<!-- and then -->
<div class="ad01">
<?php
if ( isset($mydata['ad01']) && !empty($mydata['ad01']) ) {
echo $mydata['ad01'];
} else { echo "Advertise Here"; } ?>
</div>
关于php - MySQL和PHP显示错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/26309285/