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I am using multi website for eCommerce. Since there is a field for choose website under product template form, I want to add a field with the website_domain/website_url so that I can click the url and redirect to the right website frontend. I do not like the default Publish on website button since it force me leave the backend. I am new to python. But I already done that with the web.base.url with following code:

class ProductURL(models.Model):
_inherit = 'product.template'

product_url = fields.Char(string="Product URL",compute='_computeURL')

@api.multi
def _computeURL(self):
for record in self:
base_url = self.env['ir.config_parameter'].sudo().get_param('web.base.url', default='')
if base_url:
record.product_url =base_url+""+record.website_url

So now I have a field with link web.base.url/product/product_id which I can click to go to front-end. But I have no idea how to get the website domain I selected for the product and if the website field is leaving blank then it should fall to web.base.url. Any help is appreciated. Thank you!

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Best Answer

you should see method "open_website_url" if you are on odoo 12.

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Author

I do not get it. Is this some existing function I am not aware of or some existing field. I simply tried adding <field name="open_website_url"/> to the template form but it is can not be saved. I simply need a need which can be opened in a new tap to the right website of the product. Thank you.

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