在添加新的星级之前，请重置JQuery星级

Reset the JQuery star rating before adding the new one

本文关键字：星级 JQuery 添加      更新时间：2023-09-26

我已经在Jquery中为我的网站实现了评级。当前，如果评级为4，并且同一用户添加了评级3，则值7保存在评级数据库中。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');">&#9733;</li>
  <li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">&#9733;</li>
  <li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">&#9733;</li>
  <li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">&#9733;</li>
  <li onmouseover="highlightStar(this,'rating');" onmouseout="removeHighlight();" onClick="addRating(this,'rating');">&#9733;</li>

JQuery:

function addRating(obj) {
    $('li').each(function(index) {
        $(this).addClass('selected');
        $('#rating').val((index+1));
        if(index == $("li").index(obj)) {
            return false;   
        }
    });
}

MySQL PHP代码：

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");
    }

检查用户的数据库，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";
 }
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