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基本上，我试图在我的网站上为一个将所有新数据推送到我的数据库的外部 API 组合一个实时提要，

这是我用来获取我的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();
 });

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，这对我来说似乎都是正确的，但显然第二个开发人员的眼睛有时会更好

4

1 回答 1

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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'];
}

在里面while你等待一个特定的参数改变，但你需要给它改变的机会。

变量本身不会改变，您需要刷新它们。

于 2013-07-10T17:45:02.913 回答