Question

我不知道为什么这一<代码>onChange活动没有奏效。也许我这样说,我可以看一看我的错误。我感谢您的帮助:

<td class="cellInput">
  <?php
    $options = array( 200 => 200 ,  500 => 500 ,  1000 => 1000 ,  2000 => 2000 ,  5000 => 5000 ,  Outro => Outro );
    $attributes = array( legend  => false);

    echo $form->select( capacidade , $options, array(
       class  =>   ,
       label  =>   ,
       default  =>  200 ,
       onchange  => "javascript:checkForOther(this);",
    )); 
  ?>
</td>

Answer 1

其他人建议,将属性置于正确的参数位置。另一件事是删除java书写器,在那里只有功能名称,例如:

From: onchange => "javascript:checkForOther(this);"

To: onchange => “checkFor Other(this)”

Answer 2

提出

将属性作为第四点而不是第三点:

echo $form->select( capacidade , $options, null, 
    array(
         class =>  
        , label  =>   
        , default  =>  200 
        , onchange  => "javascript:checkForOther(this);"
    )
);

Does the source of the generated page have the onChange attribute?
See the documentation

Answer 3

http://book.cakephp。关于选定方法的文件,第三种论点用于选择哪一种选择是缺席选择的。

您必须利用第四项论点将超文本属性通过<条码>。内容:

<td class="cellInput">
  <?php
  $options = array( 200 => 200 ,  500 => 500 ,  1000 => 1000 ,  2000 => 2000 ,  5000 => 5000 ,  Outro => Outro );
  $attributes = array( legend =>false);
  echo $form->select( capacidade , $options,  200 , array( class =>   ,  label  =>   ,  onchange  => "checkForOther(this);")); 
  ?>
</td>

友情链接