我有一个非常简单的表单，包含一个文本框和一个提交按钮。当用户在表单中输入内容，然后单击提交时，我想使用 PHP 和 Ajax(带有 jQuery)将表单的结果插入 MySQL 数据库。此结果应以表格形式显示在同一页面上，表格在每次插入后都会更新。
有人可以帮忙吗？
我使用的代码不起作用:
ajax.html:
<html>
<body>
<script language="javascript" type="text/javascript">
<!--
//Browser Support Code
function ajaxFunction(){
var ajaxRequest; // The variable that makes Ajax possible!
try{
// Opera 8.0+, Firefox, Safari
ajaxRequest = new XMLHttpRequest();
}catch (e){
// Internet Explorer Browsers
try{
ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
}catch (e) {
try{
ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
}catch (e){
// Something went wrong
alert("Your browser broke!");
return false;
}
}
}
// Create a function that will receive data
// sent from the server and will update
// div section in the same page.
ajaxRequest.onreadystatechange = function(){
if(ajaxRequest.readyState == 4){
var ajaxDisplay = document.getElementById('ajaxDiv');
ajaxDisplay.value = ajaxRequest.responseText;
}
}
// Now get the value from user and pass it to
// server script.
var name = document.getElementById('name').value;
var age = document.getElementById('age').value;
var wpm = document.getElementById('wpm').value;
var sex = document.getElementById('sex').value;
var queryString = "&name=" +name+ "&age=" + age ;
queryString += "&wpm=" + wpm + "&sex=" + sex;
ajaxRequest.open("GET", "ajax-example.php" +
queryString, true);
ajaxRequest.send(null);
}
//-->
</script>
<form name='myForm'>
Name: <input type='text' id='name' /><br/>
Max Age: <input type='text' id='age' /> <br />
Max WPM: <input type='text' id='wpm' />
<br />
Sex: <select id='sex'>
<option value="m">m</option>
<option value="f">f</option>
</select>
<input type='button' onclick='ajaxFunction()'
value='Query MySQL'/>
</form>
<div id='ajaxDiv'>Your result will display here</div>
</body>
</html>
ajax-example.php:
<?php
$dbhost = "localhost";
$dbuser = "demo";
$dbpass = "demo";
$dbname = "test_db";
//Connect to MySQL Server
mysql_connect($dbhost, $dbuser, $dbpass);
//Select Database
mysql_select_db($test_db) or die(mysql_error());
// Retrieve data from Query String
$age = $_GET['age'];
$sex = $_GET['sex'];
$wpm = $_GET['wpm'];
// Escape User Input to help prevent SQL Injection
$age = mysql_real_escape_string($age);
$sex = mysql_real_escape_string($sex);
$wpm = mysql_real_escape_string($wpm);
//build query
$query = "INSERT INTO form2 (name,age,sex,wpm) VALUES ('$name','$age','$sex','$wpm')";;
mysql_select_db('test_db');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
}
//Build Result String
$display_string = "<table>";
$display_string .= "<tr>";
$display_string .= "<th>Name</th>";
$display_string .= "<th>Age</th>";
$display_string .= "<th>Sex</th>";
$display_string .= "<th>WPM</th>";
$display_string .= "</tr>";
// Insert a new row in the table for each person returned
$result1=mysql_query("SELECT * FROM form2 WHERE name='$name'");
while($row = mysql_fetch_array($result1))
{
$display_string .= "<tr>";
$display_string .= "<td>$row[name]</td>";
$display_string .= "<td>$row[age]</td>";
$display_string .= "<td>$row[sex]</td>";
$display_string .= "<td>$row[wpm]</td>";
$display_string .= "</tr>";
}
$display_string .= "</table>";
echo $display_string;
?>
最佳答案
$("button_id").click(function () {
$.ajax({
url:"where you should post the data",
type: "POST",
data: the string you should post,
success: function (result) {
//display your result in some DOM element
}
});
});
当您在 php 脚本中收到数据时，查询数据库并获得结果
希望对你有帮助
关于php - 通过 Ajax 使用 PHP 在 MySQL 中插入和检索数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/8801374/