Question

基本上，我正在尝试在我的网站上放置一个实时源，以便将所有新数据推送到我的数据库中的外部API，

以下是我用来抓取 PHP 文件的当前 HTML 代码，并将所有新内容添加到数据库

$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();
 });

这是我用来处理数据库的 ajax.php 文件

$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给我，看起来对我来说都是正确的，但显然第二个开发人员眼睛看起来更好

Answer 1

这应该有效：

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'];
}

在while内等待特定参数更改，但您需要给它更改的机会。

变量本身不会改变，你需要刷新它们。

PHP中没有来自长轮询的数据

1 个答案: