我已经检查了其他问题 - 这不是重复的。我已经尝试了我能找到和实施的所有解决方案。
我正在尝试从任务发送数据.php →显示猫.php
任务.php：
<script type="text/javascript">
$(document).on("click", ".btnCat", function () {
var filter = $(this).data("id");
alert(filter);
$.ajax({
type: 'POST',
url: 'showCats.php',
data: {'filter': filter},
});
$('div.container-fluid').load('showCats.php');
});
</script>
显示猫.php：
$area = $_POST['filter'];
$sql = "select AID,name,surname,street_name,house_number, area, plz,poster,visible from addresses WHERE area LIKE '$area' AND visible LIKE 'show' ORDER BY AID DESC";
$rs = mysqli_query($con,$sql);
$str = '';
while ($res = mysqli_fetch_array($rs)) {
$str .= '
<div class="col-md-9">
<div class="task col-md-12 well" id='.$res['AID'].'>
<div>
<button class="btn btn-danger btn-xs btnDelete" id='.$res["poster"].' onclick="refresh()" data-id="'.$res['AID'].'">x</button>
</div>
<div>
<span>'. $res["name"].'</span>
<span>'. $res["surname"].'</span><br>
<span>'. $res["street_name"].'</span>
<span>'. $res["house_number"].'</span><br>
<span>'. $res["plz"].'</span>
<span>'. $res["area"].'</span>
</div>
</div>
</div>';
}
echo $str;
?>
var_dump($_POST);返回NULL，即使我可以在Chrome中的开发人员工具下看到帖子值。
我的得到：
Request URL:https://example.com/showCats.php
Request Method:GET
Status Code:200 OK
Remote Address:xxx:443
Response Headers
view source
Cache-Control:no-store, no-cache, must-revalidate, post-check=0, pre-check=0
Connection:keep-alive
Content-Encoding:gzip
Content-Type:text/html; charset=UTF-8
Date:Fri, 15 Jul 2016 18:43:56 GMT
Expires:Thu, 19 Nov 1981 08:52:00 GMT
Pragma:no-cache
Server:nginx/1.6.2
Strict-Transport-Security:max-age=31536000
Transfer-Encoding:chunked
Request Headers
view source
Accept:text/html, */*; q=0.01
Accept-Encoding:gzip, deflate, sdch, br
Accept-Language:en-US,en;q=0.8,de;q=0.6
Connection:keep-alive
Cookie:PHPSESSID=vudgbb33574tfod2vu48hst830
Host:example.com
Referer:https://example.com/tasks.php
User-Agent:Mozilla/5.0 (Macintosh; Intel Mac OS X 10_11_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/51.0.2704.103 Safari/537.36
X-Requested-With:XMLHttpRequest
我的帖子：
Request URL:https://example.com/showCats.php
Request Method:POST
Status Code:200 OK
Remote Address:xxx:443
Response Headers
view source
Cache-Control:no-store, no-cache, must-revalidate, post-check=0, pre-check=0
Connection:keep-alive
Content-Encoding:gzip
Content-Type:text/html; charset=UTF-8
Date:Fri, 15 Jul 2016 18:43:56 GMT
Expires:Thu, 19 Nov 1981 08:52:00 GMT
Pragma:no-cache
Server:nginx/1.6.2
Strict-Transport-Security:max-age=31536000
Transfer-Encoding:chunked
Request Headers
view source
Accept:*/*
Accept-Encoding:gzip, deflate, br
Accept-Language:en-US,en;q=0.8,de;q=0.6
Connection:keep-alive
Content-Length:12
Content-Type:application/x-www-form-urlencoded; charset=UTF-8
Cookie:PHPSESSID=vudgbb33574tfod2vu48hst830
Host:example.com
Origin:https://example.com
Referer:https://example.com/tasks.php
User-Agent:Mozilla/5.0 (Macintosh; Intel Mac OS X 10_11_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/51.0.2704.103 Safari/537.36
X-Requested-With:XMLHttpRequest
Form Data
view source
view URL encoded
filter:Turgi
您正在尝试从 task.php -> showCats.php 发送数据！ 您的代码通过使用以下内容可以很好地做到这一点：
$.ajax({
type: 'POST',
url: 'showCats.php',
data: {'filter': filter},
});
问题是当你这样做时：$('div.container-fluid').load('showCats.php');一个GET请求将被发送到服务器！ 所以找到var_dump($_POST)返回 NULL 是正常的。
如果要显示/获取响应，可以像这样使用 success 事件：
$.ajax({
type: 'POST',
url: 'showCats.php',
data: {'filter': filter},
//A function to be called if the request succeeds.
success: function(data) {
$('div.container-fluid').html(data)
},
//A function to be called if the request fails
error: function(xhr, status, error) {
alert('An error occurred:'+error);
}
});
设置数据类型参数会告诉您要发送的数据类型。
type: "POST",
dataType: 'json',
我同意上述评论。 再次加载程序不会报告数据库调用的值。 此外，AJAX 通常不会以 json_encode 的回显方式将值返回给调用程序，而不仅仅是将变量回显到未被查看的 PHP 页面运行中吗？