Zend Framework 将表单调用为表单内的子表单

发布于 2024-08-17 10:30:13 字数 222 浏览 10 评论 0原文

我正在尝试从另一个表单中调用一个表单。我需要它作为子表单进来。然而，我不能仅仅扩展它，因为两种形式都扩展了所需的不同内容。

这就是我的想法，但它不起作用：

    $subForm3 = new Form_CreateEventInformation();

基本上从 CreateEventInformation 表单创建一个 subform3。

原文

I am trying to call a form from within another form. I need it to come in as a subform. However, I cannot merely extend it as both forms extend different things that are required.

This is kind of what I was thinking but it doesn't work:

    $subForm3 = new Form_CreateEventInformation();

Basically create a subform3 from the form CreateEventInformation.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

难理解 2024-08-24 10:30:13

你的问题有点模糊，但我会尽力回答。

在另一种形式中调用一种形式：

class Form_WhichNeedsCreateEventInformation
extends Form_BaseClass
{
 public function init()
 { 
  $this->addSubForm(new Form_CreateEventInformation(),'createEventInformationForm');
 }
}

class Form_CreateEventInformation
extends Form_BaseClass
{
 public function init()
 {
 }
}

我认为这就是您所追求的，但您似乎已经尝试过它并提到它不起作用，什么不起作用？即到底是什么问题？

如果这是不正确的，请提供更多信息。以便我可以更有效地帮助您。 :)

your question is a little little vague, but I'll answer what I can.

Calling a form within another form:

class Form_WhichNeedsCreateEventInformation
extends Form_BaseClass
{
 public function init()
 { 
  $this->addSubForm(new Form_CreateEventInformation(),'createEventInformationForm');
 }
}

class Form_CreateEventInformation
extends Form_BaseClass
{
 public function init()
 {
 }
}

I think this is what you're after, but it seems you've tried it already and mentioned that it doesn't work what doesn't work about it? ie What exactly is the problem?

If this is incorrect, please provide more information. So that I can aid you in a more effective manner. :)

回复收藏 0 原文

~没有更多了~