Javascript 使用mongodb提交Ajax表单
我正在尝试使用ajax提交表单,但由于某些原因,它对我不起作用,请给出如何修复它的建议 我在提交时收到了警告消息,但它会将我带到另一个页面,我在ajax请求中犯了什么错误Javascript 使用mongodb提交Ajax表单,javascript,php,jquery,ajax,mongodb,Javascript,Php,Jquery,Ajax,Mongodb,我正在尝试使用ajax提交表单,但由于某些原因,它对我不起作用,请给出如何修复它的建议 我在提交时收到了警告消息,但它会将我带到另一个页面,我在ajax请求中犯了什么错误 <!DOCTYPE html> <html> <head> <title>Get Data From a MySQL Database Using jQuery and PHP</title> <script src="https://ajax.googleap
<!DOCTYPE html>
<html>
<head>
<title>Get Data From a MySQL Database Using jQuery and PHP</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
$(document).ready(function(){
// AJAX forms
$("#search_form").submit(function(e){
e.preventDefault();
//var data = $(this).serialize();
var method = $(this).attr("method");
var action = $(this).attr("action");
var username = $('#username').val();
$.ajax({
url: 'process.php',
type: 'POST',
data: { name: username },
cache: false,
success: function(data){
$('#results').html(data);
}
})
})
});
</script>
</head>
<body>
<span>Search by name: </span>
<form method="POST" action="process.php" id="search_form">
<input type="text" id="username" name="name">
<input type="submit" id="submit" value="Search">
</form>
<div id="results"></div>
</body>
</html>
使用jQuery和PHP从MySQL数据库获取数据
$(文档).ready(函数(){
//AJAX表单
$(“#搜索表格”)。提交(功能(e){
e、 预防默认值();
//var data=$(this.serialize();
var method=$(this.attr(“method”);
var action=$(this.attr(“action”);
var username=$('#username').val();
$.ajax({
url:'process.php',
键入:“POST”,
数据:{name:username},
cache:false,
成功:功能(数据){
$('#results').html(数据);
}
})
})
});
按姓名搜索:
process.php
<?php
error_reporting(E_ALL);
ini_set('display_errors', '1');
// Check if $_POST is set
if ( empty ( $_POST['name'] ) ) {
echo "Something wrong!";
exit;
}
$name = $_POST['name'];
$m = new MongoClient();
//echo "Connection to database successfully";
// select a database
$address = array(
'name'=>$name,
'city' => 'test',
'state' => 'test2',
'zipcode' => 'test3'
);
$db = $m->local;
//echo "Database mydb selected";
$collection = $db->user;
//echo "Collection selected succsessfully";
$collection->insert($address);
$user = $collection->findOne(array('name' => $name));
?>
<ul>
<li><?php echo $user['name']; ?>: <?php echo $user['city']; ?></li>
<script>
alert('test 1234');
</script>
</ul>
- :
警报(“测试1234”);
我不得不改变:
$("#search_form").submit(function(e){
致:
现在它工作得很好
<!DOCTYPE html>
<html>
<head>
<title>Get Data From a MySQL Database Using jQuery and PHP</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
$(document).ready(function() {
// AJAX forms
$(document).on('submit', '#search_form', function() {
//e.preventDefault();
//var data = $(this).serialize();
var method = $(this).attr("method");
var action = $(this).attr("action");
var username = $('#username').val();
$.ajax({
type: 'POST',
url: 'process.php',
data: {
name: username
},
cache: false,
success: function(data) {
$('#results').html(data);
}
})
return false;
});
});
</script>
</head>
<body>
<span>Search by name: </span>
<form method="POST" action="process.php" id="search_form">
<input type="text" id="username" name="name">
<input type="submit" id="submit" value="Search">
</form>
<div id="results"></div>
</body>
</html>
使用jQuery和PHP从MySQL数据库获取数据
$(文档).ready(函数(){
//AJAX表单
$(文档)。在('submit','search_form',函数()上{
//e、 预防默认值();
//var data=$(this.serialize();
var method=$(this.attr(“method”);
var action=$(this.attr(“action”);
var username=$('#username').val();
$.ajax({
键入:“POST”,
url:'process.php',
数据:{
姓名:用户名
},
cache:false,
成功:功能(数据){
$('#results').html(数据);
}
})
返回false;
});
});
按姓名搜索:
你收到了什么错误消息第一次看,这部分看起来有些奇怪数据:名称:用户名,
,你应该使用:数据:{name:username},
没有错误消息,我改成了数据:{name:username}仍然没有work@mapodevwhat你从ajaxno response@mapodev得到了什么回应
<!DOCTYPE html>
<html>
<head>
<title>Get Data From a MySQL Database Using jQuery and PHP</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script>
$(document).ready(function() {
// AJAX forms
$(document).on('submit', '#search_form', function() {
//e.preventDefault();
//var data = $(this).serialize();
var method = $(this).attr("method");
var action = $(this).attr("action");
var username = $('#username').val();
$.ajax({
type: 'POST',
url: 'process.php',
data: {
name: username
},
cache: false,
success: function(data) {
$('#results').html(data);
}
})
return false;
});
});
</script>
</head>
<body>
<span>Search by name: </span>
<form method="POST" action="process.php" id="search_form">
<input type="text" id="username" name="name">
<input type="submit" id="submit" value="Search">
</form>
<div id="results"></div>
</body>
</html>