嗨，我已经做了一些 AJAX、PHP&MySQL 排序，它在表格中给出了结果，如下面的代码所示，我的问题是 如何将 $result 引入 html div.
请帮忙
使用 PHP 代码
<?php
$q=$_GET["q"];
$con = mysql_connect('localhost', 'root', '');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("security_software", $con);
$sql="SELECT * FROM internet_security ORDER by '".$q."' DESC" ;
$result = mysql_query($sql);
echo "<table border='1'>
<tr>
<th>id</th>
<th>title</th>
<th>image</th>
<th>description</th>
<th>rating</th>
<th>download</th>
<th>buy</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['title'] . "</td>";
echo "<td>" . $row['image'] . "</td>";
echo "<td>" . $row['description'] . "</td>";
echo "<td>" . $row['rating'] . "</td>";
echo "<td>" . $row['download'] . "</td>";
echo "<td>" . $row['buy'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>
我想要这些 HTML Div 的结果
<div class="category-container">
<div class="category-image"></div>
<div class="category-link"><a href="#">#</a></div>
<div class="category-desc"><p>#</p> </div>
<div class="rating5" >Editors' rating: </div>
<div class="category-download-btn"><a href="#">Download </a></div><
<div class="category-buy-btn"><a href="#">Buy</a></div>
</div>
最佳答案
我不知道你为什么在返回ajax响应时创建表。我建议您创建 json 响应作为 ajax 的结果。使用此结果 JSON，您可以创建表，也可以在 html 中呈现它们。 在发送ajax请求的php代码中:ajax.php
<?php
$q=$_GET["q"];
$con = mysql_connect('localhost', 'root', '');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("security_software", $con);
$sql="SELECT * FROM internet_security ORDER by '".$q."' DESC" ;
$result = mysql_query($sql);
$response = array();
$i=0;
while($row = mysql_fetch_array($result))
{
$response[$i]['id'] =$row['id'];
$response[$i]['title'] = $row['title'];
$response[$i]['image'] = $row['image'];
$response[$i]['description'] = $row['description'];
$response[$i]['rating'] = $row['rating'];
$response[$i]['download'] = $row['download'];
$response[$i]['buy'] = $row['buy'];
$i++;
}
mysql_close($con);
echo json_encode($response);
?>
在你得到这个ajax响应的html文件中，我给你提示如何使用这个ajax响应:
<html>
<head>
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script type="text/javascript">
$.ajax({
url: 'ajax.php',
dataType: 'json',
success: function(response){
data = '';
$.each(response,function(i,val){
data = '<div class="category-image">'+val.image+'</div>'+
'<div class="category-link"><a href="#">'+val.id+'</a></div>'+
'<div class="category-desc"><p>'+val.description+'</p> </div>'+
'<div class="rating5" >'+val.rating+'</div>'+
'<div class="category-download-btn"><a href="'+val.download+'">Download </a></div>'+
'<div class="category-buy-btn"><a href="'+val.buy+'">Buy</a></div>';
$('<div>').attr('id',i).html(data).appendTo('#response');
});
});
}
});
</script>
</head>
<body>
<div id='response'></div>
</body>
</html>
关于php - 如何将ajax排序结果放入html div中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10596947/