Javascript 在添加新的JQuery星级之前重置JQuery星级
我已经在Jquery中为我的网站实现了评级。 当前,如果评级为4,且同一用户添加了3的评级,则值7将保存在数据库中以进行评级。 所以基本上我只想重置以前的评级,并在数据库和网页中添加当前评级(3) 我怎样才能做到这一点 代码:Javascript 在添加新的JQuery星级之前重置JQuery星级,javascript,jquery,html,css,rating,Javascript,Jquery,Html,Css,Rating,我已经在Jquery中为我的网站实现了评级。 当前,如果评级为4,且同一用户添加了3的评级,则值7将保存在数据库中以进行评级。 所以基本上我只想重置以前的评级,并在数据库和网页中添加当前评级(3) 我怎样才能做到这一点 代码: <li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★<
<li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★</li>
<li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★</li>
<li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★</li>
<li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★</li>
<li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★</li>
if( empty($_POST)){
}else if(strlen($_POST['currentReview']) >0 ){
$currentReview = $_POST['currentReview'];
$rating1 = intval($_POST['rating']) ;
echo $rating1;
$select_rating = "Select * from ratings where product_id = ". $product_id . "AND user_id = " .$user_id ;
if(!($result = mysql_query($select_rating, $databaseObj))){
print("Couldn't execute select_rating");
print $select_rating;
die("My SQL error while inserting record");
} else{
echo "RATINGS::::::" ;
$row = mysql_fetch_row($result);
print_r($row);
}
//print "Rating value is $rating1 Char value : $_POST['rating'] </br>";
$insert_query = "INSERT INTO ratings (product_id,site_id,user_id,rating, review) VALUES ($product_id,$site_id,$user_id,$rating1,\"$currentReview\") ON DUPLICATE KEY UPDATE review = \"$currentReview\",rating=$rating1;";
}
if(!($result = mysql_query($insert_query, $databaseObj))){
print("Couldn't execute insert_query");
print $insert_query;
die("My SQL error while inserting record");
}
MySQL PHP代码:
<li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★</li>
<li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★</li>
<li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★</li>
<li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★</li>
<li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">★</li>
if( empty($_POST)){
}else if(strlen($_POST['currentReview']) >0 ){
$currentReview = $_POST['currentReview'];
$rating1 = intval($_POST['rating']) ;
echo $rating1;
$select_rating = "Select * from ratings where product_id = ". $product_id . "AND user_id = " .$user_id ;
if(!($result = mysql_query($select_rating, $databaseObj))){
print("Couldn't execute select_rating");
print $select_rating;
die("My SQL error while inserting record");
} else{
echo "RATINGS::::::" ;
$row = mysql_fetch_row($result);
print_r($row);
}
//print "Rating value is $rating1 Char value : $_POST['rating'] </br>";
$insert_query = "INSERT INTO ratings (product_id,site_id,user_id,rating, review) VALUES ($product_id,$site_id,$user_id,$rating1,\"$currentReview\") ON DUPLICATE KEY UPDATE review = \"$currentReview\",rating=$rating1;";
}
if(!($result = mysql_query($insert_query, $databaseObj))){
print("Couldn't execute insert_query");
print $insert_query;
die("My SQL error while inserting record");
}
if(空($\u POST)){
}else if(strlen($_POST['currentReview'])>0){
$currentReview=$_POST['currentReview'];
$rating1=intval($_POST['rating']);
echo$rating1;
$select_rating=“select*from ratings,其中product_id=“.$product_id.”和user_id=“.$user_id;”;
if(!($result=mysql\u query($select\u rating,$databaseObj))){
打印(“无法执行选择分级”);
打印$select_评级;
die(“插入记录时我的SQL错误”);
}否则{
回声“评级:;
$row=mysql\u fetch\u row($result);
打印(行);
}
//打印“评级值为$rating1字符值:$\u POST['Rating']”;
$insert\U query=“插入到重复键更新审核中的评级(产品id、站点id、用户id、评级、审核)值($product\U id、$site\U id、$user\U id、$rating1、\“$currentReview\”),评级=“$currentReview\”;
}
if(!($result=mysql\u query($insert\u query,$databaseObj))){
打印(“无法执行insert_查询”);
打印$insert\u查询;
die(“插入记录时我的SQL错误”);
}
为用户检查数据库,如果有结果,请执行am更新而不是插入
if (mysql_query("SELECT rating FROM ratings WHERE user_id = $user_id", $databaseObj)) {
$insert_query =" UPDATE ratings SET rate = $rating1 WHERE user_id = $user_id";
}
为用户检查数据库,如果有结果,请执行am update而不是insert
if (mysql_query("SELECT rating FROM ratings WHERE user_id = $user_id", $databaseObj)) {
$insert_query =" UPDATE ratings SET rate = $rating1 WHERE user_id = $user_id";
}
不清楚您在问什么,您能澄清一下吗?我想我们需要查看您正在使用的代码,以便将评级添加到数据库中。@TonyDeStefano我已经添加了mtsql php代码:)我没有看到来自同一用户的第二个评级如何将这些数字添加到一起。我所看到的是一堆安全和弃用问题。此外,还可以看到发布到PHP脚本的代码(是AJAX吗?是表单发布吗?)。我假设您对product_id+user_id也有一个唯一的约束?不清楚您在问什么,请您澄清一下好吗?我想我们需要查看您正在使用的代码,将评级添加到数据库中。@TonyDeStefano我已经添加了mtsql php代码:)我没有看到来自同一用户的第二个评级如何将这些数字相加。我所看到的是一堆安全和弃用问题。此外,还可以看到发布到PHP脚本的代码(是AJAX吗?是表单发布吗?)。我假设您对产品id+用户id也有唯一的约束?