I have a mini program/server built on one of my computers (Machine1) and I am trying to create or overwrite a file through cURL on another computer (Machine2). So Machine2 is connected to Machine1. Ive been looking through cURL's documentation for command that will do this but have had no luck and as well on stack overflow.
https://curl.haxx.se/docs/manpage.html
I have also tried the examples on this SO post:
HTTP POST and GET using cURL in Linux
Any idea as to what the command might be through command prompt? (equivalent of a POST command). I have tried so far using -O, -K, -C and a multitude of others which have not worked.
In command line, all you need to do is using curl --form to simulate a multipart/form-data POST request:
curl --form "testfile=#thefilename.jpg" http://<Machine2>/<Path>
testfile is the field name used for form, if you don't care, just use any english word.
# is used here to make file thefilename.jpg get attached in the post as a file upload. Refer to curl man doc.
In server side, URL http://<Machine2>/<Path> should be listened. When curl send the previous POST request, server side program should get it, extract the attached file (thefilename.jpg), and save to disk.
Related
My command is:
curl -X POST --user "admin:admin" https://website.com/wp-json/wp/v2/posts/ -H "Content-Type: application/json" -d {"title":"test","content":"Content","status":"draft"}
The response I get:
{"code":"rest_invalid_json","message":"Invalid JSON body passed.","data":{"status":400,"json_error_code":4,"json_error_message":"Syntax error"}}
Thanks!
From my earlier comments, I have the feeling that this error is related to the -d possibly needing to have quotes around the data text.
So in your case, it might be worth trying to put single quotes around the -d data text because you are currently using double quotes in the json text.
Thus, your curl command might look like this:
curl -X POST --user "admin:admin" https://website.com/wp-json/wp/v2/posts/ -H "Content-Type: application/json" -d '{"title":"test","content":"Content","status":"draft"}'
This is based on how the curl docs appear to show that quotes should be used with -d related data text based on the example under the "POST (HTTP)" section of the curl documentation here. I think this is what I have done in the past, as well, but I don't have a specific personal example on me at the moment.
Update
For what it's worth, when I tried running this curl type of command on my own, I got the following error when I didn't use single quotes around the json text:
{"code":"rest_invalid_json","message":"Invalid JSON body passed.","data":{"status":400,"json_error_code":4,"json_error_message":"Syntax error"}}curl: (3) Port number ended with 'C' curl: (3) Port number ended with 'd'
When I ran the command with single quotes around the json data, like this: -d '{"title":"test","content":"Content","status":"draft"}', the command worked and a draft was added in WordPress. I also received back a json response that looks related to the new draft post.
Another thing to check might be the credentials used with the --user part. I ended up having to use an "Application Password" that I generated for a user in my WordPress account. This is different than my user's login password. If you haven't already tried using an "Application Password" in WordPress for this part, it might be worth trying to see if that fixes your issue. I found this when I:
Logged into my WordPress website
Clicked on the Users menu item
Clicked on a user
Scrolled down to the bottom of the user info page
Clicked on the Add New Application Password button
Ok, this was a windows specific problem, I guess most people out there are running unix/linux clients so this doesn't apply to them. I found the solution here: https://stackoverflow.com/a/7173011/15161479
The issue is with curl on windows, the quotes need to be escaped. I also did a couple other things like installing the "Application Passwords" plugin.
This is what my command looks like now
curl --user "user:application password" http://website.com/wp-json/wp/v2/posts/ -H "Content-Type: application/json" --data "{"""title""":"""test""","""content""":"""Content""","""status""":"""draft"""}"
Hope this helps some other people out there!
I am a in a student job where I am required to do work with a DB but it really isn't my domain.
In the Documentation it says to enter the line
GET /_cat/health?v
This returns the error
-bash: GET: command not found
It also proposes that I copy as curl. Then the command that works is
curl -XGET 'localhost:9200/_cat/health?v&pretty'
I can I make the command "GET /_cat/health?v" to work?
