我试图在不使用ajax和php刷新页面的情况下插入数据，但数据没有插入到数据库中。
<?php
require 'validation_class.php';
$obj_menufacture = new Validation_Class();
if (isset($_POST['btn'])) {
$menufacture = $obj_menufacture->menufacture_add($_POST);
}
?>
<!DOCTYPE html>
<html>
<head>
<script src='js/jquery-3.0.0.js'></script>
</head>
<body>
<div id='form_id'>
<form action='' method='post'>
<input type='text' class='m_name' name='m_name' id='m_name' onchange="names('m_name')" onkeyup="names('m_name')"/>
<span style='color:red;' id='m_name_error'></span>
<input type="submit" name='btn' value="submit" id='save'/>
</form>
</div>
<div id='cool'></div>
<style>
.border_input{
border-color:red;
}
.border_input1{
border-color:green;
}
</style>
<script>
$(document).ready(function () {
$('#save').on('click', function (e) {
e.preventDefault();
var m_name = $.trim($('#m_name').val());
var dataString = 'name='+ m_name;
if (m_name === '') {
if (m_name == '') {
$('#m_name_error').html('Please enter your name');
$('#m_name').addClass('border_input');
}
} else {
$.ajax({
url: 'validation.php',
method: 'post',
data:dataString,
success: function () {
$("#form_id").slideUp("slow");
}
});
}
});
});
function names(id) {
var val = $.trim($('#' + id).val());
if (val === '') {
$('#' + id + '_error').html('');
$('#' + id).addClass('border_input');
} else {
$('#' + id + '_error').html('');
$('#' + id).addClass('border_input1');
}
}
</script>
</body>
</html>
<?php
class Validation_Class{
public $link;
public function __construct() {
$HOST = "localhost";
$USER = "root";
$PASS = "";
$DATABASE = "store";
$this->link = mysqli_connect($HOST, $USER, $PASS, $DATABASE);
if (!$this->link) {
die('database query problem' . mysqli_error($this->link));
}
}
public function menufacture_add($data) {
$sql = "INSERT INTO menufacture(m_name)VALUES('$data[m_name]')";
if (mysqli_query($this->link, $sql)) {
$menufacture = "Menufacture insert successfully";
return $menufacture;
} else {
die('menufacruere query problem' . mysqli_error($this->link));
}
}
}
请帮我解决这个问题
答案 0 :(得分：0)
<form name="signup" id="reg" action="" method="post"enctype="multipart/form-data">
<input type='text' class='m_name' name='m_name' id='m_name' onchange="names('m_name')" onkeyup="names('m_name')"/>
<span style='color:red;' id='m_name_error'></span>
<input type="submit" name='btn' value="submit" id='save'/>
</form>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$("#save").submit(function () {
var form_data = $('#reg').serialize();
$.ajax({
type: "POST",
url: 'setting.php',
data: $('#reg').serialize(),
success: function (data) {
alert('Data save Successfully');
}, error: function () {
alert('Some Internal Error.');
}
});
});`
</script>
<强> settting.php
<?php
include("db_connection.php"); // include connection file
if(isset($_POST)) {
$fname = $_POST['m_name'];
$insert = "INSERT INTO `table_name`(`column_name`)VALUES('".$fname."')";
$insertion = mysqli_query($conn, $insert);
}
?>
答案 1 :(得分：0)
$("#save").on("click", function () {
var form_data = $('#reg').serialize();
$.ajax({
type:"POST",
url:'setting.php',
data: $('#reg').serialize(),
success: function(data){
alert("Data save Successfully");
},error: function () {
alert('Some Internal Error.');
}
});
});