php和jsCOOKIE实现前端交互

w如何精简?

 

 1 <script>
 2     document.cookie = 'wjs_cookie=' + 'amz_reviews';
 3 
 4     function w(id) {
 5         document.cookie = 'wjs_cookie=' + id;
 6         document.getElementById(id).submit();
 7     }
 8 
 9     console.log(document.cookie);
10 </script>
11 
12 <form action="" method="post" id="amz_reviews" style="display: inline;">
13     <button type="submit" name="type"
14             value="amz_reviews" <?php if ($_COOKIE['wjs_cookie'] == 'amz_reviews') echo 'style=" background-color:red;" '; ?>
15             onclick="w('amz_reviews')">差评
16     </button>
17 </form>
18 
19 <form action="" method="post" id="amz_similar_sellers" style="display: inline;">
20     <button type="submit"
21             name="type"   <?php if ($_COOKIE['wjs_cookie'] == 'amz_similar_sellers') echo 'style=" background-color:red;" '; ?>
22             value="amz_similar_sellers"
23             onclick="w('amz_similar_sellers')">跟卖
24     </button>
25 </form>
26 
27 <form action="" method="post" id="amz_listing" style="display: inline;">
28     <button type="submit"
29             name="type"  <?php if ($_COOKIE['wjs_cookie'] == 'amz_listing') echo 'style=" background-color:red;" '; ?>
30             value="amz_listing"
31             onclick="w('amz_listing')">评分、类目变化
32     </button>
33 </form>

 

posted @ 2017-01-25 12:05  papering  阅读(386)  评论(0编辑  收藏  举报