日期:2014-05-16 浏览次数:20524 次
  <script type="text/javascript">
        function CheckInput(obj,type){
            var content = type+'content';
            //alert(content);
            alert(obj.content.value);
            //问题出在这里,怎么获取不到当前表单项的值呢?
            return false;
        }
  </script>
  <form action="save.php" method="post" name="aform" onsubmit="return CheckInput(this,'a')">
      <input type="text" name="acontent" />
      <input type="submit" name="submit" value="发布" />
  </form>
  <form action="save.php" method="post" name="bform_1" onsubmit="return CheckInput(this,'b')">
      <input type="text" name="bcontent" />
      <input type="submit" name="submit" value="发布" />
  </form>
  <form action="save.php" method="post" name="bform_2" onsubmit="return CheckInput(this,'b')">
      <input type="text" name="bcontent" />
      <input type="submit" name="submit" value="发布" />
  </form>
<script type="text/javascript">
        function CheckInput(obj,type){
            var content = type+'content';
            alert(obj.content.value); //对象没有这么组合的。应该改为:obj.elements[content].value;
            
            return false;
        }
  </script>
------解决方案--------------------
<script type="text/javascript">
       function CheckInput(obj,type){
           var content = type+'content';
           //alert(content);
           alert(obj.content.value);  //alert(obj.firstChild.value);这样写可以
           //问题出在这里,怎么获取不到当前表单项的值呢?
           return false;
       }
 </script>