如何更新和发布复选框从AJAX调用的值 [英] how to update and post the value of checkbox from ajax call

问题描述

我检索到的数据从MySQL表到HTML页面即

i retrieved data from mysql table into html page i.e

$query="select * from student";
$result=mysql_query($query)or die(mysql_error());
while($rs=mysql_fetch_array($result))
{
?>
<tr>
      <td align="center"><?php echo $rs['st_id']; ?></td>
<td align="center"><?php echo $rs['name']"; ?></td>
<td align="center"><input type="checkbox" name="checked[]" <?php if($rs['checked']==1){echo "checked"; } ?> /></td>
<td align="center"><a href="delete_student.php?id="><img src="images/delete_icon.png" alt="Delete" /></a></td>
<td align="center"><a href="update_student.php?id="><img src="images/update.png" alt="Update" /></a></td>

这是给我的输出

this is gives me the output

假设ID 1的复选框被选中，并从MySQL表中检索。
现在我想做的事情是，当我检查了remaing 2复选框到此表和函数调用阿贾克斯都到PHP页面，并根据给定的ID更新检查字段的值（即2在这种情况下，3）。不刷新整个页面，也没有更新整个记录的复选框只是价值。
我是新来的阿贾克斯这样 任何帮助将大大AP preciated。

assume that checkbox of id 1 is checked and retrieved from mysql table.
Now what i want to do is that when i checked the remaing two checkbox into this table and a function is called to ajax which goes to php page and update the value of checked field according to the given id (i.e 2 and 3 in this case). Not refreshing the whole page nor updating the whole record just the value of check box.
i am new to ajax so any help would be greatly appreciated.

推荐答案

是的，探索jqGrid的是不浪费时间的。但是，因为要更新只在复选框值的变化。这里是你可以做什么。

Yes, exploring jqGrid isn't a waste of time at all. However, since you want to update only on change of checkbox value. Here's what you can do..

<input type="checkbox" name="check1"  />

和..

$().ready(function(){
  var st_id = <?php echo $rs['st_id']; ?>;
  //Or you could use a hidden field in the form

  $(':checkbox').click(function(){
    var chkName = $(this).attr('name');
    var checkVal = $(':checkbox[name='+chkName+']').attr('checked');//true or false
    $.ajax({
      url: 'MyApp/update.php?checboxName=' + checkVal,//Do update on server-side
      success: function(data) {
        alert('Updated successful.');
      }
    });
  });
});

另外，你可以做一个 $（'formID'）。序列化（） wihch返回name = value对了就可以在调用附加的所有表单元素你的PHP脚本。

Also, you can do a $('formID').serialize() wihch returns name=value pairs of all form elements that can then be appended in the call to your php script.

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