For Loop: can the end value come from the value of a variable? - for-loop

Question

I have a forvalues loop:
forvalues x = 1(1)50 {
/* Code goes here */
}
Instead of 50, ideally, I would like that value to come as follows. I have a variable name. Let length = length(name). Whatever the largest value is for length, I would like that to be in place of the 50. I could not figure how to write a forvalues loop in which the end point was not directly stated numerically.
I am thinking that I could deduce the maximum length of the variable as follows:
gen id = 1
gen length = length(name)
by id, sort: egen maxlength = max(length)
From there though I do not know how to store this value into the for loop.
Alternatively, would this be better coded by a while loop?
Something like:
gen x = 1
while (x <= maxlength) {
/* Same Code Here */
replace x = x + 1
}

Answer 1

Based on the documentation I've read, it is possible to use macros but with the caveat that changing the end of the range within the forvalues loop has no effect on the number of times the loop will occur. For instance, if length(name) is 50 when the forvalues loop starts, and you change the length of name within the loop, it will still only loop 50 times.
Technically, you'd be better off using a while loop since forvalues was intended to be used when the end of the range is a literal value. You can use a forvalues loop, but you should use a while loop.
Here's my source to back this up:
http://www.stata.com/manuals13/pforvalues.pdf
Specifically:
Technical note
It is not legal syntax to type
. scalar x = 3
. forvalues i = 1(1)x' {
2. local x =x' + 1
3. display `i'
4. }
forvalues requires literal numbers. Using macros, as shown in the following technical note, is
allowed.
And:
Using macros, as shown in the following technical note, is
allowed.
Technical note
The values of the loop bounds are determined once and for all the first time the loop is executed.
Changing the loop bounds will have no effect. For instance,
will not create an infinite loop. With `n' originally equal to 3, the loop will be performed three
times.
local n 3
forvalues i = 1(1)`n' {
local n = `n' + 1
display `i'
}
Output:
1
2
3

Answer 2

Here is the trick with Stata which I think may work for you. I am using the data auto from Stata datasets.
sysuse auto
Suppose the variable name here be price. Now you want the length of variable price.
sum price
gen length=r(N)
To see what is r(N) type return list after running the sum price.
In your loop it goes like follows: (Updated as per #Nick)
forvalues x = 1/`r(N)'{
/* Code goes here */
}
OR:
local length=r(N)
forvalue i=1/`length'{
dis "`i'"
}
Note: It is not clear why you want for loop.So my answer is restricted to what you only asked for.

Answer 3

#Metrics' first code won't quite work. Here is a better way, cutting out what I call the middle macro.
Start with something more like
. su price, meanonly
. forval j = 1/`r(N)' {

Answer 4

An equivalent approach to the one proposed by #Nick and #Metrics is the following:
sysuse auto, clear
count if !missing(price)
forvalues x = 1 / `r(N)' {
/* Code goes here */
}