I use curl to get http headers to find http status code and also return response. I get the http headers with the command
curl -I http://localhost
To get the response, I use the command
curl http://localhost
As soon as use the -I flag, I get only the headers and the response is no longer there. Is there a way to get both the http response and the headers/http status code in in one command?
This question is related to
shell
curl
http-status-code-415
Append a line "http_code:200" at the end, and then grep for the keyword "http_code:" and extract the response code.
result=$(curl -w "\nhttp_code:%{http_code}" http://localhost)
echo "result: ${result}" #the curl result with "http_code:" at the end
http_code=$(echo "${result}" | grep 'http_code:' | sed 's/http_code://g')
echo "HTTP_CODE: ${http_code}" #the http response code
In this case, you can still use the non-silent mode / verbose mode to get more information about the request such as the curl response body.
I found this question because I wanted independent access to BOTH the response and the content in order to add some error handling for the user.
You can print the HTTP status code to std out and write the contents to another file.
curl -s -o response.txt -w "%{http_code}" http://example.com
This allows you to check the return code and then decide if the response is worth printing, processing, logging, etc.
http_response=$(curl -s -o response.txt -w "%{http_code}" http://example.com)
if [ $http_response != "200" ]; then
# handle error
else
echo "Server returned:"
cat response.txt
fi
the verbose mode will tell you everything
curl -v http://localhost
For programmatic usage, I use the following :
curlwithcode() {
code=0
# Run curl in a separate command, capturing output of -w "%{http_code}" into statuscode
# and sending the content to a file with -o >(cat >/tmp/curl_body)
statuscode=$(curl -w "%{http_code}" \
-o >(cat >/tmp/curl_body) \
"$@"
) || code="$?"
body="$(cat /tmp/curl_body)"
echo "statuscode : $statuscode"
echo "exitcode : $code"
echo "body : $body"
}
curlwithcode https://api.github.com/users/tj
It shows following output :
statuscode : 200
exitcode : 0
body : {
"login": "tj",
"id": 25254,
...
}
My way to achieve this:
To get both (header and body), I usually perform a curl -D- <url>
as in:
$ curl -D- http://localhost:1234/foo
HTTP/1.1 200 OK
Connection: Keep-Alive
Transfer-Encoding: chunked
Content-Type: application/json
Date: Wed, 29 Jul 2020 20:59:21 GMT
{"data":["out.csv"]}
This will dump headers (-D
) to stdout (-
) (Look for --dump-header
in man curl).
IMHO also very handy in this context:
I often use jq to get that json data (eg from some rest APIs) formatted. But as jq doesn't expect a HTTP header, the trick is to print headers to stderr using -D/dev/stderr
. Note that this time we also use -sS
(--silent, --show-errors) to suppress the progress meter (because we write to a pipe).
$ curl -sSD/dev/stderr http://localhost:1231/foo | jq .
HTTP/1.1 200 OK
Connection: Keep-Alive
Transfer-Encoding: chunked
Content-Type: application/json
Date: Wed, 29 Jul 2020 21:08:22 GMT
{
"data": [
"out.csv"
]
}
I guess this also can be handy if you'd like to print headers (for quick inspection) to console but redirect body to a file (eg when its some kind of binary to not mess up your terminal):
$ curl -sSD/dev/stderr http://localhost:1231 > /dev/null
HTTP/1.1 200 OK
Connection: Keep-Alive
Transfer-Encoding: chunked
Content-Type: application/json
Date: Wed, 29 Jul 2020 21:20:02 GMT
Be aware: This is NOT the same as curl -I <url>
! As -I
will perform a HEAD
request and not a GET
request (Look for --head
in man curl. Yes: For most HTTP servers this will yield same result. But I know a lot of business applications which don't implement HEAD
request at all ;-P
In my experience we usually use curl this way
curl -f http://localhost:1234/foo || exit 1
curl: (22) The requested URL returned error: 400 Bad Request
This way we can pipe the curl when it fails, and it also shows the status code.
I use this command to print the status code without any other output. Additionally, it will only perform a HEAD request and follow the redirection (respectively -I
and -L
).
curl -o -I -L -s -w "%{http_code}" http://localhost
This makes it very easy to check the status code in a health script:
sh -c '[ $(curl -o -I -L -s -w "%{http_code}" http://localhost) -eq 200 ]'
This is a way to retrieve the body "AND" the status code and format it to a proper json or whatever format works for you. Some may argue it's the incorrect use of write format option but this works for me when I need both body and status code in my scripts to check status code and relay back the responses from server.
