Home Web Front-end JS Tutorial Detailed explanation of using ajax to submit a form to the database (no refresh)

Detailed explanation of using ajax to submit a form to the database (no refresh)

May 21, 2018 pm 04:17 PM
ajax form form

This article mainly introduces you to the relevant information about using ajax to submit the form to the database (without refreshing). The article introduces it in detail through the sample code. It has certain reference learning value for everyone's study or work. I feel Interested friends can come and take a look.

Everyone should know that submitting a form to the database on a static page is very simple.

<form action="test.php" method="post">
</form>
Copy after login

The disadvantage is that the page will be refreshed and the page will jump.

The technology brought to you today is the ajax form submission

The advantage is that it does not refresh the page, does not jump to the page, and is submitted silently.

As for what ajax is, go to Baidu to find out for yourself.

First we have to have a form submission page:

index.html

This page consists of two parts

1. Form control

2. jQuery ajax processing script

The jQuery script will obtain the form form data and submit it to api.php through post

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
 <title>login test</title>
 <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
 <script src="http://apps.bdimg.com/libs/jquery/2.1.4/jquery.min.js"></script>
 <script type="text/javascript">
  function insert() {
   $.ajax({
    type: "POST",//方法
    url: "api.php" ,//表单接收url
    data: $(&#39;#form1&#39;).serialize(),
    success: function () {
     //提交成功的提示词或者其他反馈代码
     var result=document.getElementById("success");
     result.innerHTML="成功!";
    },
    error : function() {
     //提交失败的提示词或者其他反馈代码
     var result=document.getElementById("success");
     result.innerHTML="失败!";
    }
   });
  }
 </script>
</head>
<body>
<p id="form-p">
 <form id="form1" onsubmit="return false" action="##" method="post">
  <p><input name="title" id="title" type="text" value="title"/></p>
  <p><input name="url" id="url" type="text" value="url"/></p>
  <p><input type="button" value="插入" onclick="insert()"></p>
  <p><p id="success"></p></p>
 </form>
</p>
</body>
</html>
Copy after login

The following is the form receiving page

api.php

This page is actually very simple

It is to connect to the database

Then insert into the database

The two values ​​inserted into the database are

title and url

<?php
$title = $_POST[&#39;title&#39;];
$url = $_POST[&#39;url&#39;];
$con = mysql_connect("localhost","root","root");
if (!$con)
 {
 die(&#39;Could not connect: &#39; . mysql_error());
 }
mysql_select_db("test", $con);
mysql_query("INSERT INTO testdata (title, url) 
VALUES (&#39;$title&#39;, &#39;$url&#39;)");

mysql_close($con);
?>
Copy after login

Then we need to create a database

The database name is test and the table name is testdata

The following is a screenshot of the database

At this point, this case is completed.

Of course, the above code simply implements the ajax submission form

But there are still many details that need to be optimized, such as form verification, data encryption, etc. You can expand, learn and improve by yourself.

The above is what I compiled for everyone. I hope it will be helpful to everyone in the future.

Related articles:

ajaxWhat are the steps required to send data to the server

ajaxImplementation of login function

How to cancel jQueryajaxRequest

The above is the detailed content of Detailed explanation of using ajax to submit a form to the database (no refresh). For more information, please follow other related articles on the PHP Chinese website!

Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn

Hot AI Tools

Undresser.AI Undress

Undresser.AI Undress

AI-powered app for creating realistic nude photos

AI Clothes Remover

AI Clothes Remover

Online AI tool for removing clothes from photos.

Undress AI Tool

Undress AI Tool

Undress images for free

Clothoff.io

Clothoff.io

AI clothes remover

Video Face Swap

Video Face Swap

Swap faces in any video effortlessly with our completely free AI face swap tool!

Hot Tools

Notepad++7.3.1

Notepad++7.3.1

Easy-to-use and free code editor

SublimeText3 Chinese version

SublimeText3 Chinese version

Chinese version, very easy to use

Zend Studio 13.0.1

Zend Studio 13.0.1

Powerful PHP integrated development environment

Dreamweaver CS6

Dreamweaver CS6

Visual web development tools

SublimeText3 Mac version

SublimeText3 Mac version

God-level code editing software (SublimeText3)

How to solve the 403 error encountered by jQuery AJAX request How to solve the 403 error encountered by jQuery AJAX request Feb 20, 2024 am 10:07 AM

Title: Methods and code examples to resolve 403 errors in jQuery AJAX requests. The 403 error refers to a request that the server prohibits access to a resource. This error usually occurs because the request lacks permissions or is rejected by the server. When making jQueryAJAX requests, you sometimes encounter this situation. This article will introduce how to solve this problem and provide code examples. Solution: Check permissions: First ensure that the requested URL address is correct and verify that you have sufficient permissions to access the resource.

