I have created a button
in jobs by using inheritance in (hr.recruitment
form view) , how could I open another module("Resumes and Letters -sub menu in Human Resource " ) form_view during buttons click event is done.my aim is that I just want open that form when this click event done.
Is it possible to solve? Need help please
yes, it is possible to open another window. you have to do like this.
@api.multi
def button_method(self):
return {
'type': 'ir.actions.act_window',
'name': 'form name',
'res_model': 'object name',
'res_id': id ,
'view_type': 'form',
'view_mode': 'form',
'target' : 'new',
}
but it is possible when record save. if you want to open wizard before save record you have to code in js like this.
in js file: openerp.module_name = function(instance) {
var QWeb = openerp.web.qweb;
_t = instance.web._t;
instance.web.View.include({
load_view: function(context) {
var self = this;
var view_loaded_def;
$('#oe_linking_e').click(this.on_preview_view_button);
//this is button class which call method for open your form.
return self._super(context);
},
//method which open form
on_preview_view_button: function(e){
e.preventDefault();
this.do_action({
name: _t("View name"),
type: "ir.actions.act_window",
res_model: "object",
domain : [],
views: [[false, "list"],[false, "tree"]],
target: 'new',
context: {},
view_type : 'list',
view_mode : 'list'
});
}
},
});
};
in xml
file add button and give id="oe_linking_e"
whatever you give in js
code.