使用ajax和php在MYSQL中插入数据
这是我的html/ajax/jquery和php文件。我必须在数据库中插入数据 但是数据没有插入到我的数据库中使用ajax和php在MYSQL中插入数据,php,jquery,mysql,Php,Jquery,Mysql,这是我的html/ajax/jquery和php文件。我必须在数据库中插入数据 但是数据没有插入到我的数据库中 <html> <head> <script src="http://code.jquery.com/jquery-1.9.1.js"></script> <script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.9/jquery.validate.min.js<
<html>
<head>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.9/jquery.validate.min.js</script>
<script>
function add() {
$("#register-form").validate({
rules: {
name: "required",
email: {
required: true,
email: true
},
Budget: {
required: true,
},
phone: "required",
budget: "required",
},
messages: {
name: "Please enter your Name",
email: "Please enter a valid Email address",
phone: "Please enter a valid Phone Number",
Budget: "Please Select a Budget",
},
submitHandler: function (form) {
//alert("success")
$.ajax({
url: "insert2.php",
type: "POST",
success: function (response) {
alert("Data Save: " + response)
}
});
}
});
}
</script>
</head>
<body>
<form action="" method="post" id="register-form" >
<div class="label">Name</div><input type="text" id="name" name="name" /><br />
<div class="label">Email</div><input type="text" id="email" name="email" /><br />
<div class="label">Phone Number</div><input type="text" id="phone" name="phone" /><br />
<div class="label">budget</div>
<select id="Budget" name="Budget">
<option value="">select</option>
<option value="1">0-100</option> <!-- first option contains value="" -->
<option value="2">100-200</option>
<option value="3">200-300</option>
</select>
<br />
<div style="margin-left:140px;">
<input type="submit" onclick="add()" name="submit" /></div>
</form>
</body>
<?php
$con=mysqli_connect("xxxxxxxx","xxx","xx","xx");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="INSERT INTO form (name, email, phone)
VALUES
('$_POST[name]','$_POST[email]','$_POST[phone]')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
使用以下命令传递表单数据
data:$('#register-form').serialize(),
$.ajax({
url: "insert2.php",
type: "POST",
data: $('#register-form').serialize(),
success: function (response) {
alert("Data Save: " + response)
}
});
<?php
$con=mysqli_connect("xxxx","xxx","xxx","xx");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$name = mysqli_real_escape_string($con, $_POST['name']); // to restrict from mysql injections
$email = mysqli_real_escape_string($con, $_POST['email']);
$phone = mysqli_real_escape_string($con, $_POST['phone']);
$sql="INSERT INTO form (name, email, phone) VALUES('$name','$email','$phone')";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
尝试在ajax中添加序列化数据：
submitHandler: function(form) {
//alert("success")
$.ajax({
url:"insert2.php",
type:"POST",
data: $('#register-form').serialize(), // <---here
success: function(response){
alert( "Data Save: " + response)
}
});
}
data: $(this).closest('#register-form').serialize(),
您得到的错误是什么？没有显示错误，但数据库id正在增加，但数据没有显示您没有将数据从ajax代码传递到服务器端PHP文件。您需要传递表单数据，如：data:$register-form.serialize；此处不显示您的ip或密码etcstring mysqli\u real\u escape\u string mysqli$link，string$escapestr
data: $(this).closest('#register-form').serialize(),