我想從PHP/AJAX基地寫的「加載更多的記錄」的代碼,但是代碼不起作用負載更不工作
PHP代碼:
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" rel="stylesheet" type="text/css" />
<link rel="stylesheet" type="text/css" href="style.css">
<script type="text/javascript">
$(document).ready(function(){
$(document).on('click','.show_more',function(){
var ID = $(this).attr('id');
$.ajax({
type:'POST',
url:'load_more.php',
data:'ID='+ID,
success:function(html){
$('#show_more_main'+ID).remove();
$('.row').append(html);
}
});
});
});
</script>
</head>
<div class="container">
<div class="row">
<?php
include_once "config.php";
$news = mysqli_query($conn,"SELECT * FROM news ORDER BY ID DESC LIMIT 2");
while($row = mysqli_fetch_assoc($news)){
$ID = $row["ID"];
$DESC = $row["news_desc"];
?>
<p class="bg-primary" style="padding: 10px;"><?php echo substr($DESC,0,342) ?> ...</p>
<?php } ?>
<div class="show_more_main" id="show_more_main<?php echo $ID; ?>">
<button class="btn btn-success show_more" id="<?php echo $ID; ?>"><i class="fa fa-circle-o-notch fa-spin"></i> ჩატვირთვა</button>
</div>
</div><!--row-->
</div><!--container-->
load_more.php
<?php
if(isset($_POST["ID"]) && !empty($_POST["ID"])){
include "config.php";
$all_query = mysqli_query($conn,"SELECT COUNT(*) as num_rows FROM news WHERE ID < ".$_POST['ID']." ORDER BY ID DESC");
$records = mysqli_fetch_assoc($all_query);
$all_records = $records["num_rows"];
$news_limit = 2;
$query = mysqli_query($conn,"SELECT * FROM news WHERE ID <".$_POST['ID']."ORDER BY ID DESC LIMIT".$news_limit);
var_dump($query);
while($row = mysqli_fetch_assoc($query)){
$DESC = $row["news_desc"];
?>
<p class="bg-primary" style="padding: 10px;"><?php echo substr($DESC,0,342) ?> ...</p>
<?php }
if($all_records > $news_limit){?>
<div class="show_more_main" id="show_more_main<?php echo $ID; ?>">
<button class="btn btn-success show_more" id="<?php echo $ID; ?>"><i class="fa fa-circle-o-notch fa-spin"></i> ჩატვირთვა</button>
</div>
<?php
}
}
?>
也許一切正常,但是當我在沒有任何按鈕,點擊會發生 我不知道什麼問題是有...幫助我解決
您的代碼易受SQL注入攻擊。您應該使用[mysqli](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php)或[PDO](http://php.net/manual/en/pdo.prepared- statement.php)按照[本文]中描述的準備語句(http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php)。 –
謝謝你可以編輯我的代碼? – DonnaTelloo
當然,我的費率是$ 125 /小時。 Plz發送bitcoinz。 –