輸入驗證JS與存放inputed用戶數據的餅乾

如果輸入名稱，newsletter類將消失，newsletter1類將出現顯示Thankyou for joining, [Nameofuser]輸入驗證JS與存放inputed用戶數據的餅乾

我有一個表格輸入姓名和電子郵件。以正確的格式輸入電子郵件時，彈出窗口出現Successful，我試圖在cookie中保存該名稱。所以，當我刷新頁面，它會顯示Thankyou for joining, [Nameofuser]

這是我迄今取得：我不能在餅乾

http://www.w3schools.com/code/tryit.asp?filename=FAEWXAJJ7VKX

https://jsfiddle.net/sachitmaskey/kav5y1ba/

保存輸入的名稱CSS

<head> 
    <style> 
     .newsletter { 
      margin - bottom: 33 px; 
     } 

     .newsletter1 { 
      z - index: 99999; 
      margin - top: -250 px; 
      color: red; 
     } 
    </style> 
</head>

HTML

<body onload = "checkCookie()"> 
    <div class="newsletter" style="float: left;"> 
     <h3>NEWSLETTER</h3> 

     Name :<input type="text" name="name"><br> 

     <form name="myForm" onsubmit="return validateForm();" method="post"> 
      Email: <input type="text" name="email"> 
      <br><br><input type="submit" value="Submit"> 
      <p id="demo"></p> 
     </form> 
    </div> 


    <div class="newsletter1" style="float: left;"> 
     <p id="thankyou"> 
     </p><br> 
    </div> 
</body> 

</html> 



**JAVASCRIPT** 
<script> 
    function validateForm() { 
     var x = document.forms["myForm"]["email"].value; 
     var name = document.forms["myForm"]["name"].value; 
     var atpos = x.indexOf("@"); 
     var dotpos = x.lastIndexOf("."); 
     if (atpos < 1 || dotpos < atpos + 2 || dotpos + 2 >= x.length) { 
      alert("Not a valid e-mail address"); 
      return false; 
     } else { 

      alert("Successful"); 
      // If name is entered,newsletter class will disapper and newsletter1 class will appear displaying "Thankyou" 


      if (y > 0) { 

       $('.newsletter1').css('display', 'block'); 
       $('.newsletter').css('display', 'none'); 


       // setting cookie, getting cookie and checking cookie if it exist or not 
       function setCookie(cname, cvalue, exdays) { 
        var d = new Date(); 
        d.setTime(d.getTime() + (exdays * 24 * 60 * 60 * 1000)); 
        var expires = "expires=" + d.toGMTString(); 
        document.cookie = cname + "=" + cvalue + ";" + expires + ";path=/"; 
       } 

       function getCookie(cname) { 
        var name = cname + "="; 
        var ca = document.cookie.split(';'); 
        for (var i = 0; i < ca.length; i++) { 
         var c = ca[i]; 
         while (c.charAt(0) == ' ') { 
          c = c.substring(1); 
         } 
         if (c.indexOf(name) == 0) { 
          return c.substring(name.length, c.length); 
         } 
        } 
        return ""; 
       } 

       function checkCookie() { 
        var user = getCookie("name"); 
        if (user != "") { 
         setCookie("name", user, 30); 
         document.getElementById("thankyou").innerHTML = "Thankyou for joining...." 
        } 
       } 
      } 
     } 
    } 
</script>

來源

2016-12-05 ktm999

這有點過時使用cookies，我會使用window.localStorage來代替。

簡單多了：

localStorage.setItem('name', "John Smith"); 
console.log(localStorage.getItem('name')); // "John Smith" 

localStorage.removeItem('name'); 
console.log(localStorage.getItem('name')); // null

來源

2016-12-05 02:01:16 K48

它是更好地爲你使用緩存，而不是餅乾，因爲可能有一些可能性，本次會議是由您的服務器，這意味着每次用戶訪問的頁面進行處理過期新會議和舊會議被銷燬。我建議您改用瀏覽器緩存。

來源

2016-12-05 01:44:15 Smit

輸入驗證JS與存放inputed用戶數據的餅乾

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