嗨,我目前正在制作评分脚本,但有一些问题
现在,脚本“ kinda”起作用-> http://xch07.wi2.sde.dk/sandbox/rating2/index.php
问题是,如果我打开2个浏览器并在两个浏览器上加载图像1并在两个浏览器上对图像进行评级,则最后发送的查询将覆盖两者之间的所有内容,因此基本上最后只提交了2个投票中的1个?
现在我的数据库看起来像这样:
id-票数-评分
目前,我只在提交投票时将投票增加1
我以本票已投票的任何价值提高袭击
有人可以告诉我该如何解决这个问题?和我的代码上的其他任何东西都非常感谢:)
OBS:有谁知道我如何检查某个人是否已经投票给定图像?
的HTML
<div class="flex">
<div class="imageWrapper mauto relative fadeInClass">
<img id="imgSrc" src="assets/img/<?php echo $id ?>.png" class="carImg">
<input id="imgValue" class="absolute displayn" type="radio" value="<?php echo $id ?>">
<input id="imgVotes" class="absolute displayn" type="radio" value="<?php echo $votes ?>">
<input id="imgRating" class="absolute displayn" type="radio" value="<?php echo $rating ?>">
<form action="" method="post" class="flex flex-drr absolute bot0 left0">
<input id="vote5" class="vote displayn" type="radio" name="vote" value="5">
<label for="vote5"></label>
<input id="vote4" class="vote displayn" type="radio" name="vote" value="4">
<label for="vote4"></label>
<input id="vote3" class="vote displayn" type="radio" name="vote" value="3">
<label for="vote3"></label>
<input id="vote2" class="vote displayn" type="radio" name="vote" value="2">
<label for="vote2"></label>
<input id="vote1" class="vote displayn" type="radio" name="vote" value="1">
<label for="vote1"></label>
<input type="submit" id="voteSubmit" class="displayn">
</form>
</div>
</div>
Javascript / Ajax
var vote = document.getElementsByClassName('vote');
var voteL = vote.length;
for (let i = 0; i < voteL; i++) {
vote[i].addEventListener('click', function () {
let imgValue = document.getElementById("imgValue");
let imgVotes = document.getElementById("imgVotes");
let imgRating = document.getElementById("imgRating");
let imgValueVal = imgValue.value;
let imgVotesVal = imgVotes.value;
let imgRatingVal = imgRating.value;
let voteValue = vote[i].value;
newImage(imgValueVal, imgVotesVal, imgRatingVal, voteValue);
});
}
function newImage(id, votes, rating, voteValue) {
var http = new XMLHttpRequest();
var url = "pages/newImage.php";
var params = "id=" + id + "&votes=" + votes + "&rating=" + rating + "&voteValue=" + voteValue;
http.open("POST", url, true);
http.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
http.onreadystatechange = function () {//Call a function when the state changes.
if (http.readyState == 4 && http.status == 200) {
alert(http.responseText);
var Data = JSON.parse(this.responseText);
alert(Data.id);
let imgSrc = document.getElementById('imgSrc');
imgSrc.src = Data.imgSrc;
let imgValue = document.getElementById('imgValue');
imgValue.value = Data.id;
let imgVotes = document.getElementById('imgVotes');
imgVotes.value = Data.votes;
console.log(Data.votes);
let imgRating = document.getElementById('imgRating');
imgRating.value = Data.rating;
}
}
http.send(params);
}
页面AJAX请求
<?php
require_once '../includes/db.php';
require_once '../includes/functions.php';
$dbCon = dbCon();
//UPDATERE DATABASED
$id = $_POST['id'];
$votes = $_POST['votes'];
$rating = $_POST['rating'];
$voteValue = $_POST['voteValue'];
$votes++;
$rating = $rating + $voteValue;
$stmt = $dbCon->prepare("UPDATE rating SET
votes = ?,
rating = ? WHERE id = " . $id);
$stmt->bind_param('ii', $votes, $rating);
$stmt->execute();
//SENDER NY QUERY AFSTED
define("SQL", "SELECT * FROM rating ORDER BY rand() LIMIT 1");
$result = $dbCon->query(SQL);
$result = $result->fetch_object();
$id = $result->id;
$votes = $result->votes;
$rating = $result->rating;
$imgSrc = "assets/img/" . $id . ".png";
$arr = array('imgSrc' => $imgSrc, 'id' => $id, 'votes' => $votes, 'rating' => $rating);
echo json_encode($arr);
最佳答案
为什么不将准备好的语句替换为以下内容:
$stmt = $dbCon->prepare("UPDATE rating SET
votes = ?,
rating = rating + ? WHERE id = " . $id);
举行og lykke :-)
关于javascript - 多重评级时评级脚本不会更新,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/42791392/