rename file with date format

[jpappire]

Can some one please advise that how to rename file name with date format.
I have below file and want to rename date with current date .

ex: support data 05122019.xlsx

i want to create file like below
support data 31032019.xlsx

[Yarnabrina]

Are you looking for something like this?

df <- iris

write.csv(x = df,
          file = "try_iris 01012019.csv")

list.files()
#> [1] "reprex_reprex.R"        "reprex_reprex.spin.R"  
#> [3] "reprex_reprex.spin.Rmd" "try_iris 01012019.csv"

file.rename(from = "try_iris 01012019.csv",
            to = paste(gsub(pattern = "\\d|.csv",
                            replacement = "",
                            x = "try_iris 01012019.csv"),
                       format(x = Sys.Date(),
                              format = "%d%m%Y"), 
                       ".csv",
                       sep = "",
                       collapse = ""))
#> [1] TRUE

list.files()
#> [1] "reprex_reprex.R"        "reprex_reprex.spin.R"  
#> [3] "reprex_reprex.spin.Rmd" "try_iris 31032019.csv"

^{Created on 2019-03-31 by the reprex package (v0.2.1)}

I've done with .csv, you can change it to .xlsx easily.

[anshuman_k]

Here's a way to do it for xlsx file.
Please check how to use the openxlsx library for more details.

wb <- openxlsx::loadWorkbook(File Name)
openxlsx::saveWorkbook (wb,paste0("support data ",Sys.Date()), overwrite = T)

[Yarnabrina]

Hi Anshuman, welcome to the community!

The method you suggested above is indeed a good one, but it doesn't really rename the file. It creates another copy of the file with a name of your choice. See below:

# creating dummy data
openxlsx::write.xlsx(x = iris,
                     file = "try_iris 01012019.xlsx")
#> Note: zip::zip() is deprecated, please use zip::zipr() instead

# name of files in working directory before
list.files()
#> [1] "reprex_reprex.R"        "reprex_reprex.spin.R"  
#> [3] "reprex_reprex.spin.Rmd" "try_iris 01012019.xlsx"

# what you (@anshuman_k) suggested
openxlsx::saveWorkbook(wb = openxlsx::loadWorkbook(file = "try_iris 01012019.xlsx"),
                       file = paste0(gsub(pattern = "\\d|.xlsx",
                                          replacement = "",
                                          x = "try_iris 01012019.xlsx"),
                                     format(x = Sys.Date(),
                                            format = "%d%m%Y"), 
                                     ".xlsx"))

# name of files in working directory after
list.files()
#> [1] "reprex_reprex.R"        "reprex_reprex.spin.R"  
#> [3] "reprex_reprex.spin.Rmd" "try_iris 01012019.xlsx"
#> [5] "try_iris 01042019.xlsx"

^{Created on 2019-04-01 by the reprex package (v0.2.1)}

[anshuman_k]

Yeah I do agree.
It will work when the file, one is working is same as the new to be created (updated).