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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;
571
Using variable from command line or terminal You don't have to use any special character before the variable name at the time of setting value in BASH like other programming languages. But you have to use '$' symbol before the variable name when you want to read data from the variable.
Command substitution allows you to capture the output of any command as an argument to another command. You can perform command substitution on any command that writes to standard output. The read special command assigns any excess words to the last variable.
Unix Command Course for Beginners In this chapter, we will learn how to use Shell variables in Unix. A variable is a character string to which we assign a value. The value assigned could be a number, text, filename, device, or any other type of data. A variable is nothing more than a pointer to the actual data.
Variables are not substituted within single quotes. You have to use double quotes in this case:
for i in {1..3}; do
HTML=$( curl -s "http://example.com/index.php?id=$i" )
done
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.'
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