Im trying to hit a rest api with token in header.
apikeyName="$(date '+%s')"
key=$(curl -k -X POST -H "Content-Type: application/json" \
-d '{"name":"'$apikeyName'", "role": "Admin"}' \
http://admin:admin#localhost:3000/api/auth/keys | jq '.key')
echo $key
# # Alerting API
curl -k -X GET 'http://localhost:3000/api/alert-notifications' -H 'Authorization: Bearer '$key'';
Terminal output
"eyJrIjoiaWJPaDFFZXZMeW1RYU90NUR4d014T3hYUmR6NDVUckoiLCJuIjoiMTY3NTM1OTc4OCIsImlkIjoxfQ=="
{"message":"invalid API key","traceID":""}
First 1 is the key printing and last one from api response. I tried to hardcode the key and it works.
Short answer: Use jq -r '.key' to extract the key from the json response without adding quotes to it.
Long answer: There is a difference between quotes on the command line and quotes embedded in a variable. Consider:
key='"abcd"'
printf '%s\n' $key "abcd"
# prints:
# "abcd"
# abcd
Quotes on the command line are bash syntax. Bash notes what is being quoted and then removes the quotes from the command line when it's done, thus printf only prints abcd in the second case above.
Quotes inside a variable are plain old data. Bash doesn't do anything with them, so they get passed through to the command like any other data and printf prints "abcd" in the first case.
In your curl case the receiver doesn't expect the key to have quotes embedded in the data. So, curl -blah "keydata" works fine because bash takes the quotes out, but curl -blah $key fails because bash does NOT remove the embedded quotes.
See also: BashParser
Related
When write bash scripts, I want to store my whole curl command in heredoc to get a better layout. The following works fine:
#/bin/bash
read -r -d '' command1 <<- MULTI_STRING_SCOPE
curl -v www.stackoverflow.com
MULTI_STRING_SCOPE
But when add some json data with the -d option, the command is executed weirdly. For example:
#/bin/bash
read -r -d '' command2 <<- MULTI_STRING_SCOPE
curl -v www.stackoverflow.com
-d '{
"hello":"world"
}'
MULTI_STRING_SCOPE
response2=$(${command2})
Wrong logs from terminal:
curl: (3) URL using bad/illegal format or missing URL
curl: (3) unmatched close brace/bracket in URL position 1:
}'
And it seems that the curl take line }' as a seperated url, and thus the json data not sent as a unit.
How to solve the problem? Any suggestions will be highly appreciated.
I learned from this post to make heredoc work with curl command.
As the comment made by #Gordon Davisson in current post, we should not mix command with data. Since the json data set to -d option is only data and other parts is command, so I decide to use heredoc to store only the json data and remain other parts to be command itself, rather than store them in string by heredoc.
The result bash script should be something like this:
#/bin/bash
response3=$(curl -v www.stackoverflow.com \
-d #- <<- MULTI_STRING_SCOPE
{
"hello":"world"
}
MULTI_STRING_SCOPE
)
Notice: heredoc indent only works with tab, not with blanks. Be careful, especially when you are working with editors like Visual Studio Code, which may have already set indent as blanks for you.
I am having some problem expanding a variable.
First I set the value of the key.
KEY=Basic YWRtaW46YWRtaW4=
Then I tried calling it in curl but it was not working. To diagnose the problem, I decided to print the actual command that was being run. This was the result
print curl -H `print "Authorization: " $KEY` "http://192.168.1.1/userRpm/WlanMacFilterRpm.htm?Page=1&exclusive=0"
>> http://192.168.1.1/userRpm/WlanMacFilterRpm.htm?Page=1&exclusive=0
It was only printing the string after the KEY. Whereas, If I replaced it myself, it was printing the correct output.
print curl -H "Authorization: Basic YWRtaW46YWRtaW4=" "http://192.168.1.1/userRpm/WlanMacFilterRpm.htm?Page=1&exclusive=0"
>> curl -H Authorization: Basic YWRtaW46YWRtaW4= http://192.168.1.1/userRpm/WlanMacFilterRpm.htm?Page=1&exclusive=0
You need to enclose the value in quotes because you have a space.
KEY="Basic YWRtaW46YWRtaW4="
Doing KEY=Basic YWRtaW46YWRtaW4= without quotes is like assigning "Basic" to "KEY" and nothing to another variable called "YWRtaW46YWRtaW4".
Then simply use it like so:
curl -H "Authorization: $KEY" "http://192.168.1.1/userRpm/WlanMacFilterRpm.htm?Page=1&exclusive=0"
I am having some trouble figuring out the expansion of a variable in my shell script. If I replace the variable with the desired string it works.
#!/bin/zsh
KEY="$(curl -Ivs -X GET "http://admin:admin#192.168.1.1" &> >(awk '/^> Authorization/{ print $3 " " $4 }'))"
# The string returned by the curl and awk command is Basic "YWRtaW46YWRtaW4=" without double quotes.
curl -H "Authorization: $KEY" "http://192.168.1.1/userRpm/WlanMacFilterRpm.htm?Page=1&exclusive=1"
# This doesn't work
curl -H "Authorization: Basic YWRtaW46YWRtaW4=" "http://192.168.1.1/userRpm/WlanMacFilterRpm.htm?Page=1&exclusive=1"
# This works
The only thing thats different in the above two lines is.
