我已经使用Ajax PHP创建了一个搜索。但是当我试图在我的搜索页面中从PHP脚本回显整个帖子时,一切都很顺利。它不起作用。在PHP中编写PHP脚本显示所有帖子而不点击特定帖子的页面。 有没有办法在不包含PHP脚本的情况下使用ajax进行回显?
这是我的 search.php :
<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('#search').keyup(function(){
var sea = $('#search').val();
$.get('test.php',{getsearch:sea},function(data){
$('#result').show().html(data);
});
});
$('#search').blur(function(){
$('#result').hide();
});
});
</script>
</head>
<body>
<div class="cover">
<div class="btns-holder">
<input type="search" name="s" placeholder="Search for something.." id="search" autocomplete="off" /><input type="submit" name="sub" value="" id="sub" />
</div>
</div>
<div id="result">
</div>
<?php
$post = @$_GET['p'];
$que = "SELECT * FROM `search` WHERE `id`='$post'";
$run = mysqli_query($con,$que);
while($row=mysqli_fetch_array($con,$que)):
?>
<img src="<?php echo $row['image']; ?>" widrh="500px" height="450px" style="float: left;"></img>
<h2 style="font-family: sans-serif;"><?php echo $row['title']; ?></h2>
<?php endwhile; ?>
</body>
这是我想要回显数据的 test.php :
<?php
$host = "localhost";
$user = "root";
$pass = "";
$db = "ajax_search";
$con = mysqli_connect($host, $user, $pass, $db);
if(mysqli_connect_errno())
{
echo "connection failed".mysqli_connect_error();
}
$search = @$_GET['getsearch'];
$que = "SELECT * FROM `search` WHERE `title` LIKE '%$search%'";
$run = mysqli_query($con,$que);
while($row=mysqli_fetch_assoc($run)):
?>
<a href="search.php?p=<?php echo $row['id']; ?>">
<div class="container">
<img src="<?php echo $row['image']; ?>" width="50px" height="45px"></img>
<h3><?php echo $row['title']; ?></h3></br>
</div>
</a>
<?php endwhile; ?>