Bash script curl

Question

#!/bin/bash

read -p 'Username: ' uservar
read -p 'Password: ' passvar

cacheVariable1="\"Content-Type:application/json"\"
cacheVariable2="\"Cache-Control:no-cache"\"
parametersVariable="'{\"username\":\"$uservar\",\"password\":\"$passvar\"}'"
echo $parametersVariable
echo $cacheVariable1 $cacheVariable2
websiteVariable="https://example.com/session"

echo $websiteVariable

entireURL="curl -X POST -H "$cacheVariable1" -H "$cacheVariable2" -d "$parametersVariable" "$websiteVariable""

echo "Entire URL IS: $entireURL"

result=`$entireURL`


echo "$result"

I want my script like this: curl -X POST -H "Content-Type: application/json" -H "Cache-Control: no-cache" -d '{"username":"[email protected]", "password":"Password123"}' "https://example.com/session"

Entire URL IS: curl -H "Content-Type:application/json" -H "Cache-Control:no-cache" -d '{"username":"zzzzzz","password":"azzzsass"}' https://cloud.tenable.com/session

But it does not execute in the bash. It gives me this error:

{"statusCode":400,"error":"Bad Request","message":"child \"username\" fails because [\"username\" is required]","validation":{"source":"payload","keys":["username"]}}

But it does not work. Can anyone help me?

Update

I solved it on my own. Everything was correct, except for executing as eval $entireURL. Because A command embedded within a parenthesis runs as a sub-shell so my environment variables were missing.

@Cyrus i checked, but i do not understand where i am going wrong. Can you help? — tech_enthusiast
– tech_enthusiast, Commented Feb 2, 2018 at 6:21
See BashFAQ #50: I'm trying to put a command in a variable, but the complex cases always fail!. Summary: don't put commands in variables, just execute them directly. (Capturing their output to variables is ok, though.) — Gordon Davisson
– Gordon Davisson, Commented Feb 2, 2018 at 6:35
@GordonDavisson I tried your way also but it did not work. thats why i went like this. It looks like there is some problem in the way i put curl script into bash, but i dont understand what. — tech_enthusiast
– tech_enthusiast, Commented Feb 2, 2018 at 6:42
I'd guess there's some problem with nested quotes in the -d parameter, but I'd have to see what you tried to be specific. In any case, putting it in a variable doesn't help any, it just adds another level of confusion. I'd post another question with the non-variable version. — Gordon Davisson
– Gordon Davisson, Commented Feb 2, 2018 at 6:48

tech_enthusiast · Accepted Answer · 2018-02-05 04:45:41Z

4

#!/bin/bash

read -p 'Username: ' uservar
read -p 'Password: ' passvar

cacheVariable1="\"Content-Type:application/json"\"
cacheVariable2="\"Cache-Control:no-cache"\"
parametersVariable="'{\"username\":\"$uservar\",\"password\":\"$passvar\"}'"
echo $parametersVariable
echo $cacheVariable1 $cacheVariable2
websiteVariable="https://example.com/session"

echo $websiteVariable

entireURL="curl -X POST -H "$cacheVariable1" -H "$cacheVariable2" -d "$parametersVariable" "$websiteVariable""

echo "Entire URL IS: $entireURL"

result=`$entireURL`

eval $entireURL

This works perfect !

answered Feb 5, 2018 at 4:45

tech_enthusiast

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