GET is a request method of the HTTP protocol. If you don't write an HTTP server or client software then you don't have to deal with it explicitly.
The command line
curl -XGET 'localhost:9200/_cat/health?v&pretty'
tells curl to request the URL http://localhost:9200/_cat/health?v&pretty using the GET request method.
GET is the default method, you don't need to specify it explicitly.
Also, the second argument you provide to curl is not an URL. curl is nice and completes it to a correct URL but other programs that expect URLs might not work the same (for various reasons). It's better to always specify complete URLs to get the behaviour you expect.
Your command line should be:
curl 'http://localhost:9200/_cat/health?v&pretty'
The apostrophes around the URL are required because it contains characters that are special to the shell (&). A string enclosed in apostrophes tells the shell to not interpret any special characters inside it.
Without the apostrophes, the shell thinks the curl command ends on & and pretty is a different command and the result is not what you expect.
Behind the scene, curl uses HTTP to connect to the server localhost on port 9200 and sends it this HTTP request:
GET /_cat/health?v&pretty
When you start working with elasticsearch, one of the first things they ask you to do to test your install is to do a GET /_cat/health?v, as shown here:
enter link description here
They fail to tell you that this will not work in a terminal, as Ravi Sharma has explained above. Maybe the elasticsearch team should clarify this a bit. At least they supply a Copy as cURL link. It is just frustrating for someone new at this.
sudo apt install libwww-perl
GET command is in package libwww-perl
In slack you can script slackbot to post messages to a channel like this:
curl --data "$msg" $'https://<yourteam>.slack.com/services/hooks/slackbot?token=<yourtoken>&channel=#random'
Now i'd like to mention a username as the first part of the message like msg="#joernhees hello self".
The problem with this is that if the --data argument of curl starts with an # sign it will interpret the string after the # as filename and post its content. Is there a way to make curl ignore the # sign and to send a literal # as the first char of a post request?
If you are on a new version of cURL you can also use the --data-raw option:
http://curl.haxx.se/docs/manpage.html#--data-raw
A word of warning is that looking my laptop it appears Yosemite ships with an older version of cURL.
In general if you're creating tools to post to Slack I'd recommend using an HTTP library in your script rather than calling out to a shell and invoking the curl command.
Actually i just found out i can do this (not sure it's the best option though):
curl --data '#-' $'https://<yourteam>.slack.com/services/hooks/slackbot?token=<yourtoken>&channel=#random' <<< "$msg"
The trick is to tell curl to read from stdin #- and then pass the message in via that.
I am in my Terminal and I want to send a POST request to a given URL. I have tested this with a REST client so I know that the parameters work.
So lets say I want to POST the following parameters:
username=tony
password=secret
To my URL: https://exmaple.com/login/
I tried the following curl command in my Terminal (I am using OSX Lion)
curl --data "username=tony&password=secret" http://exmaple.com/login/
I get an 500 Server Error back from the server so I am now thinking of something that could be different between the REST Client and the curl command.
Thanks for your help
Update: I am using a https service. Do I have to adjust my curl command to account for this?
Try this
curl -F username=tony -F password=secret http://exmaple.com/login/
-F (reference) should probably do the same as --data? Possible the problem is in the webapp.
Maybe the app you are hitting uses basic auth for authentication? Try this one:
curl --user name:password http://exmaple.com/login/
How download only exists files with curl via commandline? I have code like this:
curl http://host.com/photos/IMG_4[200-950].jpg -u user:pass -o IMG_4#1.jpg
This command download all images from IMG_4200.jpg to IMG_4950.jpg - even if they do not exist.
use -f
(HTTP) Fail silently (no output at
all) on server errors. This is mostly
done to better enable scripts etc to
better deal with failed attempts. In
normal cases when a HTTP server fails
to deliver a document, it returns an
HTML document stating so (which often
also describes why and more). This
flag will prevent curl from outputting
that and return error 22.
This method is not fail-safe and there
are occasions where non-successful
response codes will slip through,
especially when authentication is
involved (response codes 401 and 407).