CODEWITHSUNDEEP
bashshellcurlsh
havmaage
havmaage

Reputation: 603

curl command not working from within shell script but works fine out site

I am struggling with a small bash shell script that should list my csv files and upload them to a external Service using curl. when i echo my command variable out to console and execute it i works fine, but when i execute it within my script it complains about a extra qouta "

my script looks like this 

#!/usr/bin/bash
LOGDIR="/var/log/tyk/"
FILE_EXSTENSION_NAME="*.csv"
CURLCMD="curl -k -i -H"
PORT="9992"
ENDPOINT="/endpoint"
URL="https://localhost"



for i in `ls $LOGDIR$FILE_EXSTENSION_NAME`; do
    filename=`echo $i | awk -F "/" ' { print $5 }'`
    echo $filename
    CMD="$CURLCMD \"filename: $filename\" -F \"data=@$LOGDIR$filename\" $URL:$PORT$ENDPOINT"
    $CMD
done

When i run it i getfollowing output 

% Total    % Received % Xferd  Average Speed   Time    Time     Time     Current
                             Dload  Upload   Total   Spent    Left  Speed
 0     0    0     0    0     0      0      0 --:--:-- --:--:-- --:--:--0curl: (6) Could not resolve host: 2018-October-18-10.csv"; Unknown error
 0     0    0     0    0     0      0      0 --:--:-- --:--:-- --:--:--      0     
curl: (26) couldn't open file "/var/log/tyk/2018-October-18-10.csv""

If i do a echo $CMD in my script i got 

curl -k -i -H "filename: 2018-October-18-10.csv" -F "data=@/var/log/tyk/2018-October-18-10.csv" https://localhost:9992/endpoint

and that works ok

I cannot figure out what it is that i am not doing wrong

Upvotes: 3

Views: 7761

Answers (1)

Socowi
Socowi

Reputation: 27195

First of all, don't store commands in variables. Apart from that, there are two problems:

  1. Quotes inside expanded variables do not work as if you typed them in directly.
  2. The subshell $() around $($CMD) seems suspicous. Not only do you execute the command stored in $CMD, but you also execute the output of that command!

1. Quotes inside variables

When you enter the command echo "a b" then bash will process that command such that echo is executed with the argument a b. The output will be a b.

When you store that command inside a variable and expand that variable, bash will process the variable's content differently. The command echo will be executed with the arguments "a and b". The output will be "a b".

This is the reason for the error message

curl: (26) couldn't open file "/var/log/tyk/2018-October-18-10.csv""

curl is trying to open a path with an actual quote inside. Such a path does not exist on your system.

To get around this issue, you could write eval "$cmd", then the command would be executed as if you typed it in directly. However, you really shouldn't. I'd rewrite the script instead to not store commands in variables.

2. Subshell around $cmd:

cmd='echo something'
$cmd

This would print something. However, your script doesn't stop there, because you enclosed $cmd in $(). Therefore the output something gets executed too.

cmd='echo something'
$($cmd)

results in

bash: something: command not found

Upvotes: 6

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