Skip to content

ISFORMULA() not recognized / throws exception #356

@billynoah

Description

@billynoah

ISFORMULA() not recognized and throws exception. Steps to reproduce. Create a new spreadsheet. In the cell A1 enter

=ISFORMULA(B1)

The result should be FALSE and is valid. Try loading the spreadsheet in PhpSpreadsheet and read cells like this:

$spreadsheet = \PhpOffice\PhpSpreadsheet\IOFactory::load($filename);
$rows = $spreadsheet->getSheet($sheet)->toArray();

Results in an exception:

Sheet1!A1 -> undefined named range / function '_xlfn.ISFORMULA'

Related: PHPOffice/PHPExcel#1136

formula_test.xlsx

Metadata

Metadata

Assignees

No one assigned

    Type

    No type

    Projects

    No projects

    Milestone

    No milestone

    Relationships

    None yet

    Development

    No branches or pull requests

    Issue actions