xlsx.openFile: Open a file

Description Usage Arguments Author(s) References Examples

View source: R/r2excel.r

Description

Open a file from the current working directory.

Usage

1
xlsx.openFile(filename = NULL)

Arguments

filename

the name of the file. The file should be in the current working directory.

Author(s)

Alboukadel Kassambara <alboukadel.kassambara@gmail.com>

References

http://www.sthda.com

Examples

1
2
xlsx.writeFile(USArrests, file="USArrests.xlsx", sheetName="USArrests")
xlsx.openFile("USArrests.xlsx")

kassambara/r2excel documentation built on May 20, 2019, 7:40 a.m.