这是我搜索名称的index.php页面。目前，我正在警惕电影的名称和年份。我在query.php页面中静态地给出了电影名称。 但我想输入电影名称在数据库中搜索并从query.php获取结果并在我的索引页面中显示结果，这样我就不必刷新页面..
///更新了我的代码但警告没有获取任何数据
<?php
include 'dbcon.php';
?>
<!DOCTYPE html>
<html>
<head>
<title>ajax</title>
</head>
<body>
<script
src="https://code.jquery.com/jquery-2.2.4.min.js"
integrity="sha256-BbhdlvQf/xTY9gja0Dq3HiwQF8LaCRTXxZKRutelT44="
crossorigin="anonymous"></script>
<script type="text/javascript">
function loaddata()
{
var name = $('#moviename').val();
$.ajax({
type: "GET",
url: "query.php",
data: {
name:name
},
success: function (data) {
alert(data);
}
});
}
</script>
<p>Enter movie name </p>
<form action="" method="POST">
<input type="text" name="moviename" id="moviename">
<input type="submit" name="submit" value="Search" onclick="loaddata()">
</form>
<br>
<p>details here</p>
<div id="result">
</div>
</body>
</html>
这是query.php
<?php
if(isset($_POST['submit']))
{
$name = $_POST['name'];
// $name ="rango";
$query = mysqli_query( $conn,"SELECT * FROM movie WHERE name = '$name'");
while($row = mysqli_fetch_array($query))
{
echo "<p>".$row['name']."</p>";
echo "<p>".$row['year']."</p>";
}
}
?>
答案 0 :(得分：2)
请参阅下面的代码，我还包括对每次更新的评论。
here是点击事件的参考。 Sql Like语句reference。
的index.php 你不需要包含dbcon.php b / c你没有在这里使用它
<!DOCTYPE html>
<html>
<head>
<title>ajax</title>
<script
src="https://code.jquery.com/jquery-2.2.4.min.js"
integrity="sha256-BbhdlvQf/xTY9gja0Dq3HiwQF8LaCRTXxZKRutelT44="
crossorigin="anonymous"></script>
<script type="text/javascript">
// I moved the Javascript code to head section
$(function() // this function excited if the jquery is ready i mean after jquery successfully loaded
{
function loaddata()
{
var moviename= $("#moviename").val(); // read moviename value and assign;
$.ajax({
type: "GET",
url: "query.php",
data: {
name:moviename // there is no variable name so you have to assign the moveiname to name vairable ;
},
success: function (data) {
$("#result").html(data);
}
});
}
$("#submit").click(function(event) // Click Event Listener.
{
event.preventDefault();
loaddata()
});
});
</script>
</head>
<body>
<p>Enter movie name </p>
<form action="" method="POST">
<input type="text" name="moviename" id="moviename">
<input type="submit" name="submit" id="submit" value="Search"/>
<!-- if you want ot use jquery you have to use event listener. like $("#submit").click(function(event){}); code from line 31 to 35 -->
</form>
<br>
<p>details here</p>
<div id="result">
</div>
</body>
</html>
query.php文件：
<?php
include 'dbcon.php';
$name =isset($_GET['name'])?$_GET['name']:''; // Change the static value to the dynamic value you sent from ajax request
// Use Like statement to filter name
$query = mysqli_query( $conn,"SELECT * FROM movie WHERE name like '%$name%'");
while($row = mysqli_fetch_array($query))
{
echo "<p>".$row['name']."</p>";
echo "<p>".$row['year']."</p>";
}
答案 1 :(得分：0)
使用类似查询
//for checkbox btn change
$('.fa-thumbs-o-up').on('click', function() {
var LikeC = $('.imgtext').last();
var LikeA = $('.imgtext').first();
if (!LikeC.hasClass("hidechktext")) {
LikeA.removeClass('hidechktext');
}
$(this).siblings().parents('.imgtext').next().removeClass('hidechktext');
$(this).parents('.imgtext').addClass('hidechktext');
});
//for image change
$('.fa-thumbs-o-up').on('click', function() {
$('.imgall').children().siblings('.imgmain:first').removeClass('hidechktext');
$('.imgmain').prepend().addClass('hidechktext');
});
并使用post方法
$query = mysqli_query( $conn,"SELECT * FROM movie WHERE name like '%$name%'");
并在您的query.php中将名称检索为
$.ajax({
type: "POST",
url: "query.php",
data: {
name:name
},
success: function (data) {
alert(data);
}
});
答案 2 :(得分：0)
您没有将电影名称解析为<config xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns="urn:org:jgroups"
xsi:schemaLocation="urn:org:jgroups http://www.jgroups.org/schema/jgroups.xsd">
<TCP bind_port="7800"
recv_buf_size="${tcp.recv_buf_size:130k}"
send_buf_size="${tcp.send_buf_size:130k}"
max_bundle_size="64K"
sock_conn_timeout="300"
thread_pool.min_threads="0"
thread_pool.max_threads="20"
thread_pool.keep_alive_time="30000"/>
<TCPPING async_discovery="true"
initial_hosts="${jgroups.tcpping.initial_hosts:10.120.19.145[7800],localhost[7801]}"
port_range="2" />
<MERGE3 min_interval="10000"
max_interval="30000"/>
<FD_SOCK/>
<FD timeout="3000" max_tries="3" />
<VERIFY_SUSPECT timeout="1500" />
<BARRIER />
<pbcast.NAKACK2 use_mcast_xmit="false"
discard_delivered_msgs="true"/>
<UNICAST3 />
<pbcast.STABLE desired_avg_gossip="50000"
max_bytes="4M"/>
<pbcast.GMS print_local_addr="true" join_timeout="2000"
view_bundling="true"/>
<MFC max_credits="2M"
min_threshold="0.4"/>
<FRAG2 frag_size="60K" />
<pbcast.STATE_TRANSFER/>
</config>
所以首先必须将电影名称传递给ajax.php
请更新ajax.php
index.php和function loaddata()
{
var name = $('#moviename').val();/* get movie_name in name variable*/
$.ajax({
type: "GET",
url: "query.php",
data: {
name:name
},
success: function (data) {
alert(data);
}
});
}
ajax.php答案 3 :(得分：0)
首先，将您的ajax调用类型从GET更改为POST 使用ajax将要发送的数据更改为此
data: {
name1:name
},
并在php中将其作为
$name = $_POST['name1'];
以及你在php中回应的内容 将它们全部附加到一个变量上，然后回显它...类似
$query = mysqli_query( $conn,"SELECT * FROM movie WHERE name = '%$name%'");
$response = "";
while($row = mysqli_fetch_array($query))
{
$response = $response."<p>".$row['name']."</p><p>".$row['year']."</p>";
}
echo $response;