Hi I hope this is clear enough-- I have read that I should create a new directory to be able to hide the actual file name and honestly don't understand what is meant by re-creating a new directory so I am just going to try to explain this without creating a directory.
I am using a dataset and do files from some very advanced stata code writers at the World bank, trying to do a similar analysis on a sub-sample of their population (only young people 15-29 years old) so I am just trying to follow along the do file they provided and I'm stuck on the initial instruction below. Obviously that sample path doesn't work on my machine so I try to use the file path from my machine that worked to open the dataset to begin with but now tells me is invalid. I guess it's the file path and not the macro command I'm using ($in/...) that is the source of the problem? Please advise--see below:

/* Depending on the country, we merged the individual respondent data with the household roster */
* Merge Main data with Household.

*. use $in/HH_ROSTER_s10_final.dta, replace

file /HH_ROSTER_s10_final.dta not found
r(601);

So instead i used the same file path I used earlier to identify the data set: (I am trying to follow the instructions for hiding the actual file path but acknowledging I don't understand exactly how to create a new directory to accomplish this!)

use $in/C:\Users\heather.dolphin\A123456789012345678901234 5678901234567890\B12345678901234567890123456789012 34567890\C1234567890123456789012345678901234567890 \D1234567890123456789012345678901234567890\E123456 7890123456789012345678901234567890\STEP Vietnam_HH Roster_s10_final.dta, replace
Now I am told the command was invalid r(198) Can you help?

I assume the individual respondent data was in fact merged with the household roster...but perhaps you have a recommendation for how I can know for sure ?
Thanks
Heather