Openxlsx hyperlink output display in Excel - r

Question

I am trying to take in a data frame an excel sheet that has two columns,
Column A contains names of stores
Column B contains the URL of those stores.
I would like to take Column A and make it a clickable hyperlink so instead of plain text, it is a hyperlink to the store website.
I have attempted to use openxlsx package to generate the correct output.
I have attempted to use the following code snip.
x <- c("https://www.google.com", "https://www.google.com.au")
names(x) <- c("google", "google Aus")
class(x) <- "hyperlink"
writeData(wb, sheet = 1, x = x, startCol = 10)
which comes from this post of a similar nature.
https://stackoverflow.com/a/48973469/11958444
My problem however is when I replace the appropriate parts of the code e.g.:
x <- df$b
names(x) <- df$a
class(x) <- "hyperlink"
writeData(wb, sheet = 1, x = x, startCol = 10)
instead of giving me a column of hyperlinks that has the store name as the output, it gives me the entire URL as the output. Is there something I am missing from my code?
I get an output that has a clickable link, but instead of the URL appearing with the name, it instead just prints out the URL.

Answer 1

An approach using openxlsx:
library(openxlsx)
library(dplyr)
# create sample data
df <- data.frame(
site_name = c("Zero Hedge", "Free Software Foundation"),
site_url = c("https://www.zerohedge.com", "https://www.fsf.org")
)
# add new column that manually constructs Excel hyperlink formula
# note backslash is required for quotes to appear in Excel
df <- df %>%
mutate(
excel_link = paste0(
"HYPERLINK(\"",
site_url,
"\", \"",
site_name,
"\")"
)
)
# specify column as formula per openxlsx::writeFormula option #2
class(df$excel_link) <- "formula"
# create and write workbook
wb <- createWorkbook()
addWorksheet(wb, "df_sheet")
writeData(wb, "df_sheet", df)
saveWorkbook(wb, "wb.xlsx", overwrite = TRUE)

Answer 2

I like the current answer but for me it leaves the hyperlink as black. Here are three alternative solutions that I came up with...
1. mark the url column as a hyperlink
library(openxlsx)
# create sample data
df <- data.frame(
site_name = c("Zero Hedge", "Free Software Foundation"),
site_url = c("https://www.zerohedge.com", "https://www.fsf.org")
)
# create and write workbook
wb <- createWorkbook()
addWorksheet(wb, "df_sheet")
class(df$site_url)<-"hyperlink" # mark as a hyperlink
writeData(wb,"df_sheet",df$site_url,startCol = which(colnames(df)=="site_url"), startRow = 2)
writeData(wb, "df_sheet", df) # overwrite the sheet
saveWorkbook(wb, "wb.xlsx", overwrite = TRUE)
2. apply the hyperlink to the site_name column
library(openxlsx)
# create sample data
df <- data.frame(
site_name = c("Zero Hedge", "Free Software Foundation"),
site_url = c("https://www.zerohedge.com", "https://www.fsf.org")
)
# create and write workbook
wb <- createWorkbook()
addWorksheet(wb, "df_sheet")
class(df$site_url)<-"hyperlink" # mark as a hyperlink
writeData(wb,"df_sheet",df$site_url,startCol = which(colnames(df)=="site_name"), startRow = 2)
class(df$site_url)<-"character" # change back to a character which is optional
writeData(wb, "df_sheet", df) # overwrite the sheet to get the new pretty name overlaying the hyperlink
saveWorkbook(wb, "wb.xlsx", overwrite = TRUE)
3. new column with the link applied to the names from site_name
library(openxlsx)
# create sample data
df <- data.frame(
site_name = c("Zero Hedge", "Free Software Foundation"),
site_url = c("https://www.zerohedge.com", "https://www.fsf.org")
)
df$pretty_link<-df$site_name #new column for pretty link
# create and write workbook
wb <- createWorkbook()
addWorksheet(wb, "df_sheet")
class(df$site_url)<-"hyperlink" # mark as a hyperlink
writeData(wb,"df_sheet",df$site_url,startCol = which(colnames(df)=="pretty_link"), startRow = 2)
class(df$site_url)<-"character" # change back to a character which is optional
writeData(wb, "df_sheet", df) # overwrite the sheet to get the new pretty name overlaying the hyperlink
saveWorkbook(wb, "wb.xlsx", overwrite = TRUE)