Ajax 分页无法正常工作答案

【问题标题】：Ajax Pagination not working properlyAjax 分页无法正常工作
【发布时间】：2016-03-28 06:42:46
【问题描述】：

<?php include("config.php");
error_reporting(E_ALL); ini_set('display_errors', 1);
$results = mysqli_query($con,"SELECT COUNT(name) FROM user_tbl ");
$get_total_rows = mysqli_fetch_array($results); //total records
//break total records into pages
$pages = ceil($get_total_rows[0]/$item_per_page);   
?>

<script type="text/javascript">
$(document).ready(function() {
    $("#results").load("fetch_pages.php");  //initial page number to load
    $(".pagination").bootpag({
        total: <?php echo $pages; ?>,
        page: 1,
        maxVisible: 5 
    }).on("page", function(e, num){
        e.preventDefault();
        $("#results").prepend('<div class="loading-indication"><img src="ajax-loader.gif" /> Loading...</div>');
        $("#results").load("fetch_pages.php", {'page':num});
    });
});
</script>

<body>
    <div id="serch">
        <form method="post"  enctype="multipart/form-data">
            <label>SEARCH BY NAME:</label>
            <input type="text" name="name" class="name" >
            <input type="submit" name="submit" class="sbmtt" value="SEARCH" />
        </form>
    </div>
    </table>
    <div class="res">
        <div class="navi">
            <div id="results"></div>
            <div class="pagination"></div>
        </div> 
    </div>
</body>

fetch_pages.php：

<?php
include("config.php"); //include config file
if(isset($_POST["page"])) {
    $page_number = filter_var($_POST["page"], FILTER_SANITIZE_NUMBER_INT, FILTER_FLAG_STRIP_HIGH);
    if(!is_numeric($page_number)) {
        //incase of invalid page number
        die('Invalid page number!');
    } 
} else {
    $page_number = 1;
}

//get current starting point of records
$position = (($page_number-1) * $item_per_page);

//Limit our results within a specified range. 
$results = mysqli_query($con, "SELECT * FROM user_tbl WHERE name = 'aruna' ORDER BY id ASC LIMIT $position, $item_per_page");

//output results from database
echo '<ul class="page_result">';
while($row = mysqli_fetch_array($results)) {    
    echo 'Name: ' .$row['name'];
    echo '<br /><br />Contact Number: ' .$row['cont'];
    echo '<br /><br />Email ID: ' .$row['email'];
    echo '<br /><br /><a href="viewdtl.php?id='.$row['id'].'" target="_blank">View details</a>';  
    echo "<br /><hr />";
}
echo '</ul>';
?>

Config.php：

<?php
$con = mysqli_connect("localhost","root","","mydb");
$item_per_page = 3;
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>

这里的结果只显示在 div 中.. 我想在提交表单后获取结果.. 我试过 if(isset($_POST['submit'])) 但没有工作.. 应该通过从 html 表单中获取名称来完成搜索不是我在 fetch_pages.php 中给出的方式，例如'aruna' .. 它应该被$_POST['name'] 之类的表单数据获取 .. 我真的被困在这里了.. 任何帮助 plzzz .. 在此先感谢

【问题讨论】：

来自 db 的数据显示在 div 上 .. 但我想在提交表单后获取数据，其次它应该通过表单中给出的名称获取，例如 $_POST['name']
好的..我明白了。让我更新你的代码。
好的..我会非常感激.. shukriya :)
请稍候，可能需要一点时间。
没关系..非常感谢

标签： php jquery ajax pagination

【解决方案1】：

您需要传递搜索表单值。以下是完整更新的代码。

<?php include("config.php");
error_reporting(E_ALL); ini_set('display_errors', 1);
$name = "";
$where = "";
if(isset($_POST["name"])){
    $name = $_POST["name"];
    $where = " WHERE name = '".$name."' ";
}
$results = mysqli_query($con,"SELECT COUNT(name) FROM user_tbl $where");
$get_total_rows = mysqli_fetch_array($results); //total records
//break total records into pages
$pages = ceil($get_total_rows[0]/$item_per_page);   
?>

<script type="text/javascript">
$(document).ready(function() {
var name = '<?php echo $name; ?>';
$("#results").load("fetch_pages.php", {'name':name});  //initial page number to load
$(".pagination").bootpag({
   total: <?php echo $pages; ?>,
   page: 1,
   maxVisible: 5 
}).on("page", function(e, num){
    e.preventDefault();
    $("#results").prepend('<div class="loading-indication"><img src="ajax-loader.gif" /> Loading...</div>');
    $("#results").load("fetch_pages.php", {'page':num, 'name':name});
  });
});
</script>

<body>
<div id="serch">
<form method="post"  enctype="multipart/form-data">
<label>SEARCH BY NAME:</label>
<input type="text" name="name" class="name" value="<?php echo $name; ?>">
<input type="submit" name="submit" class="sbmtt" value="SEARCH" />
</form>
</div>
</table><div class="res">
<div class="navi">
<div id="results"></div>
<div class="pagination"></div>
</div> </div>
</body>

fetch_pages.php：

<?php
include("config.php"); //include config file
if(isset($_POST["page"])){
$page_number = filter_var($_POST["page"], FILTER_SANITIZE_NUMBER_INT, FILTER_FLAG_STRIP_HIGH);
if(!is_numeric($page_number)){die('Invalid page number!');} //incase of invalid page number
}else{
$page_number = 1;
}

//get current starting point of records
$position = (($page_number-1) * $item_per_page);

$name = "";
$where = "";
if(isset($_POST["name"])){
    $name = $_POST["name"];
    $where = " WHERE name = '".$name."' ";
}

//Limit our results within a specified range. 
$results = mysqli_query($con, "SELECT * FROM user_tbl $where ORDER BY id ASC LIMIT $position, $item_per_page");

//output results from database
echo '<ul class="page_result">';
while($row = mysqli_fetch_array($results))
{

echo 'Name: ' .$row['name'];
echo '<br /><br />Contact Number: ' .$row['cont'];
echo '<br /><br />Email ID: ' .$row['email'];
echo '<br /><br /><a href="viewdtl.php?id='.$row['id'].'" target="_blank">View details</a>';  
echo "<br /><hr />";

}
echo '</ul>';
?>

希望它会有所帮助。

【讨论】：