Question

当循环执行时，它生成许多按钮。如何在append（）中获得clicked按钮的值。因为我还想通过ajax发送按钮的值。纠正错误或使其正确。

    <script>

        $(document).ready(function() {  
        $("form").submit(function(evt){  
        evt.preventDefault();
        var formData = new FormData($(this)[0]);
       formData.append( 'aabb',$(".uu").click(function(){attr('value')}) );
       $.ajax({
       url: 'pin_insert.php',
       type: 'POST',
       data: formData,
       async: false,
       cache: false,
       contentType: false,
       enctype: 'multipart/form-data',
       processData: false,
       success: function (response) {
         $(".text").html(response);
       }
   });
         return false;
 });
        });

     </script>`

    <?php
 echo '<form id ="aa" method="post">';
       echo '<input id="csv" name="csv" type="file" />';
echo '</form>';
        while($row=mysql_fetch_array($result_1))
            {
                $pin=$row['board_name'];
                echo'<div>';
                echo $pin;
                echo "...............";
                echo '<button  type="submit" form="aa" class="uu" name="aabb" value='.$pin.'>Save</button>';
                echo '</div>';  
        }
?>

Answer 1

$(document).ready(function() {  
      function getformsubmit(element){
      var formData = new FormData($("form")[0]);
       formData.append( 'aabb',$(element).val());
       $.ajax({
       url: 'pin_insert.php',
       type: 'POST',
       data: formData,
       async: false,
       cache: false,
       contentType: false,
       enctype: 'multipart/form-data',
       processData: false,
       success: function (response) {
         $(".text").html(response);
       }
   });
         return false;

    }

     </script>`

    <?php
 echo '<form id ="aa" method="post">';
       echo '<input id="csv" name="csv" type="file" />';
echo '</form>';
        while($row=mysql_fetch_array($result_1))
            {
                $pin=$row['board_name'];
                echo'<div>';
                echo $pin;
                echo "...............";
                echo '<button  type="button" form="aa" class="uu" name="aabb" onclick="getformsubmit(this)" value='.$pin.'>Save</button>';
                echo '</div>';  
        }
?>

如何在append（）中获取按钮的值，其中许多按钮的类是相同的

1 个答案: