Openerp 通过on_change方法创建和编辑one2many字段的项

我有这门课（评估）

以及本课程（注：评估）

我希望用户在选择评估表单中最后一个字段（课程id）的值时，能够通过on_change方法生成一个2多个注释评估行；并使生成的行直接显示在视图中，以便他可以插入每个note_求值行的名称值（note）。拯救一切

可能吗？ 这是我当前的XML视图文件

<field name="cours_id"  context="{'evaluation_id': active_id, 'test': 1}" on_change="on_change_cours_id(examen_id,salle_de_classe_id,cours_id,aca_id)"/>
                    </group>
                    <notebook>
                        <page string="Inserer des notes">
                            <field nolabel="1" name="note_ids" context="{'evaluation_id': active_id}"/>
                        </page>

有了它，on_change方法可以在数据库中创建note_评估，但用户界面不会加载它们，one2many字段保持为空。我观察到数据库中的注释评估没有评估id

---

如何操作？

要使用onchange加载one2many数据，不必从onchange函数创建one2many数据。您只需要创建one2many字段的值。比如说

def on_change_cours_id(self,cr, uid, ids,examen_id,salle_de_classe_id,cours_id,aca_id,context=None):
    context=context or {}
    note_ids = []
    value = {}
    if ids:
        old_note_ids = self.pool.get('schoolem.note_evaluation').search(cr, uid,[('evaluation_id','in',ids)])
        self.pool.get('schoolem.note_evaluation').unlink(cr, uid, old_note_ids)
    etudiant_ids = []
    #search for etudiant_ids with the conditions examen_id,salle_de_classe_id,cours_id,aca_id etc
    for etud_id in etudiant_ids:
        note_ids.append((0,0,{'name':0,'etudiant_id':etud_id}))
    value.update(note_ids=note_ids)
    return {'value':value}

在这里，我取消了链接，因为当您保存此记录并再次以某种方式加载onchange时，将显示新的one2many列表。但是在保存时，您可以看到旧记录和新记录。因此，取消链接将删除旧记录。另外，在添加到Notes_ID时，我使用了0,0，{values}的元组，下面是对此的解释

(0, 0,  { values })    link to a new record that needs to be created with the given values dictionary
(1, ID, { values })    update the linked record with id = ID (write *values* on it)
(2, ID)                remove and delete the linked record with id = ID (calls unlink on ID, that will delete the object completely, and the link to it as well)
(3, ID)                cut the link to the linked record with id = ID (delete the relationship between the two objects but does not delete the target object itself)
(4, ID)                link to existing record with id = ID (adds a relationship)
(5)                    unlink all (like using (3,ID) for all linked records)
(6, 0, [IDs])          replace the list of linked IDs (like using (5) then (4,ID) for each ID in the list of IDs)

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我必须单击one2many的每一行进行编辑（以设置字段注释）。是否可以使one2many的整列都可编辑以避免逐行单击？如果您在主记录中设置注释，那么在从onchange加载one2many时，我们可以使用主记录中的注释感谢OMAL，我在模块中尝试了您的答案，保存后返回的值仅显示在one2many字段中。我必须在更改后立即显示，而不是在保存后显示。我该怎么做！！

def on_change_cours_id(self,cr, uid, ids,examen_id,salle_de_classe_id,cours_id,aca_id,context=None):

        context=context or {}

        #if context is None:

         #   context = {}
                for etud_id in etudiant_ids:

                    note_eval = self.pool.get('schoolem.note_evaluation')

                    if not context.get('id', False): #id de l'evaluation

                        context['id'] = context.get('evaluation_id')

                    eval_id = context.get('id', False)
                    raise osv.except_osv(('the form!'), (context.get('active_id')))

                    id = note_eval.create(cr, uid, {'name':0,'etudiant_id':etud_id,'evaluation_id':eval_id}, context=context)

def on_change_cours_id(self,cr, uid, ids,examen_id,salle_de_classe_id,cours_id,aca_id,context=None):
    context=context or {}
    note_ids = []
    value = {}
    if ids:
        old_note_ids = self.pool.get('schoolem.note_evaluation').search(cr, uid,[('evaluation_id','in',ids)])
        self.pool.get('schoolem.note_evaluation').unlink(cr, uid, old_note_ids)
    etudiant_ids = []
    #search for etudiant_ids with the conditions examen_id,salle_de_classe_id,cours_id,aca_id etc
    for etud_id in etudiant_ids:
        note_ids.append((0,0,{'name':0,'etudiant_id':etud_id}))
    value.update(note_ids=note_ids)
    return {'value':value}

(0, 0,  { values })    link to a new record that needs to be created with the given values dictionary
(1, ID, { values })    update the linked record with id = ID (write *values* on it)
(2, ID)                remove and delete the linked record with id = ID (calls unlink on ID, that will delete the object completely, and the link to it as well)
(3, ID)                cut the link to the linked record with id = ID (delete the relationship between the two objects but does not delete the target object itself)
(4, ID)                link to existing record with id = ID (adds a relationship)
(5)                    unlink all (like using (3,ID) for all linked records)
(6, 0, [IDs])          replace the list of linked IDs (like using (5) then (4,ID) for each ID in the list of IDs)