dongnianchou7047 2014-11-07 22:33 采纳率: 0%
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只有第一个表单由jQuery检查

i have one form on my page. so i want to check each input with corresponding button. so i wrote smth like this and it only checks first input even if i click on second inputs button.

<form target="myFrame" method="post" id="addProduct">
          <fieldset>
           <input type="hidden" name="iProductAdd" value="6" />
           <input type="number" onkeypress="return isNumberKey(event)" name="aProducts[6]" autocomplete="off"  value="" maxlength="4" size="4"  onclick="function()"  class="quantity" />
           <input type="submit" onclick="refreshIframe();" value="" class="addtobasket" id="addtobasket" />
         </fieldset>
          <fieldset>
           <input type="hidden" name="iProductAdd" value="7" />
           <input type="number" onkeypress="return isNumberKey(event)" name="aProducts[7]" autocomplete="off"  value="" maxlength="4" size="4"  onclick="function()"  class="quantity" />
           <input type="submit" onclick="refreshIframe();" value="" class="addtobasket" id="addtobasket" />
         </fieldset>
          <fieldset>
           <input type="hidden" name="iProductAdd" value="4" />
           <input type="number" onkeypress="return isNumberKey(event)" name="aProducts[4]" autocomplete="off"  value="" maxlength="4" size="4"  onclick="function()"  class="quantity" />
           <input type="submit" onclick="refreshIframe();" value="" class="addtobasket" id="addtobasket" />
         </fieldset>
        </form>

and i have jQuery Code:

$(document).ready(function(){
$("form").submit(function(){

// Get the Login Name value and trim it
 var name = $.trim($('.quantity').val());

// Check if empty of not
  if (name === '') {
    $("#addtobasket").css("background-image", "url('http://restorani.weby.biz/templates/default/img/tick-red.png')");
    return false;
 } else {
     $('input[type="submit"]').each(function(){
             $("#addtobasket").css("background-image", "url('http://restorani.weby.biz/templates/default/img/tick-green.png')");
     });
 }
});

});

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1条回答 默认 最新

  • dqzg62440 2014-11-07 22:39
    关注

    Your problem is that you are using ids as it were classes. An id should be unique in a document. So defining multiple objects with the same id is an error. And all functions will apply only the first they encounter. To do what you want, you have to use classes, so change your code in this way:

    $(".addtobasket").css("background-image", "url('http://restorani.weby.biz/templates/default/img/tick-green.png')");
    

    To look for all elements of that class.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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