我是Web编程的新手。我想将数据插入MySQL DB，因为我正在使用PHP，而这些数据来自Bootstrap Modal中的表单。当我单击Save按钮时，没有任何反应。我一直遵循this tutorial，并且已经完全完成了导师正在做的事情。我的代码出了什么问题？
HTML代码和JavaScript：
<body>
<p><br/></p>
<div class="container">
<p></p>
<button class="btn btn-primary" data-toggle="modal" data-target="#addData">Insert data</button>
<!-- Modal -->
<div class="modal fade" id="addData" tabindex="-1" role="dialog" aria-labelledby="addLabel">
<div class="modal-dialog" role="document">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">×</span></button>
<h4 class="modal-title" id="addlLabel">Insert Data</h4>
</div>
<form>
<div class="modal-body">
<div class="form-group">
<label for="nm">Name</label>
<input type="text" class="form-control" id="nm" placeholder="Name here">
</div>
<div class="form-group">
<label for="em">Email</label>
<input type="email" class="form-control" id="em" placeholder="Email">
</div>
<div class="form-group">
<label for="hp">Hone Number</label>
<input type="number" class="form-control" id="hp" placeholder="Your Phone">
</div>
<div class="form-group">
<label for="al">Address</label>
<textarea class="form-control" id="al" placeholder="Your address"></textarea>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button type="submit" onclick="saveData()"class="btn btn-primary">Save</button>
</div>
</form>
</div>
</div>
</div>
</div>
<!-- jQuery (necessary for Bootstrap's JavaScript plugins) -->
<script src="js/jquery-3.1.1.min.js"></script>
<!-- Include all compiled plugins (below), or include individual files as needed -->
<script src="js/bootstrap.min.js"></script>
<script>
function saveData(){
var name=$('#nm').val();
var email=$('#em').val();
var phone=$('#hp').val();
var address=$('#al').val();
$.aja({
type:"post",
url:"server.php?p=add",
data:"nm="+name+"&em="+email+"&hp="+phone+"&al="+address,
success: function(msg){
alert('Success Insert data');
}
});
}
</script>
PHP代码：
<?php
$db=new PDO('mysql:host=localhost;dbname=ajax','root','');
$page=isset($_GET['p'])?$_GET['p']:'';
if($page=='add'){
$name=$_POST['nm'];
$email=$_POST['em'];
$phone=$_POST['hp'];
$address=$_POST['al'];
$stmt=$db->prepare("insert into users values('',?,?,?,?)");
$stmt->bindParam(1,$name);
$stmt->bindParam(2,$email);
$stmt->bindParam(3,$phone);
$stmt->bindParam(4,$address);
if ($stmt->execute()) {
echo "Success";
}else{
echo "Fail".mysqli_error();
}
}else if($page=='edit'){
}if($page=='del'){
}else{
}
?>
我已使用Chrome中的开发程序，未显示任何错误。
最佳答案
错字错误。您有$.ajx({})。它必须是$.ajax({})。还以对象格式发送数据。
$.ajax({
type:"post",
url:"server.php?p=add",
data:{nm:name,em:email,hp:phone,al:address},
success: function(msg){
alert('Success Insert data');
}
});