如何在jQuery中从数据库获取星级评分的数据?

如何在jQuery中从数据库获取星级评分的数据?

问题描述:

The values do not pass for "rate_number" from database."ratingValue" alert and "id" alert are working, but "num" alert is not working.

Here is my PHP code.

$jsqla = mysql_query("select id,name,rate_score,rate_number,video_image from products where genre='$genre' limit 0,5");

if($jrowa['rate_number'] > 0){ 
   $ratea = $jrowa['rate_score'] / $jrowa['rate_number']; 
   $rateid = $jrowa['id'];
}else{ 
   $ratea = 0; 
   $rateid = $jrowa['id'];
}

Here is my HTML code.

<div class="col-sm-2 portfolio-item" style="width: 20%;">

    <input class="rating form-control input-star-rate" id="<?php echo $rateid; ?>" name="rating" value="<?php echo $ratea; ?>" data-min="0" data-max="5" data-step="0.3" data-size="xs" style="display: none; text-align: center;"/>

</div>

Here is my jQuery code.

$(function(){
      $(document).ready(function(e) {
          var $stars = $('.input-star-rate');

          $stars.bind('change', function() {
             var $this = $(this); 
             var ratingValue = $this.val();
             alert(ratingValue);

             var id = $this.attr("id");
             alert(id);

             var num = parseInt($this.attr("rate_number") + 1, 10);
             alert(num);
          });
      });
 });

值不从数据库传递“rate_number”。“ratingValue”alert和“id”alert正在工作, 但是“num”警报不起作用。 p>

这是我的PHP代码。 p>

  $ jsqla = mysql_query(“select id,name  ,rate_score,rate_number,video_image来自其中genre ='$ genre'limit 0,5“)的产品; 
 
if($ jrowa ['rate_number']&gt; 0){
 $ ratea = $ jrowa ['rate_score'  ] / $ jrowa ['rate_number'];  
 $ rateid = $ jrowa ['id']; 
} else {
 $ ratea = 0;  
 $ rateid = $ jrowa ['id']; 
} 
  code>  pre> 
 
 

这是我的HTML代码。 p>

  &lt; div class =“col-sm-2 portfolio-item”style =“width:20%;”&gt; 
 
&lt; input class =“rating form-control input-star-rate”id  =“&lt;?php echo $ rateid;?&gt;”  name =“rating”value =“&lt;?php echo $ ratea;?&gt;”  data-min =“0”data-max =“5”data-step =“0.3”data-size =“xs”style =“display:none; text-align:center;”/&gt; 
 
&lt;  / div&gt; 
  code>  pre> 
 
 

这是我的jQuery代码。 p>

  $(function(){
 $( 文件).ready(function(e){
 var $ stars = $('。input-star-rate'); 
 
 $ stars.bind('change',function(){
 var $ this  = $(this); 
 var ratingValue = $ this.val(); 
 alert(ratingValue); 
 
 var id = $ this.attr(“id”); 
 alert(id); \  n 
 var num = parseInt($ this.attr(“rate_number”)+ 1,10); 
 alert(num); 
}); 
}); 
}); 
  code  >  pre> 
  div>

Try

 var num = parseInt($this.attr("value") + 1, 10);
             alert(num);

instead of

 var num = parseInt($this.attr("rate_number") + 1, 10);
             alert(num);