我正在研究星级系统,这是我项目的一部分。每当用户通过单击星标以及用户生物数据进行评分时,该评分就会存储在我的数据库中。现在,我们知道,有时,当用户对任何产品进行评分时,他/她的想法可能会立即改变,并希望更新评分。这是我目前面临的问题。每当用户给出评分时,都会存储该评分,但是当同一用户再次给出评分时,它将再次保存该值,而不是更新现有值。请帮我。我必须在6月8日提交项目。
这是我的jQuery代码:
$(document).ready(function(){
load_business_data();
function load_business_data()
{
$.ajax({
url:"m.php",
method:"POST",
success:function(data)
{
$('#business_list').html(data);
}
});
}
$(document).on('mouseenter', '.rating', function(){
var index = $(this).data("index");
var business_id = $(this).data('business_id');
remove_background(business_id);
for(var count = 1; count<=index; count++)
{
$('#'+business_id+'-'+count).css('color', '#ffcc00');
}
});
function remove_background(business_id)
{
for(var count = 1; count <= 5; count++)
{
$('#'+business_id+'-'+count).css('color', '#ccc');
}
}
$(document).on('mouseleave', '.rating', function(){
var index = $(this).data("index");
var business_id = $(this).data('business_id');
var rating = $(this).data("rating");
remove_background(business_id);
//alert(rating);
for(var count = 1; count<=rating; count++)
{
$('#'+business_id+'-'+count).css('color', '#ffcc00');
}
});
$(document).on('click', '.rating', function(){
var index = $(this).data("index");
var business_id = $(this).data('business_id');
var username='<?php echo $_SESSION["username"];?>';
var useremail='<?php echo $_SESSION["useremail"];?>';
$.ajax({
url:"insert_rating.php",
method:"POST",
data:{index:index, business_id:business_id, username:username, useremail:useremail},
success:function(data)
{
if(data == 'done')
{
alert("There is some problem in System");
}
else
{
load_business_data();
alert("Dear "+username+",You have rate "+index +" out of 5");
}
}
});
});
});
</script>
here is my pdo code use to insert query
<?php
//insert_rating.php
session_start();
$connect = new PDO('mysql:host=localhost;dbname=final', 'root', '');
if(isset($_POST["index"], $_POST["business_id"], $_POST["username"], $_POST["useremail"]))
{
$query = "
INSERT INTO rating(business_id, rating, username,useremail)
VALUES (:business_id, :rating, :username, :useremail)
";
$statement = $connect->prepare($query);
$statement->execute(
array(
':business_id' => $_POST["business_id"],
':rating' => $_POST["index"],
':username' => $_POST["username"],
':useremail' => $_POST["useremail"],
)
);
$result = $statement->fetchAll();
if(isset($result))
{
echo 'done';
?>
首先,您需要检查用户是否已经添加了他的评分,为此,您必须根据会话用户名或ID对评分进行计数,如果他的评分已存在于数据库中,则您编写更新查询,否则编写插入查询。这是执行此操作的简单演示:
$stmt = $connect->prepare("SELECT * FROM `rating` WHERE ''");//add your conditions here according to user and that particular rating
$ifExists = $stmt->rowCount();
if ($ifExists == 0) {
//write your insert query here
} else {
//write update query here
}
我希望它能对您有所帮助,如果您需要更清楚的说明,请在下面评论