I submitted a reqest to my solr cloud server with curl:
curl http://solrserver1:8983/solr/admin/collections?action=CREATEALIAS&name=bf&collections=collection1,collection2
It return a 400 error and said: Missing required parameter: name. But you see in the command line I really provided a name parameter.
I saw several post online talking about CREATEALIAS action, so I think it must be some mistake of myself. My solr server is version 4.6.1
Could anyone know what's the reason ?
Should add quotes like this:
curl "http://solrserver1:8983/solr/admin/collections?action=CREATEALIAS&name=bf&collections=collection1,collection2"
It works and return status 0
Following commands worked:
$ bin/solr create_collection -c techproducts -d _default -n techproducts -shards 2 -replicationFactor 2 -V
The output shows the rest api url
$ curl http://localhost:7574/solr/admin/collections?action=CREATE&name=techproducts&numShards=2&replicationFactor=2&maxShardsPerNode=-1&collection.configName=techproducts
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 installed latest version of Artifactory Pro (5.8.3) on Centos7. The default admin credentials are admin/password. i want to change the password through command-line but unable to do so.
Does any one has any inputs how to do this?
Following troubleshootings i tried:
jfrog rt c rt-server-1 --url=http://domain/artifactory --user=admin --password=password ...
nothing happens , when i try to use a new password i get 401: unauthorize error
jfrog guide tells to generate security.xml and add the hash code of the new password but security.xml does not get generated even after following all their steps.
also tried to use curl commands but no use.
If any one has gone through similar issue please share your findings. let me know if you need more info.know on case
JFrog CLI currently does not support changing of a user's password. the CLI config method you were using simply lets you configure your server and credentials to be used by other CLI command later.
What you can do is use a simple curl command to invoke the change password API as described here.
specifically, in your example changing the Admin's password to "NewPassword" will be something like:
curl -X POST -u admin:password -H "Content-type: application/json" -d '{ "userName" : "admin", "oldPassword" : "password", "newPassword1" : "NewPassword", "newPassword2" : "NewPassword" }' http://yourartifactory:8081/artifactory/api/security/users/authorization/changePassword
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.
Can anyone help me to find out a unix command that is used to upload/download a file on/from an URL?
Particular URL in which i'm trying to upload/download is protected with an user id and password.
I guess curl serves this purpose but not aware of how to use it? Could you please give me sugegstions on this?
curl has a command line argument named -d (for data) and you can use it like this to send a file(you need to add a # before a file-name to have curl treat it as a file and not a value:
curl -X POST -d #myfilename http://example.com/upload
You can add multiple -d arguments if you need to send a FORM value along with your file. Like so:
curl -X POST -d #myfilename -d name=MyFile http://example.com/upload
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/