I'm trying to send a file as base64-encoded data via POST to the Bugzilla REST API as follows:
curl -X POST https://www.example.com/rest/bug/$id/attachment -H "Content-Type: application/json" \
-d "{\
\"login\" : \"$username\", \
\"password\" : \"$password\", \
\"ids\" : [ $id ], \
\"summary\" : \"...\", \
\"content_type\" : \"application/gzip\", \
\"data\" : \"$data\"\
}"
What I'm getting is an error from cURL that the argument list is too long. Presumably, this is because the file ($data) I'm trying to send is more than the shell maximum (the file is 11M). What I've seen online is that the best way to get around that is to have cURL read the data from a file using --data-binary. But since I need to send a username and password, I'd prefer not to have to save the entire file with them inside.
Is there some way to get around this maximum, or is there another way to send a large amount of data this way? I prefer native Linux tools, as I want this script to be portable.
You can try using a file like this, which is the recommended way.
curl -i \
-H 'Accept:application/json' \
-H 'Authorization:Basic $username:$password' \
-X POST -d #datafile.txt https://www.example.com/rest/bug/$id/attachment
Related
According to the document Get Azure AD tokens for service principals:
curl -X POST -H 'Content-Type: application/x-www-form-urlencoded' \
https://login.microsoftonline.com/<tenant-id>/oauth2/v2.0/token \
-d 'client_id=<client-id>' \
-d 'grant_type=client_credentials' \
-d 'scope=2ff814a6-3304-4ab8-85cb-cd0e6f879c1d%2F.default' \
-d 'client_secret=<client-secret>'
Now, I could get the correct output,like:
The Azure AD access token is in the access_token value within the output of the call.
What I want is that I need the get the value of the access_token and set it to the variable, so that I could use it in next REST API scripts.
But I'm not very familiar with Bash and curl, can anyone offer advice?
use jq to extract access_token from the json, and VAR=$(...) to store it in a variable,
ACCESS_TOKEN=$(curl -X POST -H 'Content-Type: application/x-www-form-urlencoded' \
https://login.microsoftonline.com/<tenant-id>/oauth2/v2.0/token \
-d 'client_id=<client-id>' \
-d 'grant_type=client_credentials' \
-d 'scope=2ff814a6-3304-4ab8-85cb-cd0e6f879c1d%2F.default' \
-d 'client_secret=<client-secret>' \
| jq -r .access_token )
then you can use ACCESS_TOKEN like
curl -d access_token="$ACCESS_TOKEN"
but be wary, bash is a shitty scripting language, you should not attempt to use bash for complex logic, you should probably switch to a better scripting language like Python, Perl, or PHP, rather than implementing complex logic in Bash. (same goes for Windows's cmd and PowerShell. all 3 are languages unsuitable, but not incapable, of complex logic)
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 'd like to post directly a json object from a url(json) to another url
so the command goes as follows:
curl "<resource_link>.json" -o sample.json
curl -X POST "<my_link>" "Content-type: application/json" -d #sample.json
I 'd like to avoid this, so what is the solution? Is it something like that?
curl -X POST "<my_link>" "Content-type: application/json" -d "curl <resource_link>.json"
But it does not work? Also, this one post Stream cURL response to another cURL command posting the result
does not explain thouroughly and it is not working
Yes,
curl
manual explains the '#' but it does not explain about using another curl
Alternatievely, if I could save somewhere temporarily the 1st cURL response and use it in the other command(but not in a file)
You don't want -x POST in there so let's start with dropping that.
Send the results from the first transfer to stdout by not using -o, or telling -o to use stdout with -o-, and
Make sure your second transfer accepts the data to send on stdin, by using -d#-.
curl "<link>.json" | curl "<link2>" -H "Content-type: application/json" -d #-
With curl 7.82.0 and later
Starting with curl 7.82.0 you can do it even easier with the new --json option:
curl "<link>.json" | curl "<link2>" --json #-
I would like to create a curl output live in a single shell command, to log a output from an Ansible job in realtime filling a log file.
I've tried this command:
curl -f -k -N -H 'Content-Type: application/json' -XPOST \
--user admin:awxsecret \
http://192.168.42.100/api/v2/jobs/1620/
...but it only returns the output generated thus far, not waiting for newly-generated content.
As #charles-duffy said: "AWX does support websockets" I will work with this solution.
I'm using curl to create several classifications. I have written the json for the many classifications and they are in one folder. I would like to create all the classifications in one go. But using curl I can only create them one at a time. How could I make them in one request?
curl -u admin:admin -H "Content-Type: application/json" -X POST -d #pii.json http://127.0.0.1:21000/api/atlas/v2/types/typedefs
The curl manual for -d says 'Multiple files can also be specified'. How can I do this? All my attempts have failed.
Do I need a bash script instead? If so, could you help me - I'm not a coder and I'm struggling without an example!
Thanks in advance.
You probably don't want to use multiple -d with JSON data since curl concatenates multiple ones with a & in between. As described in the man page for -d/--data:
If any of these options is used more than once on the same command
line, the data pieces specified will be merged together with a
separating &-symbol. Thus, using '-d name=daniel -d skill=lousy' would
generate a post chunk that looks like 'name=daniel&skill=lousy'.
You can however easily and conveniently pass several files on stdin to let curl use them all in one go:
cat a.json b.json c.json | curl -d#- -u admin:admin -H "Content-Type: application/json" http://127.0.0.1:21000/api/atlas/v2/types/typedefs
(please note that -X POST has no place on a command line that uses -d)
I found the following to work in the end:
<fileToUpload.dat xargs -I % curl -X POST -T "{%}" -u admin:admin -H "Content-Type: application/json" http://127.0.0.1:21000/api/atlas/v2/types/typedefs
Where fileToUpload.dat contained a list of the .json files.
This seemed to work over Daniel's answer, probably due to the contents of the files. Hopefully this is useful to others if Daniel's solution doesn't work for them.
I needed to upload all the *.json files from a folder via curl and I made this little script.
nfiles=*.json
echo "Enter user:"
read user
echo "Enter password:"
read -s password
for file in $nfiles
do
echo -e "\n----$file----"
curl --user $user:$password -i -X POST "https://foo.bar/foo/bar" -H "Content-Type: application/json" -d "#$file"
done
Maybe fits your needs.