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  • 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>
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  • 原文地址:https://www.cnblogs.com/rsapaper/p/6349294.html
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