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/
Related
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
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.
I'm working on OAUTH 2.0 stuff using following curl command which is working fine in my terminal.
command -
curl -u testclient1:testpass1 http://localhost/oauth2-server/token.php -d 'grant_type=password&username=bshaffer&password=brent123'
I want to know what's equivalent HTTP request for above CURL command so that, i can use guzzle(comparatively easier) to make HTTP request to get the token. I've tried a lot of combination but not getting the right way to do it.
finally after a lot of googling I managed to find out the equivalent HTTP request of that CURL command.
Here is the command -
http://testclient1:testpass1#localhost/oauth2-server/token.php?grant_type=password&username=bshaffer&password=brent123
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 need to call a HTTP request in ubuntu how do i do it? I can't seem to find an answer around on how to to do it?
How do run the following url without calling a browser like lynx to do it?
http://www.smsggglobal.com/http-api.php?action=sendsms&user=asda&password=123123&&from=123123&to=1232&text=adsdad
in your command prompt, run the following:
curl http://www.smsggglobal.com/http-api.php?action=sendsms&user=asda&password=123123&&from=123123&to=1232&text=adsdad
the curl command executes an http request for a given url and parameters.
if you need to specify another HTTP method, use curl -X <TYPE> <URL>, like this:
curl -X POST http://www.smsggglobal.com/http-api.php?action=sendsms&user=asda&password=123123&&from=123123&to=1232&text=adsdad
curl documentation: http://curl.haxx.se/docs/manpage.html
to display the results:
curl http://www.smsggglobal.com/http-api.php?action=sendsms&user=asda&password=123123&&from=123123&to=1232&text=adsdad
or
to save the results as a file
wget http://www.smsggglobal.com/http-api.php?action=sendsms&user=asda&password=123123&&from=123123&to=1232&text=adsdad