插入并通过Ajax的使用PHP检索MySQL数据 [英] Inserting and retrieving data in MySQL using PHP through Ajax
问题描述
我有一个非常简单的形式,包含一个文本框和一个提交按钮。当用户输入一些东西到表单,然后点击提交,我想使用PHP和Ajax(使用jQuery)插入表单结果到MySQL数据库。这个结果应在其中每个插入后更新的表的形式被显示在同一页上。
任何人都可以请帮助?
在code我用的是不工作:
ajax.html
:
< HTML>
<身体GT;
< SCRIPT LANGUAGE =JavaScript的类型=文/ JavaScript的>
<! -
//浏览器支持code
功能ajaxFunction(){
VAR ajaxRequest; //让Ajax成为可能的变量!
尝试{
//歌剧8.0+,火狐,Safari浏览器
ajaxRequest =新XMLHtt prequest();
}赶上(五){
// Internet Explorer浏览器
尝试{
ajaxRequest =新的ActiveXObject(MSXML2.XMLHTTP);
}赶上(五){
尝试{
ajaxRequest =新的ActiveXObject(Microsoft.XMLHTTP);
}赶上(五){
// 出了些问题
警报(您的浏览器打破了!);
返回false;
}
}
}
//创建将接收数据的功能
//从服务器发送和将更新
//格在同一页面节。
ajaxRequest.onreadystatechange =功能(){
如果(ajaxRequest.readyState == 4){
变种ajaxDisplay =的document.getElementById('ajaxDiv');
ajaxDisplay.value = ajaxRequest.responseText;
}
}
//现在得到用户的价值,并把它传递给
//服务器脚本。
。VAR名称=的document.getElementById('名')值;
VAR年龄=的document.getElementById('年龄')值。
。VAR WPM =的document.getElementById('WPM')值;
。VAR性=的document.getElementById('性')值;
VAR查询字符串=&放大器;名称=+姓名+&放大器;年龄=+年龄;
查询字符串+ =&放大器; WPM =+ WPM +&放大器;性别=+性;
ajaxRequest.open(GET,Ajax的使用example.php+
查询字符串,真实);
ajaxRequest.send(空);
}
// - >
< / SCRIPT>
<表格名称=myForm的'>
名称:<输入类型='文本'ID ='名称'/>< BR />
最大年龄:其中;输入类型='文本'ID ='年龄'/> < BR />
最大WPM:LT;输入类型='文本'ID ='WPM'/>
< BR />
性别:<选择一个id ='性'>
<期权价值=M> M< /选项>
<期权价值=F> F< /选项>
< /选择>
<输入类型=按钮的onclick ='ajaxFunction()'
值='查询MySQL的/>
< /形式GT;
< DIV ID ='ajaxDiv'>您的结果将显示此处< / DIV>
< /身体GT;
< / HTML>
AJAX-使用example.php
:
< PHP
$ DBHOST =localhost的;
$ DBUSER =演示;
$ DBPASS =演示;
$ DBNAME =TEST_DB;
//连接到MySQL服务器
的mysql_connect($ DBHOST,$ DBUSER,$ DBPASS);
//选择数据库
mysql_select_db($ TEST_DB)或死亡(mysql_error());
//从查询字符串数据
$年龄= $ _GET ['年龄'];
$性爱= $ _GET ['性'];
$ WPM = $ _GET ['WPM'];
//逃逸用户输入,以帮助prevent SQL注入
$年龄= mysql_real_escape_string($岁);
$性爱= mysql_real_escape_string($性);
$ WPM = mysql_real_escape_string($ WPM);
//生成查询
$查询=INSERT INTO窗口2(姓名,年龄,性别,WPM)VALUES('$名字','$时代,$性','$ WPM');;
mysql_select_db('TEST_DB');
$ RETVAL =的mysql_query($的SQL,$康恩);
如果(!$ RETVAL)
{
死亡('不能进入数据:'mysql_error());
}
//生成结果字符串
$ display_string =<表>;
$ display_string =&其中; TR>中;
$ display_string =<第i个姓名和LT; /第i个;
$ display_string =<第i个年龄< /第i个;
$ display_string =<第i个性别和LT; /第i个;
$ display_string =&其中;第i个的WPM&所述; /第i个;
$ display_string =。&所述; / TR>中;
//插入新行的表中返回的每个人
$ RESULT1 =请求mysql_query(SELECT * FROM窗口2,其中name ='$名字');
而($行= mysql_fetch_array($ RESULT1))
{
$ display_string =&其中; TR>中;
$ display_string =< TD> $行[名]< / TD>中;
$ display_string =< TD> $行[年龄] LT; / TD>中;
$ display_string =< TD> $行[性]< / TD>中;
$ display_string =< TD> $行[WPM] LT; / TD>中;
$ display_string =。&所述; / TR>中;
}
$ display_string =< /表>;
回声$ display_string;
?>
$(button_id)。点击(函数(){
$阿贾克斯({
网址:你应该张贴数据,
键入:POST,
数据:你应该张贴的字符串,
成功:函数(结果){
//在一些DOM元素显示你的结果
}
});
});
当您收到的PHP脚本创建查询的数据到数据库中,并得到你的结果。
希望这将有助于
I have a very simple form, containing a textbox and a submit button. When the user enters something into the form, then clicks submit, I would like to use PHP and Ajax (with jQuery) to insert the result of the form into a MySQL database. this result should be displayed on the same page in the form of a table which is updated after every insert.
Can anyone please help?
The code I have used that isn’t working:
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;
?>
$("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
}
});
});
When you receive the data in the php script make query to the database and get your result
hope this would help
这篇关于插入并通过Ajax的使用PHP检索MySQL数据的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!