Use Variable in Command Substitution

Question

I need some help with the following simple bash script, where the variable i does not seem to get substituted when running curl (causing an error).

(This is just a simple abstraction of the actual script)

for i in {1..3}
do
  HTML=$(curl -s 'http://example.com/index.php?id=$i')
done;

try enclosing in double quotes instead of single quotes

amadain
– amadain

2013-08-09 06:48:50 +00:00
Commented Aug 9, 2013 at 6:48 — amadain
– amadain, Commented Aug 9, 2013 at 6:48

nosid · Accepted Answer · 2013-08-09 06:47:44Z

19

Variables are not substituted within single quotes. You have to use double quotes in this case:

for i in {1..3}; do
    HTML=$( curl -s "http://example.com/index.php?id=$i" )
done

answered Aug 9, 2013 at 6:47

nosid

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amadain · Accepted Answer · 2013-08-09 06:51:27Z

2

From http://tldp.org/LDP/abs/html/varsubn.html

Enclosing a referenced value in double quotes (" ... ") does not interfere with variable substitution. This is called partial quoting, sometimes referred to as "weak quoting." Using single quotes (' ... ') causes the variable name to be used literally, and no substitution will take place. This is full quoting, sometimes referred to as 'strong quoting.'

A

answered Aug 9, 2013 at 6:51

amadain

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