I have an input form where user passes in a URL. Currently there is no validation but now need one to verify if URL is a really valid web link or not. I couldn't find any built-in URL validators in play framework. If inputted URL isn't valid then throw an error message to the user. How to go about doing this in play?
Use Java URL class and catch MalformedURLException to detect an invalid URL. Simply instantiate the class with passed string new URL(formUrl)
http://docs.oracle.com/javase/6/docs/api/java/net/URL.html
Information about how to implement a custom form validation can be found here:
http://www.playframework.com/documentation/2.2.x/ScalaCustomValidations
Related
In my Flutter (Android/iOS) app I am using Firebase Dynamic Links for Patreon.com OAuth2 apis.
My dynamic link is https://myappname.page.link/patreon
The deep link is https://myappname.net/patreon
Patreon is using the https://myappname.page.link/patreon as a redirect_url , and is supposed to append some parameters to it, so it looks like
https://myappname.net/patreon?code=xxx
However, all I receive inside my app is the naked url https://myappname.net/patreon
There are no parameters attached to it.
So how can I tell Firebase to preserve the query parameters Patreon is attaching to the redirect_url?
As an alternate question, is there a better way to listen for incoming response inside of a Flutter app, without the use of Dynamic Links?
You loose all parameters by using that.
If you're relying on Patreon to send back that parameter I'd suggest to generate a small proxy where you can redirect your calls to the dynamic link by generating it on the fly.
So:
Patreon shares www.myhost.com/supah-link?p1=aaa&p2=bbb
Your micro-service which runs on www.myhost.com/supah-link receives the call
You generate a dynamic link like the following:
https://example.page.link/?link=https://www.example.com/someresource&apn=com.example.android&amv=3&ibi=com.example.ios&isi=1234567&ius=exampleapp&p1=aaa&p2=bbb
NOTE: Pay attention to the &p1=aaa&p2=bbb parameters added
Return a 302 and redirect to your newly generated link
Based on how you configure it from the console this link can redirect to the store or your app, in your app you can listen for the link as follows:
FirebaseDynamicLinks.instance.onLink(
onSuccess: (dynamicLink) async => handleDeepLink(dynamicLink?.link),
);
In handleDeepLink you can parse your custom query parameters.
NOTE: The URL parameter you pass via the dynamic link HAS TO BE ENCODED! Which means your link will look more like this:
https://example.page.link/?link=https%3A%2F%2Fwww.example.com%2Fsomeresource%26apn%3Dcom.example.android%26amv%3D3%26ibi%3Dcom.example.ios%26isi%3D1234567%26ius%3Dexampleapp%26p1%3Daaa%26p2%3Dbbb
I want to upload some XML files to Alfresco, so the create con tent form has an input file form element.
I need to check if the XML is well-formed, and I already have the backend validation functions triggered on ResourceBehavior.onContentUpdate. If the XML is malformed, I want to notify the user with a dialog window.
So far, I can prevent the user to submit malformed XML by throwing an exception when the XML is malformed, but I can't figure out how to have share to display an error message.
I have been looking at all the validation JS in share, but remember, file input forms need to be submitted first so that you can have a look at its content, thus the validation has to be server-sided.
Any pointers on where should I begin?
The problem you are going to have is that your backend behaviour is not aware of the specific client session that made the changes and what client session it is that needs to be notified.
If you want to display a useful message then you are going to have to write some additional Share customisation. Some options which you can explore are having an action or webscript that returns whether the XML is valid or not and customising the Share upload form to execute this action/webscript after the file has been uploaded and then return the relevant message to the user.
You'll find a pretty detailed post on modifying the upload form here:
http://www.ixxus.com/blog/2011/09/customising-upload-files-dialog-alfresco-share
If you're feeling lazy then I'd consider just aborting the file creation if the XML is invalid during an onCreate behaviour and then the user will see an 'Internal Error'.
I just started to learn Grails and my question of this could be dumb. So apologies if it is dumb. I have a client program and a REST web service - both coded by me. The client program calls the RESTful service using POST (to add record to a database). When I call object.save() there are some validation errors that are returned.
