I am trying to get the response from a curl url only for a particular value. For example
i am using the command
URLS=$(curl -g -H "Authorization: ${abc}" "https://api.buildkite.com/v2/organizations/org/agents?meta_data=[queue=dev]")
echo "${URLS}"
The metadata is actually as below:
"meta_data": [
"queue=dev"
]
The above curl command is giving the response for all agents in all queues and not able to get the required ones specific to queue=dev.
What is the correct way to pass url with brackets?
Related
I need to make a POST request via cURL from the command line. Data for this request is located in a file. I know that via PUT this could be done with the --upload-file option.
curl host:port/post-file -H "Content-Type: text/xml" --data "contents_of_file"
You're looking for the --data-binary argument:
curl -i -X POST host:port/post-file \
-H "Content-Type: text/xml" \
--data-binary "#path/to/file"
In the example above, -i prints out all the headers so that you can see what's going on, and -X POST makes it explicit that this is a post. Both of these can be safely omitted without changing the behaviour on the wire. The path to the file needs to be preceded by an # symbol, so curl knows to read from a file.
I need to make a POST request via Curl from the command line. Data for this request is located in a file...
All you need to do is have the --data argument start with a #:
curl -H "Content-Type: text/xml" --data "#path_of_file" host:port/post-file-path
For example, if you have the data in a file called stuff.xml then you would do something like:
curl -H "Content-Type: text/xml" --data "#stuff.xml" host:port/post-file-path
The stuff.xml filename can be replaced with a relative or full path to the file: #../xml/stuff.xml, #/var/tmp/stuff.xml, ...
If you are using form data to upload file,in which a parameter name must be specified , you can use:
curl -X POST -i -F "parametername=#filename" -F "additional_parm=param2" host:port/xxx
Most of answers are perfect here, but when I landed here for my particular problem, I have to upload binary file (XLSX spread sheet) using POST method, I see one thing missing, i.e. usually its not just file you load, you may have more form data elements, like comment to file or tags to file etc as was my case. Hence, I would like to add it here as it was my use case, so that it could help others.
curl -POST -F comment=mycomment -F file_type=XLSX -F file_data=#/your/path/to/file.XLSX http://yourhost.example.com/api/example_url
I was having a similar issue in passing the file as a param. Using -F allowed the file to be passed as form data, but the content type of the file was application/octet-stream. My endpoint was expecting text/csv.
You are able to set the MIME type of the file with the following syntax:
-F 'file=#path/to/file;type=<MIME_TYPE>
So the full cURL command would look like this for a CSV file:
curl -X POST -F 'file=#path/to/file.csv;type=text/csv' https://test.com
There is good documentation on this and other options here: https://catonmat.net/cookbooks/curl/make-post-request#post-form-data
I had to use a HTTP connection, because on HTTPS there is default file size limit.
https://techcommunity.microsoft.com/t5/IIS-Support-Blog/Solution-for-Request-Entity-Too-Large-error/ba-p/501134
curl -i -X 'POST' -F 'file=#/home/testeincremental.xlsx' 'http://example.com/upload.aspx?user=example&password=example123&type=XLSX'
I have problem to execute curl request in the directory has comma in the name using bash command line.
curl --request POST --form "file=#$PWD/input_file" http://HOSTURL.com > output_file
if the directory name is
"test" works
"test test" works
"test, test" doesn't work.
I tried many ways to escape characters like quotations, back slush, changing IFS... but still getting error "failed creating formpost data".
Could someone advise how I should treat such directory names?
This looks like a case curl isn't designed to handle. However, by passing the filename on stdin, you can avoid needing it to correctly parse that value at all.
curl --request POST --form "file=#-" http://HOSTURL.com <input_file >output_file
I have a python flask server running and the following http GET works on my browser, http://example.com/v1/api?lng=18.565810740668912&lat=-33.93153605161741 , however when I use command curl http://example.com/v1/api?lng=18.565810740668912&lat=-33.93153605161741 it doesn't work. How to correct my query to work for curl?
You have to pass URL as a string.
It's because for shell a lot of characters used in URL have special meaning (ie = is an assignment, & makes command to run in background, etc). So
curl "http://example.com/v1/api?lng=18.565810740668912&lat=-33.93153605161741"
should work as expected.
To pass the query-parameters separated and let cURL build the query-string we can also use option -G together with usual -d like
curl -G -d 'lng=18.565810740668912' -d 'lat=-33.93153605161741' http://example.com/v1/api
See also: Construct a Query String (TLDR: Use -G argument)
I'm having a problem with the YouTube API. I am trying to make a bash application that will make watching YouTube videos easy on command line in Linux. I'm trying to take some video search results through cURL, but it returns an error: curl: (16) HTTP/2 stream 1 was not closed cleanly: error_code = 1
the cURL command that I use is:
curl "https://ww.googleapis.com/youtube/v3/search" -d part="snippet" -d q="kde" -d key="~~~~~~~~~~~~~~~~"
And of course I add my YouTube data API key where the ~~~~~~~~ are.
What am I doing wrong?
How can I make it work and return the search attributes?
I can see two things that are incorrect in your request:
First, you mistyped "www" and said "ww". That is not a valid URL
Then, curl's "-d" options are for POSTing only, not GETting ,at least not by default. You have two options:
Add the -G switch to url, which lets curl re-interpret -d options as query options:
curl -G https://www.googleapis.com/youtube/v3/search -d part="snippet" -d q="kde" -d key="xxxx"
Rework your url to a typical GET request:
curl "https://www.googleapis.com/youtube/v3/search?part=snippet&q=kde&key=XX"
As a tip, using bash to interpret the resulting json might not be the best way to go. You might want to look into using python, javascript, etc. to run your query and interpret the resulting json.
I've below code in shell script
#!/bin/bash
oauth_consumer_key='sdfsfsd'
oauth_consumer_key_secret='1sdfsdfs1'
oauth_token='5wrwerwr476a1737fe09de2e4ew'
oauth_token_secret='ec2231779e4'
url='https url goes here'
token=$(./oauth $oauth_consumer_key $oauth_consumer_key_secret $oauth_token $oauth_token_secret GET $url)
curl_path='/usr/bin/curl'
curl_args="-H 'Authorization: $token'"
resp=$($curl_path $curl_args $url)
echo $resp
Here, I'm first running oauth.sh from this program by passing params & capturing output in token which works as expected. And this output looks as shown below:
OAuth oauth_consumer_key="45435", oauth_token="4r43", oauth_signature_method="HMAC-SHA1", oauth_timestamp="1415328827", oauth_nonce="4535345", oauth_version="1.0", oauth_signature="bKeewO%2BTJ7IHjurhtaftn9dNfxA%3D"' 'my url goes here'
Now, I need to invoke curl command from this program by passing above as param as shown below:
curl -H 'Authorization: OAuth oauth_consumer_key="45435", oauth_token="4r43", oauth_signature_method="HMAC-SHA1", oauth_timestamp="1415328827", oauth_nonce="4535345", oauth_version="1.0", oauth_signature="bKeewO%2BTJ7IHjurhtaftn9dNfxA%3D"' 'my url goes here'
When I run above curl command from terminal console, it works but from my shell script it gives error couldn't resolve OAuth
Can anyone help me out in fixing this issue?
Thanks!
It looks like this line is looking for "oauth" in the current directory.
token=$(./oauth
I would try putting the absolute path to oauth instead of ./oauth and see if that works.
If that doesn't work, I would strip it down a little and comment out everything after that line and just echo $token and see if it's showing the expected results and go from there.