提交表单并在同一页面上显示结果而不刷新

non*_*ipt 2 php mysql ajax jquery google-visualization

可能重复:
如何让用户在表单中提交文本,如何让AJAX进入数据库并显示在同一页面上?

我想做的事:

index.html - >表单提交到some.php - >进程数据(来自index.html)并将数据发送到server.php - >将结果返回到index.html div.

我读过ajax,jQuery,我在这个网站上看到了数百个问题,但我还是想不通.

index.html的:

<form action="some.php"  method="post">
    Start date: <br/> <input name="idate" id="firstdate" type="text" /><br />
    End date: <br /> <input name="fdate" id="seconddate" type="text" /><br />
    <br />
    <input type="button" id="searchForm" onclick="SubmitForm();" value="Send" />
</form>
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some.php:

<?php
session_start();
$_SESSION['data1'] = $_POST['firstdate'];
$_SESSION['data2'] = $_POST['seconddate'];
?>

 function drawChart() {
    var jsonData = $.ajax({
        url: "server.php",
        dataType: "json",
        async: false
    }).responseText;

    var obj = jQuery.parseJSON(jsonData);
    var data = google.visualization.arrayToDataTable(obj);

(...)
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server.php:

$SQLString = "SELECT    
            count(score) as counts,
            DATE(date),
            SUM(CASE WHEN gender = 1 then 1 ELSE 0 END) Male,
            SUM(CASE WHEN gender = 2 then 1 ELSE 0 END) Female,
            AVG(age) as age, score
            FROM persons  
            WHERE date > '".$_SESSION['date1']."' AND date < '".$_SESSION['date2']."' 
            GROUP BY DATE(date) 
            ORDER BY DATE(date) asc";   

(...) 
$data[0] = array('day','counts','Male','Female','Age','Score');     
(...)
echo json_encode($data);
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kap*_*zak 7

HTML:

<form id="my_form">
    Start date: <br/> <input name="idate" id="firstdate" type="text" /><br />
    End date: <br /> <input name="fdate" id="seconddate" type="text" /><br />
    <input id="submit_form" type="submit" value="Submit">
</form>

<div id="update_div"></div>
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jQuery的:

var submit_button = $('#submit_form');

submit_button.click(function() {

    var start_date = $('firstdate').val();
    var end_date = $('seconddate').val();

    var data = 'start_date=' + start_date + '&end_date=' + end_date;

    var update_div = $('#update_div');

    $.ajax({
        type: 'GET',
        url: 'proccess_form.php',
        data: data,   
        success:function(html){
           update_div.html(html);
        }
    });
});
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proccess_form.php:

<?php
    $date1 = GET_['start_date'];
    $date2 = GET_['end_date'];

    // PERFORM THE SQL QUERY //
?>
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