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我一直从我的 jquery 获取值并将其传递给我的 PHP。我得到了正确的值，问题是我的 highchart 没有使用我从 jQuery 获得的值显示数据。这是我的脚本。

<?php
    include ('connect.php');
    $years = array(2011=>'Select  year','2012','2013','2014','2015','2016','2017','2018','2019','2020','2021','2022','2023','2024','2025','2026','2027','2028','2029','2030');
    $year_count= count($years);
    $year = $year_count + 2010;
?>

<script>
$(document).ready(function()
{
        $("#years").change(function(event)
   {
      var y=$(this).val();
      if(y!='2011' || y!='NULL')
      {
          alert(y);
          $("#memcount").load('reports/yearly_sales.php', {"y":y});
         $("#top10").load('reports/yearly_customers.php', {"y":y});
      }
   });
});


Year: <select id="years">
<?php
for($yr=2011; $yr<=$year; $yr++)
{
    echo "<option value='".$yr."'>".$years[$yr]."</option>";
}
?>

这是我的 php

<?php
if(isset($_POST['y'])) 
 {
$y = $_POST['y'];

 ?>

<div id='top10' style='width: 680px; height: 400px; margin: 0 auto'></div>

    <?php
     $months = array();
     $jan_tot = 0;
     $feb_tot = 0;
    //jan
    $jan = mysql_query("SELECT count(customer_id) AS count_id, date_added FROM `customer` WHERE date_added LIKE '$y-01-%'");
    while($janfetch = mysql_fetch_array($jan))
    {
        $jan_tot=(int)$janfetch['count_id'];
    }

    //feb
    $feb = mysql_query("SELECT count(customer_id) AS count_id FROM `customer` WHERE date_added LIKE '$y-02-%'");
    while($febfetch = mysql_fetch_array($feb))
    {
        $feb_tot=(int)$febfetch['count_id'];
    }
     }
    ?>

感谢那些会帮助我的人。我非常需要这个。

4

1 回答 1

0

尝试这个：

$query = "SELECT * FROM `order`  WHERE  `order_status_id`='5' AND
                                        `date_added` LIKE '".$year."-01-%'";

$jan = mysql_query($query) or die($query."<br/><br/>".mysql_error());
于 2012-04-08T10:05:12.197 回答