如何在ajax或php中通过检查文本框值来更新数据库值 Ajax将允许您通过javascript使用php脚本，而无需重新加载页面
假设您有3个文件： application.js如何在ajax或php中通过检查文本框值来更新数据库值 Ajax将允许您通过javascript使用php脚本，而无需重新加载页面,php,jquery,mysql,ajax,Php,Jquery,Mysql,Ajax,假设您有3个文件： application.js <?php include('db_connect.php'); $id = $_GET["id"]; if(isset($_POST['name'])) { $name=mysql_real_escape_string($_POST['name']); $query=mysql_query("SELECT * FROM brands WHERE name = '$name
<?php
include('db_connect.php');
$id = $_GET["id"];
if(isset($_POST['name']))
{
$name=mysql_real_escape_string($_POST['name']);
$query=mysql_query("SELECT * FROM brands WHERE name = '$name'",$link);
$row = mysql_num_rows($query);
if($row==0)
{
echo
"
<script type='text/javascript'>
$('#logo').animate({marginLeft:'20px'},100);
$('#logo').animate({marginLeft:'0px'},100);
</script>
";
}
else
{
$database = mysql_query("SELECT * FROM brands",$link);
$logo = $_POST['name'];
$logo_n = $row["name"];
if ($logo == $logo_n || $logo == strtolower($logo_n) || $logo == strtoupper($logo_n)) {
mysql_query("UPDATE brands SET level_br = '$id' WHERE id = $id",$link);
}
else {
echo "<script>alert('NO');</script>";
echo "<script>console.log('NO');</script>";
}
}
}
?>
$(document).ready(function(){
// add the click event to the checkbox
$('#TheCheckbox').click(function() {
// Looking if your checkox is checked
var Am_I_Checked = !$(this).is(':checked')
// Use javascrip to send an ajax call to your php file
var remoteFile = 'www.yourdomain.com/db_update.php';
// Data to send to your php file, The php file will see
// those vas in the $_GET array ...
// This is json format use all the time in javascript if you need more info
// google it :) .
// http://www.w3schools.com/json/json_intro.asp
var dataToSend = {
'Am_I_Checked' : Am_I_Checked,
'variable1' : 'the value',
'variable2' : 'the value'
// etc...
}
// This is the fonction that will be executed once the php call is done
// I know its look like a variable, but its a lamba style function, that
// also exist in php is you dont know
//
// If you notice, the data parameter is what will be return by php ( echoed by ...)
// In this example that will be recognize as json
var oneAjaxIsDone = function(data) {
//example
alert(data.whatever); // in this example, that will alert the string : "you want to return"
}
$.ajax({
dataType: "json",
url: remoteFile, /* the remote script to call */
data: dataToSend, /* the json data of element that you want to send to your script */
success: oneAjaxIsDone /* there you pass your lamba */
});
})
})
<html>
<header>
<script src="jquery.js">
<script src="application.js">
</header>
<body>
<input id="TheCheckbox" type="checkbox">
</body>
</html>
ajax：如果你知道的话，就把它看作是你的脑袋和服务器脚本之间的出租车司机吧，你的问题到底是什么？看起来你是在要求我们为你写一篇ajax教程的介绍。已经有很多了。试试谷歌。我使用ajax，但通过重新加载页面来更新数据库首先，你的代码真的很混乱，混合使用php和html总是一个坏主意，你在上面添加javascript！请给我举个例子Fredic Naultt非常感谢我会尝试的
<?php
$ajaxVars = $_GET[];
// Do your database stuff .....
// Now we will build the return, remeber the ajaxIsDone function above...
$return = array('whatever' => 'you want to return');
// When you send data from javascript to php via ajax that was in json format
// we have to do the sameting here
$jsonString = json_encode($return);
echo $jsonString;
?>