Question

但是，您在哪里添加指向数据库中插入数据的PHP文件的链接？

Answer 1

这是需要做的事情：

$('input.wow').rating({
 callback: function(value){
            $.ajax({
        type: 'GET',
        url: "http://www3.inrees.com/rating_films_ajax.php?action=voter",
data:"films_id=<?php echo $id; ?>&rating="+value,
        success: function(){
         alert(value);
                 }
              })

  }
})

问题回答，我的中指指向插件的构想者

Answer 2

检查页面源代码：似乎没有文件指定：它是一个简单的HTML表单：

<form name="api-select">
    <span class="star-rating-control">
        <div class="rating-cancel"><a title="Cancel Rating"></a></div>
        <div class="star-rating rater-0 star star-rating-applied star-rating-live star-rating-on"><a title="A">A</a></div>
        <div class="star-rating rater-0 star star-rating-applied star-rating-live star-rating-on"><a title="B">B</a></div>
        <div class="star-rating rater-0 star star-rating-applied star-rating-live star-rating-on"><a title="C">C</a></div>
        <div class="star-rating rater-0 star star-rating-applied star-rating-live"><a title="D">D</a></div>
        <div class="star-rating rater-0 star star-rating-applied star-rating-live"><a title="E">E</a></div>
    </span>
    <input style="display: none;" class="star star-rating-applied" name="api-select-test" value="A" type="radio">
    <input style="display: none;" class="star star-rating-applied" name="api-select-test" value="B" type="radio">
    <input style="display: none;" class="star star-rating-applied" name="api-select-test" value="C" type="radio">
    <input style="display: none;" class="star star-rating-applied" name="api-select-test" value="D" type="radio">
    <input style="display: none;" class="star star-rating-applied" name="api-select-test" value="E" type="radio">
    <input value="Submit »" onclick="$(this).next().html( $(this.form).serialize() || '(nothing submitted)' );" type="button">
</form>

因此，我认为您应该只在表单元素上设置action和method属性。

jQuery星级评分插件

2 个答案: