ira*_*ira 6 forms jquery silverstripe data-objects
示例:我有会员(可以登录和更新他们的数据),他们有一个或多个资格.所以我有一个DataObject'Members'和一个带有has_one/has_many关系的DataObject'Qualification'.
像这样的东西:
class Qualification extends DataObject {
private static $db = array (
'Title' => 'Text',
'From' => 'Date',
'Till' => 'Date'
);
private static $has_one = array (
'Member' => 'Member',
);
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...
class Member extends DataObject {
...
private static $has_many = array (
'Qualifications' => 'Qualification',
);
...
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现在我想在前端构建一个表单,允许成员一次添加许多资格,并以相同的形式更新现有资格.
它可能看起来像这样
Qualifikation One
标题:xxx(textfield)来自:xxx(datefield)直到:xxx(datefield)
Qualifikation二
标题:xxx(textfield)来自:xxx(datefield)直到:xxx(datefield)
+添加资格
最好的方法是什么?
我可以使用jQuery动态添加字段,如下所示:http://jsfiddle.net/nzYAW/
但是我如何处理更新并将它们添加到数据库中.我尝试的一切都非常复杂和混乱,所以我想也许其他人有一个我现在看不到的想法.我们将不胜感激!
我用3dgoo的解决方案解决了我的问题。我在前端表单中使用 GridField 以及GridField 扩展模块和组件GridFieldEditableColumns以及GridFieldAddNewInlineButton. 这是一个例子:
public function MyForm() {
$config = GridFieldConfig::create();
$config->addComponent(new GridFieldButtonRow('before'));
$config->addComponent(new GridFieldEditableColumns());
$config->addComponent(new GridFieldAddNewInlineButton());
$gridField = GridField::create('Qualifications', 'Qualifications', Qualification::get()->filter(array('MemberID' => Member::currentUserID()))),$config);
$fields = new FieldList(
.... here goes some other Fields like Textfields ...
TextField::create('MyTextField'),
CheckboxField::create('MyCheckboxField'),
$gridField,
);
$actions = new FieldList(
FormAction::create('myAction','save'),
FormAction::create('myOtherAction','save and next')
);
$form = new Form($this, __FUNCTION__, $fields, $actions);
$form->loadDataFrom(Member::get()->byID(Member::currentUserID()));
return $form;
}
public function myAction($data, $form) {
$member = Member::get()->byId(Member::currentUserID());
$form->saveInto($member);
$member->write();
}
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我还必须将 canView、canEdit、canCreate 和 canDelete 函数添加到 Qualification DataObject 以允许编辑和显示它。
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