我是 AJAX 编程的新手和 PHP 编程的初学者。我不知道为什么,但是当用户点击箭头以“Upvote”帖子反复且非常快速时,PHP login_check 决定用户不再登录。如果我以正常速度点击箭头,程序将工作,但是当我快速射击时,它变得很奇怪。
PHP代码:
<?php
include "db_connect.php";
include "functions.php";
sec_session_start();
我想知道这是否是竞争条件的明确情况以及我能做些什么来防止它 -
AJAX 代码:
$(document).ready(function() {
$("#upvotearrow").click(function() {
setTimeout(function() { }, 500);
$resdiv=$("#upvotedownvote_resultalert");
$content=$("#upvotedownvote_resultalert_content");
$.ajax({
type: "POST",
url: "../secure/process_upvotedownvote.php",
data: { vote: "upvote", poemid: $("#poemidfield").val() },
dataType:"HTML"
})
.done(function(param) {
if (param=="true_upvote") {
$content.html("Upvote registered!");
$resdiv.css("visibility", "visible");
}
else {
$content.html("Invalid request");
$resdiv.css("visibility", "visible");
}
});
});
$("#downvotearrow").click(function() {
setTimeout(function() { }, 500);
$resdiv=$("#upvotedownvote_resultalert");
$content=$("#upvotedownvote_resultalert_content");
$.ajax({
type: "POST", //POST data
url: "../secure/process_upvotedownvote.php", //Secure upvote/downvote PHP file
data: { vote: "downvote", poemid: $("#poemidfield").val() }, //Get type of vote and poem_id in URL
dataType:"HTML" //Set datatype as HTML to send back params to AJAX function
})
.done(function(param) { //Param- variable returned by PHP file
if (param=="true_downvote") {
$content.html("Downvote registered!");
$resdiv.css("visibility", "visible");
}
else {
$content.html("Invalid request");
$resdiv.css("visibility", "visible");
}
});
});
});
可以在此处查看带有现场演示的网站。
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提前感谢您的任何建议!