响应后附加在 HTML 中的未知文本
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【中文标题】响应后附加在 HTML 中的未知文本【英文标题】：Uknown text appending in HTML after response 【发布时间】：2019-06-21 04:22:39 【问题描述】：我一直在尝试实现无限滚动，在 Ajax 中成功调用之后，当我将结果附加到 html 时，它总是给我一个不必要的数字 1，如图所示，我不知道在哪里它来自
Pagination.php
<?php
include 'connection.php';
// header('Content-Type: application/json');
$pageno = $_POST['pageno'];
$no_of_records_per_page = 5;
$offset = ($pageno-1) * $no_of_records_per_page;
$sql = "SELECT *,
posts.id AS post_id,
posts.created_at AS post_created_at
FROM posts
LEFT JOIN users_account
ON posts.user_id = users_account.id
LEFT JOIN users_personal_info
ON users_account.id = users_personal_info.user_id
ORDER BY post_id DESC LIMIT $offset, $no_of_records_per_page";
$stmt = $connect->prepare($sql);
$stmt->execute();
$rows = $stmt->fetchAll(PDO::FETCH_OBJ);
$post_count = $stmt->rowCount();
if ($post_count > 0)
$results = include('data.php');
echo $results;
else
echo 'maxed';
Index.php
<?php
include 'connection.php';
?>
<!DOCTYPE html>
<html lang="en">
<head>
<?php include 'mda_header.php' ?>
<style>
#loader
display: block;
margin: auto;
</style>
</head>
<body>
<div class="container">
<?php
$sql = "SELECT *,
posts.id AS post_id,
posts.created_at AS post_created_at
FROM posts
LEFT JOIN users_account
ON posts.user_id = users_account.id
LEFT JOIN users_personal_info
ON users_account.id = users_personal_info.user_id
ORDER BY post_id DESC LIMIT 5";
$stmt = $connect->prepare($sql);
$stmt->execute();
$rows = $stmt->fetchAll(PDO::FETCH_OBJ);
?>
<?php include 'data.php' ?>
<div id="response"></div>
</div>
<input type="hidden" id="pageno" value="1">
<img id="loader" src="https://i.redd.it/ounq1mw5kdxy.gif">
<script src="jquery.inview.min.js"></script>
<script>
$(document).ready(function()
$('#loader').on('inview', function(event, isInView)
if (isInView)
var nextPage = parseInt($('#pageno').val()) + 1;
console.log(nextPage);
var total = parseInt($('#total').val());
$.ajax(
type: 'POST',
url: 'pagination.php',
data: pageno: nextPage, total: total ,
success: function(response)
if (response != 'maxed')
$('#response').append(response);
$('#pageno').val(nextPage);
else
$('#loader').hide();
);
);
</script>
</body>
</html>
这是我获取结果的地方，我在代码中没有看到任何数字
data.php
<?php foreach ($rows as $row): ?>
<div class="media">
<img class="align-self-start mr-3 rounded-circle" src="uploads/<?= $row->image_upload ?>" >
<div class="media-body">
<h5 class="mt-0">
<a href="mdaProfileView.php?user=<?= $row->username ?>"
data-post-author="<?= $row->username ?>"
data-placement="right"
data-container="body"
data-toggle="popover"
data-trigger="hover"
data-html="true"
title="<b><?= ucfirst($row->fname) ?> <?= ucfirst($row->lname) ?>, <?= $row->age ?></b>"
data-content="<p class='text-muted'><?= empty($row->description) ? 'No description' : $row->description ?></p>"
>
<?= $row->username ?>
</a>
</h5>
<small class="text-muted">
<i>
<i class="fa fa-clock-o"></i>
<span class="time" data-time-post-id="<?= $row->post_id ?>"><?= time_elapsed($row->post_created_at) ?></span>
</i>
</small>
<p><small class="text-muted post-body"><?= $row->body ?></small></p>
<?php
$like_stmt = $connect->prepare("SELECT *, COUNT(*) AS total_likes FROM likes WHERE post_id = $row->post_id");
$like_stmt->execute();
$like_row = $like_stmt->fetch(PDO::FETCH_OBJ);
$comment_stmt = $connect->prepare("SELECT *, COUNT(*) AS total_comments FROM comments WHERE post_id = $row->post_id");
$comment_stmt->execute();
$comment_row = $comment_stmt->fetch(PDO::FETCH_OBJ);
?>
<ul class="list-inline">
<li class="list-inline-item">
<small class="text-muted">
<span class="likes_total_<?= $row->post_id ?>">
<?= $like_row->total_likes ?></span><?= $like_row->total_likes > 1 ? ' Likes' : ' Like' ?>
</small>
</li>
<li class="list-inline-item">
<small class="text-muted">
<span class="comments_total_<?= $row->post_id ?>">
<?= $comment_row->total_comments ?>
</span>
<?= $comment_row->total_comments > 1 ? ' Comments' : ' Comment' ?>
</small>
</li>
<li class="list-inline-item float-right">
<small class="text-muted">25 Shares</small>
</li>
</ul>
<ul class="list-inline">
<li class="list-inline-item">
<small>
<?php
$user_id = user()->user_original_id;
$isUserLiked_stmt = $connect->prepare("SELECT * FROM likes WHERE post_id = $row->post_id AND user_id = $user_id");
$isUserLiked_stmt->execute();
?>
<a
class="social-icon text-xs-center like like_button_<?= $row->post_id ?><?= $isUserLiked_stmt->rowCount() ? ' bold' : '' ?>"
data-user-id="<?= user()->user_original_id ?>"
data-author="<?= $row->username ?>"
data-post-id="<?= $row->post_id ?>"
data-type="post"
href="javascript:void(0)">
<i class="fa fa-thumbs-up up"
></i> Like</a>
</small>
</li>
<li class="list-inline-item">
<!-- comment_button -->
<small><a class="social-icon text-xs-center comment"
href="javascript:void(0)"
data-post-id="<?= $row->post_id ?>"
data-toggle="modal"
data-target="#exampleModal"><i class="fa fa-comment"></i> Comment</a></small>
</li>
<li class="list-inline-item float-right">
<small><a class="social-icon text-xs-center" href="javascript:void(0)"><i class="fa fa-share"></i> Share</a></small>
</li>
</ul>
</div>
</div>
<?php endforeach; ?>
【问题讨论】：
如果将<input type="hidden" id="pageno" value="1"> 替换为<input type="hidden" id="pageno" value="2"> 会发生什么情况现在数字1 变成2了吗？
【参考方案1】：
您只需要一个包含 - 您也不需要回显结果，您可能会发现 $results 是 1 表示 include() 成功。改变
$results = include('data.php');
echo $results;
到
include('data.php');
如果包含的脚本使用return 传回数据，则只需要赋值和echo。
【讨论】：
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