Dynamic Name For xlsx Report

Question

Hi All,
I'm working on module that generates salary report based on a criteria.I have Created a wizard namely *my_module_name.wizard, *It's of 'models.TransientModel'.My Wizard has a field named *file_name *i want this field as my xlsx file's name.
I have already tried this , It doesn't work , it always prints the file attribute's value as file name
<report
id= "mymodule_xlsx"
model= "mymodule.wizard"
string= "SIF Report"
report_type= "xlsx"
name= "mymodule_xlsx.xlsx"
file= "mymodule_xlsx"
print

Gustavo Hinojosa · Accepted Answer

modifying the object from python.
@ api.multi
def print_xlsx(self):
report = self.env ['ir.actions.report'] ._ get_report_from_name ('module.ng_hemployee_xlsx')
# en v11 ##################
report.report_file = "new name"
return self.env.ref ('module.ng_hemployee_xlsx').    report_action (self)
# en v9
report.name = "new name"
return self.env ['reporte'].      get_action (self, 'module.ng_hemployee_xlsx')

Reydi Hernandez · Answer

Hi guys, I solved the problem as follows:
#V12
@api.multi
​def action_print(self):
self.env.ref(report_template_xlsx).report_file = "New_name.xlsx"
return self.env.ref(report_template_xlsx).report_action(self)

Krupesh Laiya · Answer

for above you have to override odoo's report method from report  module

Arif · Answer

I have solved it by the following way. Hope it helps
# in your wizard model's action method
report_obj = self .env.ref( 'module.your_report_id' )
report_obj.name = f"your_file_prefix_{self .read()[ 0 ].get( 'your_input_filename' ) }_postfix.xlsx"
return report_obj.report_action( self, data =data)

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