How to solve jQuery AJAX request 403 error How to solve jQuery AJAX request 403 error Feb 19, 2024 pm 05:55 PM

jQuery is a popular JavaScript library used to simplify client-side development. AJAX is a technology that sends asynchronous requests and interacts with the server without reloading the entire web page. However, when using jQuery to make AJAX requests, you sometimes encounter 403 errors. 403 errors are usually server-denied access errors, possibly due to security policy or permission issues. In this article, we will discuss how to resolve jQueryAJAX request encountering 403 error

PHP and Ajax: Building an autocomplete suggestion engine PHP and Ajax: Building an autocomplete suggestion engine Jun 02, 2024 pm 08:39 PM

Build an autocomplete suggestion engine using PHP and Ajax: Server-side script: handles Ajax requests and returns suggestions (autocomplete.php). Client script: Send Ajax request and display suggestions (autocomplete.js). Practical case: Include script in HTML page and specify search-input element identifier.

How to solve the problem of jQuery AJAX error 403? How to solve the problem of jQuery AJAX error 403? Feb 23, 2024 pm 04:27 PM

How to solve the problem of jQueryAJAX error 403? When developing web applications, jQuery is often used to send asynchronous requests. However, sometimes you may encounter error code 403 when using jQueryAJAX, indicating that access is forbidden by the server. This is usually caused by server-side security settings, but there are ways to work around it. This article will introduce how to solve the problem of jQueryAJAX error 403 and provide specific code examples. 1. to make

How to get variables from PHP method using Ajax? How to get variables from PHP method using Ajax? Mar 09, 2024 pm 05:36 PM

Using Ajax to obtain variables from PHP methods is a common scenario in web development. Through Ajax, the page can be dynamically obtained without refreshing the data. In this article, we will introduce how to use Ajax to get variables from PHP methods, and provide specific code examples. First, we need to write a PHP file to handle the Ajax request and return the required variables. Here is sample code for a simple PHP file getData.php:

Tips for using Laravel form classes: ways to improve efficiency Tips for using Laravel form classes: ways to improve efficiency Mar 11, 2024 pm 12:51 PM

Forms are an integral part of writing a website or application. Laravel, as a popular PHP framework, provides rich and powerful form classes, making form processing easier and more efficient. This article will introduce some tips on using Laravel form classes to help you improve development efficiency. The following explains in detail through specific code examples. Creating a form To create a form in Laravel, you first need to write the corresponding HTML form in the view. When working with forms, you can use Laravel

PHP vs. Ajax: Solutions for creating dynamically loaded content PHP vs. Ajax: Solutions for creating dynamically loaded content Jun 06, 2024 pm 01:12 PM

Ajax (Asynchronous JavaScript and XML) allows adding dynamic content without reloading the page. Using PHP and Ajax, you can dynamically load a product list: HTML creates a page with a container element, and the Ajax request adds the data to that element after loading it. JavaScript uses Ajax to send a request to the server through XMLHttpRequest to obtain product data in JSON format from the server. PHP uses MySQL to query product data from the database and encode it into JSON format. JavaScript parses the JSON data and displays it in the page container. Clicking the button triggers an Ajax request to load the product list.

PHP and Ajax: Ways to Improve Ajax Security PHP and Ajax: Ways to Improve Ajax Security Jun 01, 2024 am 09:34 AM

In order to improve Ajax security, there are several methods: CSRF protection: generate a token and send it to the client, add it to the server side in the request for verification. XSS protection: Use htmlspecialchars() to filter input to prevent malicious script injection. Content-Security-Policy header: Restrict the loading of malicious resources and specify the sources from which scripts and style sheets are allowed to be loaded. Validate server-side input: Validate input received from Ajax requests to prevent attackers from exploiting input vulnerabilities. Use secure Ajax libraries: Take advantage of automatic CSRF protection modules provided by libraries such as jQuery.

See all articles