-H "Authorization: $KEY"
-H "Authorization: Basic YWRtaW46YWRtaW4="
HTTP generally uses DOS style CR+LF line terminators. Whenever you parse data out of curl, you have to account for this.
To check if this is the problem, run your script with bash -x yourscript or zsh -x yourscript to see trace output that shows otherwise invisible carriage returns:
var=$'value\C-M' # zsh
var=$'value\r' # bash
(Dash and ash/busybox unfortunately doesn't highlight this problem, so try with one of the above shells)
To strip them, pipe your data through tr -d '\r'.
#!/bin/bash
while IFS='' read -r line || [[ -n "$line" ]]; do
echo "Text read from file: $line"
curl 'https://shoesworkshop.net/libraries/ajax/ajax.invoice.php?act=viewallinvoice&invoiceid="${line}"&sEcho=1&iColumns=8&iDisplayStart=0&iDisplayLength=20&bRegex=false&bRegex_0=false&bSearchable_0=true&bRegex_1=false&bSearchable_1=true&bRegex_2=false&bSearchable_2=true&bRegex_3=false&bSearchable_3=true&bRegex_4=false&bSearchable_4=true&bRegex_5=false&bSearchable_5=true&bRegex_6=false&bSearchable_6=true&bRegex_7=false&bSearchable_7=true&iSortCol_0=0&sSortDir_0=asc&iSortingCols=1&bSortable_0=true&bSortable_1=true&bSortable_2=true&bSortable_3=true&bSortable_4=true&bSortable_5=true&bSortable_6=true&bSortable_7=true' -H 'Host: shoesworkshop.net'| sed 's/^[^[[]*:/:/'
done < "$1"
inside $line there is a value like this
AAAAA
SSSSS
DDDDD
and i want to pass $line into curl command
can someone help me how?
i tried "'${line}'" and '${line}' and it still not working
i want to make a repeat call using curl get request from the url using variable from $line
For simple URLs, one way is to just use double quotes for the complete URL, including your variable expansion, ${line}, like this:
curl "https://shoe...&invoiceid=${line}&sEcho=1&iCo...table_7=true"
(Under single quotes, your shell variable line is not expanded.)
If your URL contains shell-special characters like $, it's best to combine both single and double quotes (and concatenate several strings, like explained here). For Example:
curl 'https://shoe...&invoiceid='"$line"'&sEcho=1&iCo...table_7=true'
# ^------ fixed part ------^ ^var^ ^------- fixed part ------^
However, if your variable contains characters that have to be URL-encoded (like space, &, ?, etc.) it's best to let curl handle that with --data-urlencode option. When called with this option, curl will default to POST method, but you can override this with -G, in which case your parameters will be appended to URL query. For example:
line="1&2?3 4"
curl "http://httpbin.org/get?x=1&y=2" --data-urlencode z="$line" -G
produces the right URL:
http://httpbin.org/get?x=1&y=2&z=1%262%3F3%204
Your script, fixed:
#!/bin/bash
while IFS='' read -r line || [[ -n "$line" ]]; do
echo "Text read from file: $line"
curl --data-urlencode invoiceid="$line" -G 'https://shoesworkshop.net/libraries/ajax/ajax.invoice.php?act=viewallinvoice&sEcho=1&iColumns=8&iDisplayStart=0&iDisplayLength=20&bRegex=false&bRegex_0=false&bSearchable_0=true&bRegex_1=false&bSearchable_1=true&bRegex_2=false&bSearchable_2=true&bRegex_3=false&bSearchable_3=true&bRegex_4=false&bSearchable_4=true&bRegex_5=false&bSearchable_5=true&bRegex_6=false&bSearchable_6=true&bRegex_7=false&bSearchable_7=true&iSortCol_0=0&sSortDir_0=asc&iSortingCols=1&bSortable_0=true&bSortable_1=true&bSortable_2=true&bSortable_3=true&bSortable_4=true&bSortable_5=true&bSortable_6=true&bSortable_7=true' -H 'Host: shoesworkshop.net' | sed 's/^[^[[]*:/:/'
done < "$1"
I need to use the curl command:
curl -d '{ "auth_token": "YOUR_AUTH_TOKEN", "text": "Hey, Look what I can do!" }' \http://localhost:3030/widgets/welcome
in a bash script but instead of "Hey, Look what I can do!" after the "YOUR_AUTH_TOKEN"; I need a variable $p .
This is my first time trying a bash script and read the tutorials on quotes and such but I still am not able to make it work.
The easiest thing to do is to read the data from a here document (a type of dynamically created temporary file), rather then trying to quote the entire thing as a string:
curl -d#- http://localhost:3030/widgets/welcome <<EOF
{ "auth_token": "YOUR_AUTH_TOKEN",
"text": "$p" }
EOF
If the argument to -d begins with a #, the remainder of the argument is taken as the name of a file containing the data. A file name of - indicates standard input, and the standard input to curl is supplied by the lines between the EOF tokens.
The alternative is to double-quote the string and escape all the embedded double quotes. Yuck.
... -d "{\"auth_token\": \"YOUR_AUTH_TOKEN\", \"text\": \"$p\"}" ...
An alternative to chepner's very good answer:
data=$( printf '{"auth_token":"YOUR_AUTH_TOKEN","text":"%s"}' "$p" )
curl -d "$data" http://localhost:3030/widgets/welcome