我的问题是基于这个写得很好的教程：http://www.sanwebe.com/2013/03/loading-more-results-from-database
演示：http://www.sanwebe.com/downloads/43-load-more-results
本质上，我正在努力实现这个+搜索。从某种意义上说，结果将基于用户的搜索查询来填充。
让我用代码详细说明：
这是主页
<?php
include("includes/db.php");
$results = mysqli_query($con,"SELECT COUNT(*) FROM courses");
$get_total_rows = mysqli_fetch_array($results); //total records
//break total records into pages
$item_per_page = 10;
$total_pages = ceil($get_total_rows[0]/$item_per_page);
?>
<html>
<head>
<script type="text/javascript" src="js/jquery-1.9.0.min.js"></script>
<script>
$(document).ready(function() {
var track_click = 0; //track user click on "load more" button, righ now it is 0 click
var total_pages = <?php echo $total_pages; ?>;
$('#results').load("fetch_pages.php", {'page':track_click}, function() {track_click++;}); //initial data to load
$(".load_more").click(function (e) { //user clicks on button
$(this).hide(); //hide load more button on click
$('.animation_image').show(); //show loading image
if(track_click <= total_pages) //user click number is still less than total pages
{
//post page number and load returned data into result element
$.post('fetch_pages.php',{'page': track_click}, function(data) {
$(".load_more").show(); //bring back load more button
$("#results").append(data); //append data received from server
//scroll page smoothly to button id
$("html, body").animate({scrollTop: $("#load_more_button").offset().top}, 500);
//hide loading image
$('.animation_image').hide(); //hide loading image once data is received
track_click++; //user click increment on load button
}).fail(function(xhr, ajaxOptions, thrownError) { //any errors?
alert(thrownError); //alert with HTTP error
$(".load_more").show(); //bring back load more button
$('.animation_image').hide(); //hide loading image once data is received
});
if(track_click >= total_pages-1) //compare user click with page number
{
//reached end of the page yet? disable load button
$(".load_more").attr("disabled", "disabled");
}
}
});
});
</script>
</head>
<body>
<center>
<form method="get" action="testSearch.php" enctype="multipart/form-data" autocomplete="off">
<input type="text" class="search" id="searchid" name="user_query" id="searchBar" placeholder="Search for courses" />
<input type="submit" id="searchButton" name="search" value="search" class="btn btn-danger" autocomplete="off"/>
<div id="results"></div>
<div align="center">
<button class="load_more" id="load_more_button">load More</button>
<div class="animation_image" style="display:none;"><img src="ajax-loader.gif"> Loading...</div>
</div>
</body>
</html>
这就是fetch_page.php，基本上是从这里填充内容的：
<?php
include("includes/db.php");
//sanitize post value
$page_number = filter_var($_POST["page"], FILTER_SANITIZE_NUMBER_INT, FILTER_FLAG_STRIP_HIGH);
//throw HTTP error if page number is not valid
if(!is_numeric($page_number)){
header('HTTP/1.1 500 Invalid page number!');
exit();
}
$item_per_page = 10;
//get current starting point of records
$position = ($page_number * $item_per_page);
//Limit our results within a specified range.
$results = mysqli_query($con,"SELECT course_title, course_date1, course_provider,course_sdesc FROM courses ORDER BY course_date1 DESC LIMIT $position, $item_per_page");
//output results from database
echo '<ul class="page_result">';
while($row = mysqli_fetch_array($results))
{
echo '<li id="item_'.$row["course_title"].'"><span class="page_name">'.$row["course_date1"].') '.$row["course_provider"].'</span><span class="page_message">'.$row["course_sdesc"].'</span></li>';
}
echo '</ul>';
?>
现在深入了解详细信息：
以下是允许进行搜索的表单（在索引页面的内容中）：
<form method="get" action="testSearch.php" enctype="multipart/form-data" autocomplete="off">
<input type="text" class="search" id="searchid" name="user_query" id="searchBar" placeholder="Search for courses" />
我一直在想做以下事情，但我一直没有这么成功：在主页的标题中（在计算分页的页面代码中）：我想做以下更改：
if(isset($_GET['user_query']))
{
$search_query = $_GET['user_query'];
$results = mysqli_query($con,"SELECT COUNT(*) FROM courses where course_title like '%$search_query%'");
$get_total_rows = mysqli_fetch_array($results); //total records
//break total records into pages
$item_per_page = 10;
$total_pages = ceil($get_total_rows[0]/$item_per_page);
}
?>
如果试图在提取页面中也这样做，请将代码调整为以下内容：
<?php
include("includes/db.php");
if(isset($_GET['user_query']))
{
$search_query = $_GET['user_query'];
//sanitize post value
$page_number = filter_var($_POST["page"], FILTER_SANITIZE_NUMBER_INT, FILTER_FLAG_STRIP_HIGH);
//throw HTTP error if page number is not valid
if(!is_numeric($page_number)){
header('HTTP/1.1 500 Invalid page number!');
exit();
}
$item_per_page = 10;
//get current starting point of records
$position = ($page_number * $item_per_page);
//Limit our results within a specified range.
$results = mysqli_query($con,"SELECT course_title, course_date1, course_provider,course_sdesc FROM courses ORDER BY course_date1 where course_title like '%$search_query%' DESC LIMIT $position, $item_per_page");
//output results from database
echo '<ul class="page_result">';
while($row = mysqli_fetch_array($results))
{
echo '<li id="item_'.$row["course_title"].'"><span class="page_name">'.$row["course_date1"].') '.$row["course_provider"].'</span><span class="page_message">'.$row["course_sdesc"].'</span></li>';
}
echo '</ul>';
}
?>
然后什么也没发生。
我认为问题在于我试图从另一个页面的url中检索一个变量。换句话说，fetch_page.php正在尝试从testSearch.php中检索user_query=，我不知道如何解决它
新方法：我把它放在现场，这样你就可以知道我想做什么：http://testapplication220-env.elasticbeanstalk.com/testSearch.php
在testSearch.php（主页面上，我这样做是为了获取用户搜索查询并稍后检索它）。
session_start();
$search_query = $_GET['user_query'];
$_SESSION['userSearch'] = $search_query;
$results = mysqli_query($con,"SELECT COUNT(*) FROM courses where course_title like '%$search_query%'");
然后是获取页面（函数，我尝试了以下操作）：
session_start();
//sanitize post value
$page_number = filter_var($_POST["page"], FILTER_SANITIZE_NUMBER_INT, FILTER_FLAG_STRIP_HIGH);
//throw HTTP error if page number is not valid
if(!is_numeric($page_number)){
header('HTTP/1.1 500 Invalid page number!');
exit();
}
$item_per_page = 10;
//get current starting point of records
$position = ($page_number * $item_per_page);
//Limit our results within a specified range.
$results = mysqli_query($con,"SELECT course_title, course_date1, course_provider,course_sdesc FROM courses ORDER BY course_date1 where course_title like '%$_SESSION['userSearch']%' DESC LIMIT $position, $item_per_page");
我想我得到了答案
如果页面变量不包含数字，则抛出500错误。
使用filter_var函数从post数组中转义页面变量
但根据php.net的说法，filter_var只会验证数据，而不会将其转换为no。我已经在phpfiddle中测试了它，它还会返回字符串。
var_dump(filter_var('23',FILTER_SANITIZE_NUMBER_INT, FILTER_FLAG_STRIP_HIGH));
因此，在用is_numeric比较页面之前，您必须将其转换为整数，这可以用以下方法完成
$page = (int) $page;