如何让AJAX知道我已选中复选框?我有一个从数据库中选择的类别列表。那么如何让AJAX知道我已经检查了什么?
这是我的搜索PHP:
<div class ="search-category-container">
<div class ="search-category-header featured-header">
<label class ="featured-font">category</label>
</div>
<div class ="search-category-content">
<?php
$result=mysqli_query($connection,"SELECT * FROM category");
while($row= mysqli_fetch_array($result)) { ?>
<label class="checkbox category-list ">
<input type="checkbox" name="category_list[]" value="<?php echo $row['name']; ?>" form="search-form"><?php echo $row['name']; ?>
</label>
<?php
}
?>
</div>
</div>
这是我在没有AJAX的情况下用于搜索的搜索功能。现在我正尝试使用AJAX来获取数据,我可以使用该功能吗?
<?php
if($filter == "post" && $time == "all" && $status == "all" && isset ($_POST['category_list'])) {
foreach ($_POST['category_list'] as $category) {
$result = mysqli_query($connection, "SELECT * FROM category WHERE name IN ('$category')")or die(mysqli_error($connection));
while($row= mysqli_fetch_array($result)) {
$getCategory = $row['id'];
$getPostIDRow = mysqli_query($connection, "SELECT * FROM post_category WHERE category_id = '$getCategory'") or die(mysqli_error($connection));
while($row2= mysqli_fetch_array($getPostIDRow)) {
$getPostID = $row2['post_id'];
$result2 = mysqli_query($connection,"SELECT * FROM post WHERE title LIKE '%$search%' AND id = '$getPostID'") or die(mysqli_error($connection));
$count2 = mysqli_num_rows($result2);
if($count2>0) {
while($row2= mysqli_fetch_array($result2)) {
$postID = $row2['id'];
$result3 = mysqli_query($connection, "SELECT * FROM user_post WHERE post_id = '$postID'") or die(mysqli_error($connection));
while($row3 = mysqli_fetch_array($result3)) {
$getUserName = mysqli_query($connection, "SELECT * FROM user WHERE id = '".$row3['user_id']."'")or die(mysqli_error($connection));
while($row4 = mysqli_fetch_array($getUserName)) {?>
<div class ="post-container" id="search-container">
<div class ="post-header-container">
<div class ="post-header">
<a href ="post.php?id=<?php echo urlencode($row2['id']);?>&user=<?php echo $row3['user_id']; ?>">
<p class ="post-header-font"><?php echo ($row2['title']); ?></p>
</a>
</div>
<div class ="post-user">
<p class ="faded-font">by : <a href="user.php?user=<?php echo $row3['user_id']; ?>"><?php echo $row4['username']; ?></a></p>
</div>
</div>
<div class ="post-content-container">
<p class ="post-content-font">
<?php echo ($row2['summary']); ?>
</p>
</div>
<div class ="post-info-container">
<div class ="post-info">
<span class ="glyphicon glyphicon-eye-open"> views: <?php echo ($row2['views']);?></span>
</div><div class ="post-info">
<span class ="glyphicon glyphicon-pencil"> answers:</span>
</div><div class ="post-info">
<span class ="glyphicon glyphicon-ok"> status: <?php echo ($row2['status']);?></span>
</div>
</div>
</div><?php
}
}
}
}
}
}
}
?>
这是AJAX搜索功能
$(document).ready(function(){
function search() {
var searchWord = $("#search").val();
var filter = $("#filter:checked").val();
var time = $("#time:checked").val();
var status = $("#status:checked").val();
$.ajax({
type:"post",
url:"searchFunction.php",
data:"search="+searchWord+"&filter="+filter+"&time="+time+"&status="+status,
success:function(data) {
$("#searchContainer").html(data);
$("#search").val("");
}
});
}
$("#searchButton").click(function(){
search();
});
$("#search").keyup(function(e){
if(e.keyCode == 13) {
search();
}
});
});
为了回答您的Ajax问题,我强烈建议更改您的代码以使用“表单” DOM来获得用户体验和更轻松的维护,只需fyi并使用序列化功能,该功能还将发送“已选中”复选框。
https://api.jquery.com/serialize/
function search() {
var postData = $('myForm').serialize(); // i.e <form id="myForm">
$.ajax({
type:"post",
url:"searchFunction.php",
data: postData,
success:function(data) {
$("#searchContainer").html(data);
$("#search").val("");
}
});
}
这将自动为您完成所有工作,而不必运行一堆jQuery选择调用并将您自己的HTTP查询放在一起。如果您需要知道您的Ajax正在运行。在浏览器的检查器模式下,找到“网络”选项卡,单击该选项卡,您应该看到该搜索文件的ajax调用,其中包含您需要了解的所有内容。什么是HTTP请求和响应标头以及正文。
Ps确保在Submit事件上返回false,并且HTML表单上的字段名称与Ajax的$ _POST关键字名称匹配。
$("#myForm").on('submit', function(){
search();
return false;
});
祝你好运!
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