我正在尝试构建一个多用户聊天系统，其中数据保存在数据库中并根据请求提取。下面的代码只是一个示例，我知道其中有很多很好的 SQL 注入(inject)，一旦整个事情运行良好，我将对其进行更改。另外，将 php 服务器代码保留在同一个文件中只是为了将代码保持在一个位置整齐，当服务器代码位于不同的 php 文件中时，我将在这里解释的问题仍然存在。
代码运行良好，但问题是我的数据库在一段时间后停止响应。如果有更多用户同时聊天(此脚本的多个实例)，那么数据库很快就会停止响应一段时间，并在一段时间后再次开始响应。 请建议我哪里出错了。
<?php
if($_POST){
$db=mysqli_connect("localhost","***","****","****");
if ( isset($_POST['update']) ){
$q = 0;
$lasttime = isset($_POST['timestamp']) ? $_POST['timestamp'] : 0;
while (1){
sleep(3);
$mresult = mysqli_query($db,"SELECT * FROM tblchat WHERE msg_datetime > $lasttime");
$wresult = mysqli_query($db,"SELECT writer_alias FROM tblwriter WHERE writer_isactive=1 AND (UNIX_TIMESTAMP(NOW())-last_activity)<10");
if (mysqli_num_rows($mresult)){ $msgs = array(); while ($row = mysqli_fetch_object( $mresult )) { $msgs[] = $row; } mysqli_free_result($mresult); echo json_encode(array("writers"=>$writers,"msgs"=>$msgs)); flush(); break; }
if (mysqli_num_rows($wresult)){ $writers = array(); while ($row = mysqli_fetch_object( $wresult )) { $writers[] = $row; } mysqli_free_result($wresult); echo json_encode(array("writers"=>$writers,"msgs"=>$msgs)); flush(); break; }
++$q;
if ($q>15){ break; }
}
}elseif ( isset($_POST['save']) ){
$msg = isset($_POST['msg']) ? $_POST['msg'] : '';
if ($msg != ''){
$from = $_POST["from"];
$to = $_POST["to"];
mysqli_query($db,"INSERT INTO tblchat VALUES('".$to."','".$from."','".$msg."','".time()."')");
echo json_encode(array("success"=>"1"));
flush();
}
}
mysqli_close($db);
exit();
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>testing comet</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
</head>
<body>
<p>
<input type="text" name="word" id="word" value="" />
<input type="button" name="send" value="Send" id="mybutton"/>
</p>
<div id="content"></div>
<script type="text/javascript">
var lastime = 0;
$("#mybutton").click(function(){
$.post(location.href, {save:"1",from:"1",to:"5",msg:$("#word").val()}, function(data){
console.log(data);
$("#word").val("");
}, "json");
});
$(document).ready( function(){ update(); } );
function update(){
console.log("update called");
$.ajax({ type: "POST", url: location.href, data: {update:"1",timestamp:lastime}, success: function(data1){
console.log(data1);
lastime = handleDATA(data1.msgs);
}, dataType: "json", complete: update, timeout: 60000 });
}
function handleDATA (data){
for(i=0;i<data.length;i++){
$("#content").append(data[i].msg_from +": "+ data[i].msg+"</br>");
}
return data[data.length-1].msg_datetime;
}
</script>
</body>
</html>
最佳答案
虽然这可能是也可能不是绝对的解决方案，但让我们用您的代码修复一些问题(以及一些格式以帮助更好地说明):
旧部分(第 7 - 15 行):
while (1){
sleep(3);
$mresult = mysqli_query($db,"SELECT * FROM tblchat WHERE msg_datetime > $lasttime");
$wresult = mysqli_query($db,"SELECT writer_alias FROM tblwriter WHERE writer_isactive=1 AND (UNIX_TIMESTAMP(NOW())-last_activity)<10");
if (mysqli_num_rows($mresult)){ $msgs = array(); while ($row = mysqli_fetch_object( $mresult )) { $msgs[] = $row; } mysqli_free_result($mresult); echo json_encode(array("writers"=>$writers,"msgs"=>$msgs)); flush(); break; }
if (mysqli_num_rows($wresult)){ $writers = array(); while ($row = mysqli_fetch_object( $wresult )) { $writers[] = $row; } mysqli_free_result($wresult); echo json_encode(array("writers"=>$writers,"msgs"=>$msgs)); flush(); break; }
++$q;
if ($q>15){ break; }
}
顶部新功能:
// This is separate of the while below, put this
// function at top of your script (less code later)
function toArray( myqli_result $result )
{
while( $row = mysqli_fetch_object( $result ) )
{
$array[] = $row;
}
mysqli_free_result( $result );
return $array;
}
新的 while 语句:
//Replacement While Function
while( $q < 15 )
{
sleep(3);
$mresult = mysqli_query( $db, "SELECT * FROM tblchat WHERE msg_datetime > $lasttime" );
$wresult = mysqli_query( $db, "SELECT writer_alias FROM tblwriter WHERE writer_isactive = 1 AND ( UNIX_TIMESTAMP( NOW() ) - last_activity ) < 10" );
//If you need both results, you must check for both results
//And do what you need to do, BEFORE break;
if ( mysqli_num_rows( $mresult ) > 0 && mysqli_num_rows( $wresult ) > 0 )
{
$msgs = toArray( $mresult );
$writers = toArray( $wresult );
echo json_encode( array( "writers" => $writers, "msgs" => $msgs ) );
flush();
break;
}
++$q;
}
干净的代码让程序员感到高兴，也让以后的调试变得更加容易。
如果您需要两个结果集(看起来您确实如此)，那么您将无法像现在的代码一样获得这两个结果集，当您返回第一个结果集时，您将摆脱它。这将允许在 json_encode 之前处理两个查询。
这里，toArray( mysqli_result $result ) 是一个函数，它只接受 mysqli_result 对象生成并返回数组，同时释放结果。
如果您发现自己使用不同的变量多次编写相同的代码行，您可能需要为其创建一个函数。
关于php - 长轮询导致数据库错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/18540252/