Incrementing variable inside the curl URL parameter

Question

What I want to do it use curl with bash variable to retrieve content length from URL with incrementing parameters.

for ((i=2561;i<=2563;i++))
do 
(curl -i -s -k  -X 'GET' 'http:someurl.asp?q=$i)
done

When I do like this I always get the same content length from HTTP response. Now if I take same curl command and run outside the loop, like this:

curl -i -s -k  -X 'GET' 'http:someurl.asp?q=2562

it works perfectly. I tried both for and while loops, also tried reading from file with cat, but it's always the same content length. What am I missing here?

You aren't closing the quotes so i don't know how it isn't causing an error. Also it's in single quotes so the parameter isn't expanded, also $2562 will be blank unless you pass 2562 parameters to your script. — 123
– 123, Commented Jun 30, 2016 at 13:37
You're quoting the url in single-quotes, the $i won't be expanded. Moreover, if you want the result to contain a $, you have to escape it, ie use "http:someurl.asp?q=\$$i". — Aaron
– Aaron, Commented Jun 30, 2016 at 13:41
Thanks guys. @Aaron Yes, double quotes indeed fixed that! I spend good hour trying to figure this out. P.s. that $ before number was just a copy mistake. — ChildinTime
– ChildinTime, Commented Jun 30, 2016 at 13:49
@123 Actually, $2562 is interpreted as ${2}562 by bash. — chepner
– chepner, Commented Jun 30, 2016 at 14:09

chepner · Accepted Answer · 2016-06-30 14:08:00Z

2

A corrected code would be

for ((i=2561; i<=2563; i++))
do 
    curl -i -s -k  -X GET "http:someurl.asp?q=$i"
done

You need double quotes to allow $1 to be expanded to the current value of i.

answered Jun 30, 2016 at 14:08

chepner

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sjsam · Accepted Answer · 2016-06-30 13:49:11Z

0

Try

curl -i -s -k  -X 'GET' "http:someurl.asp?q=$i"

Using Single quotes ie - 'http:someurl.asp?q=$i' usually means

no variable expansion
no command substitution
you get the exact stuff you put inside ''.

Also (process) runs the process in a subshell which is unnecessary here.

Sidenote

The C style forloop sonstruct doesn't require you to precede variables with $. However when using the variables you need precede them $.

answered Jun 30, 2016 at 13:49

sjsam

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