我正在为评论系统制作一些脚本,幸运的是一切正常,除了在ajax调用期间困扰我的一个小问题。
当我调用使用ajax进行实时数据搜索的页面时,复选框看起来不应该是这样。但是当我输入javascript代码时,复选框看起来应该是这样,但是然后函数没有&#39工作。
这是代码的样子:
Ajax调用
<script>
$(document).ready(function(){
load_data('');
function load_data(query, typehead_search = 'yes')
{
$.ajax({
url:"fetc.php",
method:"POST",
data:{query:query, typehead_search:typehead_search},
success:function(data)
{
$('#email_data').html(data);
}
});
}
$('#email_search').typeahead({
source: function(query, result){
$.ajax({
url:"fetc.php",
method:"POST",
data:{query:query},
dataType:"json",
success:function(data){
result($.map(data, function(item){
return item;
}));
load_data(query, 'yes');
}
});
}
});
$(document).on('click', 'li', function(){
var query = $(this).text();
load_data(query);
});
});
</script>
执行实时搜索数据的页面
<?php
require_once("private/initialize.php");
//fetch.php
if(isset($_POST["query"])){
$request = mysqli_real_escape_string($dbc, $_POST["query"]);
$query = "SELECT * FROM comments WHERE deleted = 2 AND comment_email LIKE '%".$request."%' ORDER BY comment_id DESC";
$result = mysqli_query($dbc, $query);
$data =array();?>
<?php
$html = '';
$html .= '
<table class="table table-hover table-mail">
<tbody>';
if(mysqli_num_rows($result) > 0){
while($comment = mysqli_fetch_array($result)){
$data[] = $comment["comment_title"];
$data[] = $comment["comment_email"];
$html .= '
<tr class="read" id="'.$comment["comment_id"].'">
<td class="check-mail">
<input type="checkbox" name="comment_id[]" class="i-checks delete_customer" value="'.$comment["comment_id"].'">
</td>
<td class="mail-ontact"><a href="mail_detail.php?comment='.$comment['comment_id'].'">'.$comment['comment_title'].'</a></td>
<td class="mail-subject"><a href="mail_detail.php?id='.$comment['comment_id'].'">'.$comment['comment_email'].'</a></td>
<td class=""></td>
<td class="text-right mail-date">8.22 PM</td>
</tr>
';
}
}
else{
$data = 'No Data Found';
$html .= '
<tr>
<td colspan="3">No Data Found</td>
</tr>
';
}
$html .= '
</tbody>
</table>';
if(isset($_POST['typehead_search']))
{
echo $html;
}
else
{
$data = array_unique($data);
echo json_encode($data);
}
}
?>
// the javascript that is making the problem
<script>
$(document).ready(function(){
$('.i-checks').iCheck({
checkboxClass: 'icheckbox_square-green',
radioClass: 'iradio_square-green',
});
});
</script>
事先,我非常感谢任何给予的帮助。