In my gsp I have fieldError tag coded to read the error message for each field in the bean and show it in the screen. I'm assuming I needed to pass the domain instance that failed the validation from RESTful service to the client so the client could inturn send it to gsp which will automatically show errors. Correct me if this is wrong. however I don't know how to pass the domain instance object as XML from the web service. When the validations are successful though, I get the object like "render object as XML". However when validations fail, I don't know how to pass the entire failed domain instance object back as xml.
I tried to code,
if (student.save()){
render student as XML
} else {
student.errors.each(){
println it
}
def errmsg = student.errors.allErrors.collect { g.message(error:it) }
render(contentType:"text/xml") {
respstud {
for(err in errmsg) {
message(err)
}
}
}
}
This returned the specific error message back to the client as xml, but I'm needing the entire student object to be sent back to the client when save() fails also. Can someone please help?
let me know if you need more info. This is my first post in stackoverflow so I don't really know if I have to provide more details. Any help is greatly appreciated.
Thanks,
Prem
Because the request is being made via ajax, you will need to handle it the Ajax way.
Your gsp is compiled on the server and sent as html to the client when the page is first requested.
One way to handle this would be to set an error on the ajax response and return a snippet of the code you would like to render.
The response may then be handled in the error section of your calling JavaScript and the code snippet rendered on the page where you wish.
The question is generally language/framework agnostic but if it matters I work with Grails and most interested in grails specific solution if such exists.
There's a form mapped to URL: /foo/create. When user type in this URL to his browser the form is shown.
Form action attribute directs to /foo/save and has method POST. If saving is successful, then standard post-redirect-get pattern is applied, and user is redirected to /foo/show.
But, if user specified incorrect data, they should see the same form again with error messages and all their data preserved. To implement this behavior, I do forward to the controller which produces the form (the same is mapped to /foo/create).
After that user sees the form with data and error messages, but URL field is changed in browser to /foo/save. And if user change focus to URL field and press enter - 404 will be shown (because nothing is mapped to /foo/save + method=GET pair).
The long story short: URL /foo/save is shown in a browser (as there were no redirection after form was submitted) but it directs to nowhere if accessed by HTTP GET method.
How to deal with this situation? Surely, I can map something to /foo/save but I wonder if there's a way not to change URL shown in a browser after form with wrong data was submitted?
Two approaches:
The form submits to itself, i.e. /foo/create submits to /foo/create, only if successful the page is redirected to /foo/show. This should use a post-redirect-get cycle as well and store the submitted data in the session, but could be a simple POST without redirect.
/foo/save always redirects again, either to /foo/create if the data was invalid or to /foo/show if the data was valid. This will always use a post-redirect-get cycle with the data saved in the session.
I am using T4MVC to redirect to another action return RedirectToAction(MVC.MyController.MyAction());.
In result it is doing get request.
Is there any way to make post request from controller. I want to keep all the same but only make post instead get. I cant find any methods for that. I found one post helper here http://geekswithblogs.net/rakker/archive/2006/04/21/76044.aspx but i cant pass any values i need using this post helper. I was trying to pass values through TempData but they are not coming when i using this helper. May be some one have any ideas?
The reason i want to do this because when user come from one controller to another and then if user click update or just click enter in browser address bar, page will break.
Should i use session for that reason?
A RedirectToAction will always perform a GET, never a POST (it returns a HTTP 302 to the browser, which will then issue a GET request).
To persist data across the redirect, if it is data that can be easily represented as a string and stored in the query string, then you can just add it to the route values of the redirect.
e.g.
return RedirectToAction("Search", new { searchString = "whatever" });
If it is a complex type, then you will need to store it in TempData. A number of other questions on StackOverflow (such as this one) give details on how.
If repeatedly storing to and reading from TempData across your application offends your code-sense, then you can encapsulate this by using the PassParametersDuringRedirect attribute and generic RedirectToAction available in the MvcContrib project. Some details on this technique are available here.
only way of doing post is by having a form and doing submit on that form, either with a submit button or with javascript, any info you want passed to that action must be in that form and you will find everything posted in FormCollection(hope I spelled it right).