在PHP中没有来自长轮询的数据

基本上，我正试图在我的网站上建立一个实时提要，以获得一个外部API，将所有新数据推送到我的数据库中

下面是我用来抓取PHP文件并将任何新内容添加到数据库中的当前HTML代码

$min_id_result = $DB->select("SELECT `id` FROM `api_media` ORDER BY `id` DESC",true);
//Basically all this is doing is returning the latest ID in the database

var Id = "<?=$last_id_result['id'];?>";
function waitForPics() {
    $.ajax({
        type: "GET",
        url: "/ajax.php?id="+Id,

        async: true,
        cache: false,

        success: function(data){
            var json = eval('('+ data +')');
            var img = json['img'];
            if(json['img'] != "") {
                $('<li/>').html('<img src="'+img+'" />').appendTo('#container')
            }
            Id = json['id'];
            setTimeout(waitForPics,1000);
        },
     });
}

$(document).ready(function() {
    waitForPics();
 });

$min_id_result = $DB->select("SELECT * FROM `api_media` ORDER BY `id` DESC",true);
$next_min_id = $min_id_result['id'];

$last_id = $_GET['id'];

while($next_min_id <= $last_id) {
    usleep(10000);
    clearstatcache();
}

$qry_result = $DB->select("SELECT * FROM `api_media` ORDER BY `id` DESC");

$response = array();
foreach($qry_result as $image) {
    $response['img'] = $image['image'];
    $response['id'] = $image['id'];
}
echo json_encode($response);

下面是我用来处理数据库的ajax.php文件

$min_id_result = $DB->select("SELECT `id` FROM `api_media` ORDER BY `id` DESC",true);
//Basically all this is doing is returning the latest ID in the database

var Id = "<?=$last_id_result['id'];?>";
function waitForPics() {
    $.ajax({
        type: "GET",
        url: "/ajax.php?id="+Id,

        async: true,
        cache: false,

        success: function(data){
            var json = eval('('+ data +')');
            var img = json['img'];
            if(json['img'] != "") {
                $('<li/>').html('<img src="'+img+'" />').appendTo('#container')
            }
            Id = json['id'];
            setTimeout(waitForPics,1000);
        },
     });
}

$(document).ready(function() {
    waitForPics();
 });

$min_id_result = $DB->select("SELECT * FROM `api_media` ORDER BY `id` DESC",true);
$next_min_id = $min_id_result['id'];

$last_id = $_GET['id'];

while($next_min_id <= $last_id) {
    usleep(10000);
    clearstatcache();
}

$qry_result = $DB->select("SELECT * FROM `api_media` ORDER BY `id` DESC");

$response = array();
foreach($qry_result as $image) {
    $response['img'] = $image['image'];
    $response['id'] = $image['id'];
}
echo json_encode($response);

我遇到的问题是，它没有从请求中返回任何数据，对我来说似乎都是正确的，但显然，开发人员的第二眼有时会更好，这应该是可行的：

while($next_min_id <= $last_id) {
    usleep(10000);
    clearstatcache();
    $min_id_result = $DB->select("SELECT * FROM `api_media` ORDER BY `id` DESC",true);
    $next_min_id = $min_id_result['id'];
}

当您等待某个特定参数更改时，您需要在内部为其提供更改的机会

变量本身不会改变，

---

您需要刷新它们。

您永远不会更新while循环中的任何内容，因此它永远不会给您提供新数据。你经常使用usleep的while循环来检查信息，在你的循环中，你所做的就是告诉统计缓存清除。你看过网络流量firebug、google chrome工具、ngrep、tcpdump。。。你能得到任何回应吗？@patrick，这就是我想要它做的，检查两个ID是否相同，然后清除stat cache以了解CPU原因等等。。如果不是，它将跳出while循环并执行文件的其余部分。一旦进入循环，它将在脚本/请求之前不会中断timeout@Curtis，那么您只需要一个if语句，它总是反复检查相同的值，如果$next_min_id小于或等于$last_id，它将永远循环，然后像DanFromGermany提到的那样超时。