curl -X GET -w "%{stderr}{\"status\": \"%{http_code}\", \"body\":\"%{stdout}\"}" -s -o - “https://github.com” 2>&1
run the code above and you should get back a json in this format:
{
"status" : <status code>,
"body" : <body of response>
}
with the -w write format option, since stderr is printed first, you can format your output with the var http_code and place the body of the response in a value (body) and follow up the enclosing using var stdout. Then redirect your stderr output to stdout and you'll be able to combine both http_code and response body into a neat output
while : ; do curl -sL -w "%{http_code} %{url_effective}\\n" http://host -o /dev/null; done
The -i
option is the one that you want:
curl -i http://localhost
-i, --include Include protocol headers in the output (H/F)
Alternatively you can use the verbose option:
curl -v http://localhost
-v, --verbose Make the operation more talkative
I have used this :
request_cmd="$(curl -i -o - --silent -X GET --header 'Accept: application/json' --header 'Authorization: _your_auth_code==' 'https://example.com')"
To get the HTTP status
http_status=$(echo "$request_cmd" | grep HTTP | awk '{print $2}')
echo $http_status
To get the response body I've used this
output_response=$(echo "$request_cmd" | grep body)
echo $output_response
A one-liner, just to get the status-code would be:
curl -s -i https://www.google.com | head -1
Changing it to head -2
will give the time as well.
If you want a while-true loop over it, it would be:
URL="https://www.google.com"
while true; do
echo "------"
curl -s -i $URL | head -2
sleep 2;
done
Which produces the following, until you do cmd+C
(or ctrl+C
in Windows).
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:38 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:41 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:43 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:45 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:47 GMT
------
HTTP/2 200
date: Sun, 07 Feb 2021 20:03:49 GMT
This command
curl http://localhost -w ", %{http_code}"
will get the comma separated body and status; you can split them to get them out.
You can change the delimiter as you like.
Some good answers here, but like the OP I found myself wanting, in a scripting context, all of:
curl
exit status codeThis is difficult to achieve with a single curl
invocation and I was looking for a complete solution/example, since the required processing is complex.
I combined some other bash recipes on multiplexing stdout/stderr/return-code with some of the ideas here to arrive at the following example:
{
IFS= read -rd '' out
IFS= read -rd '' http_code
IFS= read -rd '' status
} < <({ out=$(curl -sSL -o /dev/stderr -w "%{http_code}" 'https://httpbin.org/json'); } 2>&1; printf '\0%s' "$out" "$?")
Then the results can be found in variables:
echo out $out
echo http_code $http_code
echo status $status
Results:
out { "slideshow": { "author": "Yours Truly", "date": "date of publication", "slides": [ { "title": "Wake up to WonderWidgets!", "type": "all" }, { "items": [ "Why <em>WonderWidgets</em> are great", "Who <em>buys</em> WonderWidgets" ], "title": "Overview", "type": "all" } ], "title": "Sample Slide Show" } }
http_code 200
status 0
The script works by multiplexing the output, HTTP response code and curl
exit status separated by null characters, then reading these back into the current shell/script. It can be tested with curl
requests that would return a >=400 response code but also produce output.
Note that without the -f
flag, curl
won't return non-zero error codes when the server returns an abnormal HTTP response code i.e. >=400, and with the -f
flag, server output is suppresses on error, making use of this flag for error-detection and processing unattractive.
Credits for the generic read
with IFS processing go to this answer: https://unix.stackexchange.com/a/430182/45479 .
Wow so many answers, cURL devs definitely left it to us as a home exercise :) Ok here is my take - a script that makes the cURL working as it's supposed to be, i.e.:
Save it as curl-wrapper.sh
:
#!/bin/bash
output=$(curl -w "\n%{http_code}" "$@")
res=$?
if [[ "$res" != "0" ]]; then
echo -e "$output"
exit $res
fi
if [[ $output =~ [^0-9]([0-9]+)$ ]]; then
httpCode=${BASH_REMATCH[1]}
body=${output:0:-${#httpCode}}
echo -e "$body"
if (($httpCode < 200 || $httpCode >= 300)); then
# Remove this is you want to have pure output even in
# case of failure:
echo
echo "Failure HTTP response code: ${httpCode}"
exit 1
fi
else
echo -e "$output"
echo
echo "Cannot get the HTTP return code"
exit 1
fi
So then it's just business as usual, but instead of curl
do ./curl-wrapper.sh
:
So when the result falls in 200-299 range:
./curl-wrapper.sh www.google.com
# ...the same output as pure curl would return...
echo $?
# 0
And when the result is out of in 200-299 range:
./curl-wrapper.sh www.google.com/no-such-page
# ...the same output as pure curl would return - plus the line
# below with the failed HTTP code, this line can be removed if needed:
#
# Failure HTTP response code: 404
echo $?
# 1
Just do not pass "-w|--write-out" argument since that's what added inside the script
Source: Stackoverflow.com