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xlsxreportnameodoo11community
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Muhammed Nishad

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_report_name="(object.file_name)"
    menu="False"
    attachment_use="0"/>

Please Tell me how do we do this?

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Gustavo Hinojosa
Jawaban Terbai

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')

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Reydi Hernandez
Jawaban Terbai

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)

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Krupesh Laiya
Jawaban Terbai

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

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Muhammed Nishad
Penulis

The report is printing fine.The Problem is with the file Name.

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Arif
Jawaban Terbai

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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