To access function from an library you have either load and attach it:

library(stringr) # Or require(stringr)
str_replace(x, "XXXX", "XXxxx")

or use double colon operator:

stringr::str_replace(x, "XXXX", "XXxxx")

Unfortunately :: is quite expensive so if you prefer to keep your namespace clean you should consider creating a local binding:

str_replace <- stringr::str_replace
str_replace(x, "XXXX", "XXxxx")

On a side not using internal API with ::: is probably not the best idea. Ignoring good practices it is simply far to slow to be useful in practice.

Edit: It appears @akrun has deleted his answer, so it is reproduced below. I made the (perhaps faulty) assumption you wished to transform the whole dataframe, and that the dataframe was appropriately formatted. These issues are avoided with clearer questions, and sample data.

It's a little hard to tell with no clue as to the values of your global variables and no data (you can generate fake data, btw, as long as it presents the same issue), but my guess is below.

When piping, the previous result is piped in as the first argument of the next function. You can see this in the error message: Error in str_replace_all(., string, pattern, replacement) -- the ., shows the piped in argument. Here, the first argument is "string". Therefore, the piped in result is being used as string, "string" is used as pattern, "pattern" is being used for replacement, and the "" you put in for replacement is left as an unused argument, causing your error.

Might help to use str_replace_all(pattern, replacement), or specify arguments: str_replace_all(pattern = pattern, replacement = replacement)

Ex.

data <- as.data.frame(matrix(ncol=2, nrow=2))
data$V1 <- c("  NA", "foo")
data$V2 <- c("bar", "boo")
data %>%
    str_replace_all("oo", "xx")

If you only wish to transform one column (from @akrun): Simply use mutate to create a new column based off of the preexisting column. If you wish to replace the column, give it the same name:

Ex.

data <- as.data.frame(matrix(ncol=2, nrow=2))
data$V1 <- c("  NA", "foo")
data$V2 <- c("bar", "boo")
data
    V1  V2
1   NA bar
2  foo boo

#new column
data %>%
    mutate(new = str_replace_all(V1, "oo", "xx"))

    V1  V2  new
1   NA bar   NA
2  foo boo  fxx

#column replacement
data %>%
    mutate(V1 = str_replace_all(V1, "oo", "xx"))

    V1  V2
1   NA bar
2  fxx boo

Escaping the parentheses does it...

str_replace(fruit,"\\(\\)","")
# [1] "goodapple"

You may also want to consider exploring the "stringi" package, which has a similar approach to "stringr" but has more flexible functions. For instance, there is stri_replace_all_fixed, which would be useful here since your search string is a fixed pattern, not a regex pattern:

library(stringi)
stri_replace_all_fixed(fruit, "()", "")
# [1] "goodapple"

Of course, basic gsub handles this just fine too:

gsub("()", "", fruit, fixed=TRUE)
# [1] "goodapple"

Check and make sure that your source files are encoded in UTF-8 as well. This can be easily avoided in the future by changing the default encoding to UTF-8 within your text editor of choice.

As @mrlew pointed out, str is result of parseInt and therefore, it's a number. replace() is a string method, so it will not work on a number.

If I understood correctly, you would like to replace a string, retrieve a code and then generate an image tag with the new string and code.

I'd go with...

<input type="text" id="text" />
<input type="button" value="See Code" onclick="myFunction();">
<input type="text" id="code" name="code">

<script>
function myFunction() {
    //changed
    var str = document.getElementById("text").value; //get text
    var res = str.replace("ftpadress", "htmladress"); //replace
    var code = parseInt(str).toString(); //get code and cast it back to a string
    document.getElementById("code").value = code; //insert code
    var withTag = code.concat("<img src='", res, "' width='100%'>"); //generate tag
}
</script>