【问题标题】:How to filter data using dropdown option如何使用下拉选项过滤数据
【发布时间】:2014-09-22 06:21:48
【问题描述】:

我尝试使用下拉选项过滤一些数据。当用户选择年份/部门并单击过滤按钮时,它假设从数据库中读取包含该值的所有行。这是我能做到的,但它没有用。真的希望有人能帮助我。

 if (isset($_GET['do']))
{
    if($_GET['do'] == 'filter')
    {
        if( empty($_POST['key_year']))
        {
            redirect("list.php");
        }
        else
        {   
            $_SESSION['key_year'] = $_POST['key_year'];         
            $_SESSION['key_division'] = $_POST['key_division'];         
        }
    }
    else if($_GET['do'] == 'clear')
    {
        unset($_SESSION['key_year']);
        redirect("list.php");
    }
    else
    {}  
}
$limit = 10;    
$page = ( isset($_GET['page']) ) ? $_GET['page'] : 1;   
$end = ( isset($_GET['page']) ) ? ($limit * $_GET['page']) : $limit;
$offset = ( isset($_GET['page']) ) ? ($limit * $_GET['page']) - $limit : 0;
$start = $offset +1;
$next = $page + 1;
$prev = $page - 1;
$key_year = ( isset($_SESSION['key_year']) ) ? $_SESSION['key_year'] : '';
$key_year = ( isset($_SESSION['key_division']) ) ? $_SESSION['key_division'] : '';

$sql  = "SELECT * FROM tbl_perfomance ";
$sql .= "ORDER BY `p_aid` ASC";
$sql .= " LIMIT $offset, $limit";
$rs = mysql_query($sql);

$sql2="SELECT COUNT(p_aid) AS total FROM tbl_perfomance ";
if(isset($key_year) && isset($key_division))
{
    $sql2 .= "WHERE p_year LIKE '%$key_year%' AND p_division LIKE '%$key_division%'";
}
$rs2=mysql_query($sql2);
$getRec = mysql_fetch_assoc($rs2);
//total
$rowCount=$getRec['total'];
$end = ($end > $rowCount) ? $rowCount : $end;
$total_page = ceil($rowCount/$limit);

$for_start = ( $page <= 2 ) ? 1 : $page - 2;
$for_stop_tmp = $page + 2;
$for_stop  = ( $for_stop_tmp > $total_page ) ? $total_page : $page + 2; 

这是表格。

            <form name="filter" action="list.php?do=filter" method="GET">
                    </label>
                    <label style="color:darkgreen;float:left;width:11%" align="left">
                        Division
                        <select name="key_division">
                            <option value="All">All</option>
                            <option value="ITNT">ITNT</option>
                           <option value="ITP">ITP</option>
                        </select>
                    </label>     -->                
                    <label style="color:darkgreen;float:left;width:11%" align="left">                           
                        Year        
                        <select name="key_year">
                            <option value="all">All</option>
                            <option value="2014">2014</option>
                            <option value="2015">2015</option>
                        </select>
                    </label>
                    <label style="color:darkgreen;float:left;width:11%" align="left">
                    </label><br>
                        <input class="button_green" type="submit" value="Filter &raquo;" /> 

            </form> 

【问题讨论】:

  • 尝试回显你的 sql 查询并在 phpmyadmin 中测试它

标签: php mysql filter html.dropdownlistfor


【解决方案1】:

我感觉你只是为分页sql而不是为内容sql添加条件

尝试像这样更改您的代码,以便您拥有正确的$sql1 和$sql2。

$sql  = "SELECT * FROM tbl_perfomance ";

$sql2="SELECT COUNT(p_aid) AS total FROM tbl_perfomance ";
if(isset($key_year) && isset($key_division))
{
    $sql2 .= "WHERE p_year LIKE '%$key_year%' AND p_division LIKE '%$key_division%'";
    $sql1 .= "WHERE p_year LIKE '%$key_year%' AND p_division LIKE '%$key_division%'";
}

$sql .= "ORDER BY `p_aid` ASC";
$sql .= " LIMIT $offset, $limit";
$rs = mysql_query($sql);

$rs2=mysql_query($sql2);
$getRec = mysql_fetch_assoc($rs2);

一旦你这样做了而不是回显两个查询并直接在 php myadmin 或任何其他 mysql 工具中运行它们以检查它们是否正常工作。

【讨论】:

    猜你喜欢
    • 1970-01-01
    • 2021-08-01
    • 1970-01-01
    • 2017-10-21
    • 1970-01-01
    • 2014-11-06
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    相关资源
    最近更新 更多