Php 如何在jQuery中从数据库中获取数据？

“id”和“rate_number”的值未从数据库中传递。“ratingValue”正在工作，但“id”警报和“num”警报不工作

这是我的PHP代码

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

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

下面是HTML代码

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

    <input class="rating form-control input-star-rate" 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>

---

您没有在

中设置id属性

以及：

$(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;
            alert(num);
        });
    });
});

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

<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;"/>