Question

我有一个使用方法post的表单。表单包含一个提交按钮，单击它时，我的数据库记录将显示在html表中。我想限制每页显示的行数（5），但我不想使用GET。有什么方法可以用我的post方法做到这一点吗？

// My form using post method
<form action = "" name = "dealer_call_log.php" id = "dealer_call_log.php" method = "post">


//Data displayed in the table below from post

 $record_per_page = 5;
$page = '';
 if(isset($_GET['page'])){
     $page = $_GET['page'];
 }
 else{
     $page = 1;
 }

 $start_from = ($page - 1) * $record_per_page;

if(isset($_POST['submit'])){

    echo '<center>';

    echo '<br>';

    $sql = "SELECT SFID, Comment, Time FROM tbl_call_log_detail 
    WHERE
     (dealer_id = '$call_id' AND '$endDate'='1970-01-01' AND '$startDate' ='1970-01-01')
  OR ( Time <= '$endDate' AND Time >= '$startDate'  
                    AND (dealer_id = '$call_id' OR'$call_id'='' ))    
  OR ('$endDate'='1970-01-01' AND '$startDate' ='1970-01-01'  AND '$call_id'='')
     ORDER BY Time DESC LIMIT $start_from, $record_per_page" ;
    $result = mysqli_query($conn, $sql);
    $rows = mysqli_num_rows($result);
    $all_property = array();
    echo "<table class = 'data-table' border = '1' cellpadding = '9' bgcolor = '#CCCCCC'>
            <tr class = 'data-heading'>";
    while($property = mysqli_fetch_field($result)){
        echo '<td><b> '. $property ->name. ' </b></td>';
        array_push($all_property, $property ->name);

    }
    echo '</tr>';
    }
    while ($row = mysqli_fetch_array($result)){
        echo '<tr>';
        foreach($all_property as $item){
            echo '<td> '. $row[$item] . ' </td>';
        }
        echo '</tr>';
        echo '</center>';
            }
        }
$page_query = "SELECT * FROM tbl_call_log_detail ";
$page_result = mysqli_query($conn, $page_query);
$total_records = mysqli_num_rows($page_result);
$total_pages = ceil($total_records/$record_per_page);
$start_loop = $page;
$difference = $total_pages - $page;

if($difference <= $total_pages){
    $start_loop = $total_pages - $difference;
}
$end_loop = $start_loop + 2;
if($difference > $total_pages){
    $end_loop = $total_pages;
}
if($page > 1){
    echo "<a href= 'dealer_call_log.php?page=1'>First</a>";
    echo "<a href= 'dealer_call_log.php?page=".($page - 1)."'><<</a>";
}
for ($i = $start_loop; $i <= $end_loop; $i++){
    echo "<a href= 'dealer_call_log.php?page=".$i."'>".$i."</a>";
}
if($page <= $end_loop){
    echo "<a href= 'dealer_call_log.php?page=".($page + 1)."'>>></a>";
    echo "<a href= 'dealer_call_log.php?page=".$total_pages."'>Last</a>";
}
 if($page < 1){
$page = 1;
 }
    echo '</table>';

任何帮助将不胜感激。谢谢！

Answer 1

您可以使用LIMIT 20 OFFSET 0获取前20个结果。然后，对于下一页，您可以使用LIMIT 20 OFFSET 20获取第二组20个结果。有关此

的更多信息，请参阅W3Schools

您还可以使用帖子跟踪您的页面编号，但是您必须将其发布到输入字段中。您可以使用隐藏的输入类型：<input type="hidden" name="pageNumber" value="0">并在每次切换页面时更改值。

然后你可以这样做：

    $limit = 20;
    if(isset($_POST['pageNumber']) {
        $page = $_POST['pageNumber'];
    } else {
        $page = 0;
    }

    $sql = "SELECT sfid, comment, time_stamp, time_of_submission FROM tbl_call_log_detail 
    WHERE
     (dealer_id = '$call_id' AND '$endDate'='1970-01-01' AND '$startDate' ='1970-01-01')
  OR ( time_stamp <= '$endDate' AND time_stamp >= '$startDate'  
                    AND (dealer_id = '$call_id' OR'$call_id'='' ))    
  OR ('$endDate'='1970-01-01' AND '$startDate' ='1970-01-01'  AND '$call_id'='')
     ORDER BY time_stamp DESC
  LIMIT " . $limit . " OFFSET " . $page * $limit;

echo '<input type="hidden" name="pageNumber" value="' . $page + 1 . '">'

我们可以使用POST在多个页面中显示表记录吗？

1 个答案: