Hello,

For Example:- The action is defined like this,

 <record id="calculate_marks_action" model="ir.actions.act_window">
	<field name = "name">Result</field>
	<field name = "res_model">calculate.marks</field>
	<field name = "view_type">form</field>
	<field name = "view_mode">form,tree</field>
</record>

Then to trigger this action and open its view you need to define the button as:-

<button type='action' name='%(calculate_marks_action)d' string="Button A" />

Hence the general syntax for the button to open a new view can be:-

<button type='action' name='%(your_action_id)d' string="Button Text" />

Hope this helps,

Thank You.

button name="%(action_account_invoice_payment)d" type="action" states="open" string="Register Payment" groups="account.group_account_invoice" class="oe_highlight"

Thanks a lot for your answer. But, my question was how to create a button that make you open a new view defined in an other class!!

Create Forms for Your Model and add like below code in xml

<record id="view_employee_form_loan" model="ir.ui.view">

<field name="name">hr.employee.loan.form</field>

<field name="model">hr.employee</field>

<field name="inherit_id" ref="hr.view_employee_form"/>

<field name="arch" type="xml">

<data>

<button name="%(act_hr_employee_loan_request)d" string="Check Loan" class="oe_inline oe_stat_button" icon="fa-book" type="action" groups="base.group_hr_user"/>

</data>

</field>

</record>

<record id="act_hr_employee_loan_request" model="ir.actions.act_window">

<field name="name">Loans</field>

<field name="type">ir.actions.act_window</field>

<field name="res_model">hr.employee.loan</field>

<field name="view_type">form</field>

<field name="view_mode">tree,form</field>

<field name="context">{'search_default_employee_id': [active_id], 'default_employee_id': active_id}</field>

</record>