Output of cURL is corrupted when stored to a bash variable

Question

This is what console outputs when I execute cURL directly from the terminal:

/# curl -ksi http://localhost/ 
HTTP/1.1 302 Moved Temporarily
Server: nginx/1.6.2
Date: Sun, 14 Jan 2018 11:49:38 GMT
Content-Type: text/html
Content-Length: 160
Connection: keep-alive
Location: http://localhost/welcome/default
Cache-Control: private, max-age=0

<html>
<head><title>302 Found</title></head>
<body bgcolor="white">
<center><h1>302 Found</h1></center>
<hr><center>nginx/1.6.2</center>
</body>
</html>

But if I try to save the output of the cURL to a variable the content of it ends up looking corrupted:

/# VAR=`curl -ksi http://localhost/`
/# echo $VAR
 </html>nter>nginx/1.6.2</center>er>d>fault

What is it that I am doing wrong in this case?

You are not quoting variables, that's what's wrong. Always say `"$VAR"` unless you have a specific reason not to. — n. 1.8e9-where's-my-share m., Jan 14 '18 at 12:00
@Cyrus \r's are not a problem. \r's not followed by \n's are a problem. No quoting means all the \n's are turned into spaces. Just use the quotes to not let it happen. — n. 1.8e9-where's-my-share m., Jan 14 '18 at 12:27
@Cyrus may I ask you why the question is marked as duplicate? If I would know quoting variables is the issue I would probably search for "when should I use quoting?". The reason of downvoting is completely out of my understanding :) Well done community! — wizardzloy, Jan 14 '18 at 16:40

score 0 · Answer 1 · answered Jan 14 '18 at 11:58

0

Use an array

var=( "$(curl -ksi http://localhost/)" )
echo "${var[@]}"

^{Side-Note: Don't use all uppercase variables like VAR as user variables as they are reserved for the shell environment}

answered Jan 14 '18 at 11:58

sjsam

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The array will only contain a single element. It brings no value over a simple string variable. – tripleee Jan 14 '18 at 12:24
I agree.. I was thinking about a specific use case, which in fact I saw in some answer in SO.. But I couldn't link it though. – sjsam Jan 14 '18 at 12:28
thanks for the side note! The answer is not entirely right though, not quoting the variable was the issue – wizardzloy Jan 14 '18 at 16:47
Indeed, but you got the point right? :-) So all, good – sjsam Jan 14 '18 at 16:48

Output of cURL is corrupted when stored to a bash variable

1 Answers1